理论力学英文版试卷A(附参考答案)

本科理论力学期末考试卷及答案3套

诚信应考，考试作弊将带来严重后果！本科生期末测试1《理论力学 I 》2020-2021（1）注意事项：1. 开考前请将密封线内各项信息填写清楚； 2. 所有答案请直接答在试卷上； 3．考试形式：（闭）卷；4. 本试卷共（六）大题，满分100分，考试时间120分钟。

题号一二三四五六总分得分一、判断题 (正确打“√”，错误打“×”，将答案填在下表中，每小题1分，共10分) 题号 1 2 3 4 5 6 7 8 9 10 答案 XXX√√XX√√√1. 力在两个坐标轴上的投影与力沿这两个坐标轴方向进行分解得到的分力的意义是相同的。

2. 力偶无合力的意思是说力偶的合力为零。

3. 质点系惯性力系的主矢与简化中心的选择有关，而惯性力系的主矩与简化中心的选择无关。

4. 平面力系向某点简化之主矢为零，主矩不为零。

则此力系可合成为一个合力偶，且此力系向任一点简化之主矩与简化中心的位置无关。

5. 某瞬时刚体上各点的速度矢量都相等,而各点的加速度矢量不相等,因此该刚体不是作平动。

6. 两齿轮啮合传动时，传动比等于主动轮与从动轮的转速比，若主动轮转速增大，则传动比也随之增大。

7. 若刚体内各点均作圆周运动，则此刚体的运动必是定轴转动。

8. 不管质点做什么样的运动，也不管质点系内各质点的速度为何，只要知道质点系的质量，质点系质心的速度，即可求得质点系的动量。

9. 质点系的内力不能改变质点系的动量与动量矩。

10. 刚体受到一群力作用，不论各力作用点如何，此刚体质心的加速度都一样。

姓名学号学院专业班级座位号( 密封线内不答题 ) ……………………………密………………………………………………封………………………………………线……………………………………得分oyxFF'c 二、单项选择题（8小题，每题2分，共16分，将答案填在下表中。

）题号 1 2 3 4 5 6 7 8 答案 ADAADBBC1. 二力平衡条件的使用范围是（）A . 刚体 B. 刚体系统 C. 变形体 D. 任何物体或物体系统 2. 不经计算，可直接判定出图示桁架中零力杆的根数为（）。

初一英语物理原理单选题40题

初一英语物理原理单选题40题1.There is a book on the table. If we push it gently, it will start to move. This is because of _____.A.gravityB.frictionC.forceD.magnetism答案：C。

本题考查物理现象对应的英语表达。

A 选项gravity 是重力的意思，书在桌子上移动不是因为重力。

B 选项friction 是摩擦力的意思，这里轻轻推动书刚开始移动不是摩擦力的作用。

C 选项force 是力的意思，推动书是因为施加了力。

D 选项magnetism 是磁力的意思，题干中没有涉及磁力。

2.When we drop a ball, it falls to the ground. This is mainly due to _____.A.electricityB.gravityC.magnetismD.pressure答案：B。

A 选项electricity 是电的意思，球落地不是因为电。

B 选项gravity 重力，球落地是因为重力作用。

C 选项magnetism 磁力，这里没有磁力。

D 选项pressure 是压力的意思，球落地不是因为压力。

3.A car can stop suddenly because of _____.A.gravityB.frictionC.forceD.magnetism答案：B。

A 选项gravity 重力不能让车突然停下。

C 选项force 力比较宽泛，这里具体是摩擦力让车停下。

B 选项friction 摩擦力，车刹车靠的是摩擦力。

D 选项magnetism 磁力，这里和磁力无关。

4.If we slide a box on the floor, the force that opposes its motion is _____.A.gravityB.frictionC.forceD.magnetism答案：B。

理论力学A--参考答案答案

理论力学A 卷参考答案一、选择题1-6 DABADC二、填空题(1) 绝对速度相对速度牵连速度(2) 平动转动 (3) 1442222=+b y b x ，2bk ，2bk 2 (4) l M2，l M(5) 2p ml ω=，（→）； 2(65/24)o L m l ω=，逆时针方向(6) mL ω/2，mL 2ω/3，mL 2ω2/6三、计算题1、解：（1）设圆柱O 有向下滚动趋势，取圆柱O0=∑A M ----0sin max 1T =-⋅-⋅M R F R θP0=∑y F ----0cos N =-θP F又N max F M δ= ……………………2分设圆柱O 有向上滚动趋势，取圆柱O0=∑A M ----0sin max 2T =+⋅-⋅M R F R θP0=∑y F ----0cos N =-θP F ……………………2分又N max F M δ=……………………2分系统平衡时： ……………………2分)cos (sin 1T θR θP F δ-=（2）设圆柱O 有向下滚动趋势.0=∑C M 0m a x =-⋅M R F s0=∑y F 0c o s N =-θP F 又θδcos P R F s = 所以θδcos P R F s = (2)只滚不滑时，应有 θP f F f F s s s cos N =≤ R f s δ≥同理，圆柱O 有向上滚动趋势时得 R f s δ≥只滚不滑时，Rf s δ≥ (2)2、解：速度分析：1. 杆AB 作平面运动，基点为B.……………………….1分2.动点：滑块 A ，动系：OC 杆……………………….2分沿v B 方向投影：……………………….1分……………………….1分沿vr 方向投影：……………………….1分加速度分析：n t ABAB B A a a a a ++=……………………….2分沿v C 方向投影：……………………….2分……………………….1分……………………….1分3、解：突然解除约束瞬时，杆OA 将绕O 轴转动，不再是静力学问题。

理论力学英文版试卷B(附参考答案和评分标准)

XXX工业大学2017 ～2018 学年第二学期期末考试试卷学院_________________ 班级__________ 姓名__________ 学号___________ 《理论力学》课程试卷B（附参考答案和评分标准）Instructor: Lin Zongxi题号 1 2 3 4 5 6 总得分题分20 15 15 15 15 20得分1．Choose the correct answers with proper justification (10×2=20, 20%)1 2 3 4 5 6 7 8 9 10C B A B B A C CD C(1) Which one of the following is a scalar quantity? .A) Force B) Position C) Mass D) Velocity(2) If a particle starts from rest and accelerates according to the graph shown, the particle’svelocity at t = 20 s is ____ ______.A) 200 m/sB) 100 m/sC) 0D) 20 m/s(3) A particle has an initial velocity of 3 m/s to the left at s0 = 0 m. Determine its position whent= 3 s if the acceleration is 2 m/s2 to the right. .A) 0 m B) 6.0 m C) 18.0 m D) 9.0 m(4) The directions of the tangential acceleration and velocity are always .A) perpendicular to each other. B) collinear.C) in the same direction. D) in opposite directions.(5) Calculate the impulse due to the force F. .A) 20 kg·m/sB) 10 kg·m/sC) 5 N·sD) 15 N·s(6) As shown in the right figure, two blocks are interconnected by a cable. Which of thefollowing is correct? .A) v A= - v BB) (v x)A= - (v x)BC) (v y)A= - (v y)BD) All of the above.(7) If a particle moves in the x-y plane, its angular momentum vector is in the .A) x direction. B) y direction. C) z direction. D) x-y direction.(8) If the position, s, is given as a function of angular position, θ, by s=10 sin2θ, the velocity, v,is . (Noting that θ=ωt, ω is the angular velocity at time t).A) 20 cos 2θB) 20 sin 2θC) 20 ω cos 2θD) 20 ω sin 2θ(9) If v A=10 m/s, determine the angular acceleration,α, of the rod whenθ=30︒. .A) 0 rad/s2B) -50.2 rad/s2C) -112 rad/s2D) -173 rad/s2(10) A slender rod (mass =M) is at rest. If a bullet (mass = m) is fired with a velocity of v b, theangular momentum of the bullet about A just before impact is .A) 0.5 m v b2B) m v bC) 0.5 m v bD) zeroDetermine the magnitude of the resultant force acting on the screw eye and its direction measured clockwise from the x axis.Solution:Using the vector analysis, we have the x- and y-components of forces 6kN and 2kN as()()()()o1o1o2o26kN cos603kN6kN sin6033 5.196kN2kN cos452 1.414kN2kN sin452 1.414kNxyxyFFFF=-=-=========(4%)Hence, the force vectors are obtained as1112223 5.1961.414 1.414x yx yF FF F=+=-+=+=+F i j i jF i j i j(4%)According to the vector addition rule, the resultant force is expressed in the vector form()()1212121.586 6.61R x x y yRx RyF F F FF F=+=+++=+=-+F F F i ji j i j(3%)So, the magnitude and direction of the resultant force are obtained as()()()()2222o1.586 6.61 6.80kN6.61arccos arccos103.496.80R Rx RyRyRF F FFFθ=+=-+=⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭(4%)Determine the magnitude of the moment of the 200N force about the x axis. Solve the problem using both a scalar and a vector analysis.Solution:(1) Scalar analysis.The x, y and z-components of the force F can be expressed as()()()ooocos200N cos120100Ncos200N cos60100Ncos200N cos45173.2NxyzF FF FF Fαβγ===-======(3%)Thus, the magnitude of the moment of the force F about the x axis is obtained as()()()()23100N0.25m173.2N0.3m17.43N mx y zM F d F d=-+=-+=(4%)(2) Vector analysis.First, we establish a position vector from origin O to point A on the force line of action:0.30.25OA A==+r r j k(2%)And the vector of force F is100100173.2x y zF F F=++=-++F i j k i j k(2%)Hence, the moment of force F about point O yields00.30.2517.432530100100173.2O==-+-i j kM i j k(2%)And the magnitude ofOM about the x-axis is obtained as()17.432530N m17.43N mx OM==-+=i M i i j k(2%)The overhanging beam is supported by a pin at A and the two-force strut BC. Determine the horizontal and vertical components of reaction at A and the reaction at B on the beam.Solution:(1) Free-body diagram (6%)We draw a free-body diagram for the overhanging beam as below:(2) Equations of equilibrium (6%)Consider the counterclockwise moment of the force positive. According to the free-body diagram, we have x- and y-component equations and moment about point A of equilibrium as follows()()()()()()()20,02.51.50,600N800N02.51.50,600N1m2m800N4m900N m02.5x Ax By Ay BA BF F FF F FM F=-==+--==-+--=∑∑∑After solving, we have (3%)3133.33N,950N,3916.67NAx Ay BF F F==-=The negetive sign indicates that the direction sence is oppisite to that shown in the free-body diagram.A 50-lb bar is rotating downward at 2 rad/s. The spring has an un-stretched length of 2 ft and a spring constant of 12 lb/ft. Determine the angle (measured down from the horizontal) to which the bar rotates before it stops its initial downward movement. (The acceleration of gravity g=32.2 ft/s2).(1%) Solution:Potential Energy:Let’s put the datum in line with the rod when θ= 0. Then, at position 1, the gravitational potential energy is zero and the elastic potential energy will beV1 = ½ k (s1)2= ½ (12) (4 - 2)2(2%)Gravitational potential energy at position 2: - (50) (3 sin θ) (2%)Elastic potential energy at position 2: (2%)½ (12) {4 + (6 sin θ) - 2}2 . So, V2 = - (50) (3sin θ) + ½ (12) {4 + (6 sin θ) - 2}2Kinetic Energy:At position 1 (when θ = 0), the rod has a rotational motion about point A.T1 = ½ I A (w2 ) = ½{1/3 (50/32.2) 62} (22 ) (2%)At position 2, the rod momentarily has no translation or rotation since the rod comes to rest. Therefore, T2 = 0. (2%)Now, substitute into the conservation of energy equation.T1 + V1 = T2 +V2 (2%)½{1/3 (50/32.2) 62}( 22 ) + ½ (12) (4 - 2)2= 0.0 - (50)(3 sin θ) + ½(12){4 + (6 sin θ) - 2}2 Solving for sin θ yields (2%)sin θ = 0.4295. Thus, θ= 25.4 deg.6. (20%)The disk is rotating with ω =3 rad/s, α = 8 rad/s 2 at this instant. Determine the acceleration at point B , and the angular velocity and acceleration of link AB .(5%)Solution:Draw the kinematic diagram and then apply the relative-acceleration equation:2//B A AB B A AB B A ω=+⨯-a a αr r (4%)where()()()()222o o /o o o o0.2*3 1.8m/s0.2*8 1.6m/s 0.4cos300.4sin 30,01.6 1.80.4cos300.4sin 301.60.4sin 30 1.80.4cos30n A t A B A AB AB AB B AB ABABa αωααα====↓=-===-+⨯-=++-+a a r i j αk i i j k i j i j(5%)Equating the i and j components, we haveo o 1.60.4sin300 1.80.4cos30B AB AB a αα=+=-+ (2%)After solving, we have222.61m/s 5.2rad/s (counterclockwise)B AB a α=→= (4%)。

北航《理论力学(第二学期)》考题整理

A
（a）
B
A
(b)
B
A:情况（a）时 AB 杆的角加速度大； B:情况（b）时 AB 杆的角加速度大； C：条件不足，不能确定。二、填空题（第 8 小题 5 分，其余每空 4 分，共 65 分, 将计算的最简结果填写在空格上） 1、定轴转动刚体绕 O 轴在铅垂面内运动，若其质量为 2m，对转轴的转动惯量为 J, 质心到转轴的距离 OC=b,根据系统的广义坐标（如图 3 所示），建立其运动微分方程。答:运动微分方程为： 2、匀质三角板用圆柱铰链与铅垂的 AB 杆和 CD 杆连接，如图 4 所示。已知
=
O
。
已知图示瞬时刚体的角速度为（方向由 O 点指向 A B 点），角加速度为（方向由 O 点指向 B 点）。求正
4

A
图7
北京航空航天大学交通科学与工程学院《理论力学（第二学期）》复习资料
方体上顶点 B 速度 vB 和加速度 aB 。答：
B 点速度的大小 vB = B 点加速度的大小 aB =
4、两个相同的均质杆 AC、BC（各质量均为 m 长为 L）由铰链 C 连接在图示平面内运动，已知图示瞬时铰链 C 的速度大小为ｕ，杆的角速度的大小为ω，方向如图 3A-D 所示，则该瞬时图 3_______所示情况，系统得动能最大。
AB=2L,CD=AC=L,各刚体的质量分别为 m1 ， m2 ， m3 ，若图示瞬时 AB 杆
3
北京航空航天大学交通科学与工程学院《理论力学（第二学期）》复习资料
的角速度为 (方向如图)。该瞬时系统动量的大小 P=
。
B
O
m2
D

【含答案解析】人教版高一英语物理学概念练习题50题

【含答案解析】人教版高一英语物理学概念练习题50题1. In the experiment of free - fall motion, an object is dropped from a certain height. The _____ of the object keeps increasing as it falls.A. forceB. velocityC. accelerationD. mass答案解析：B。

本题考查物理概念中的速度概念以及英语词汇的辨析。

选项A“force”（力），在自由落体运动中，物体受到的力是重力，保持不变，不是随着下落而持续增加的，所以A错误。

选项B“velocity”（速度），在自由落体运动中，物体由于受到重力加速度的作用，速度会不断增加，符合题意。

选项C“acceleration”（加速度），在自由落体运动中加速度为重力加速度g，是不变的常量，C错误。

选项D“mass”（质量），物体的质量是物体的固有属性，不会随着下落而改变，D 错误。

2. When a car is moving on a straight road at a constant _____, it means that the car is not accelerating.A. speedB. directionC. distanceD. time答案解析：A。

本题考查速度概念和英语词汇。

选项A“speed”（速度），当汽车以恒定速度行驶时，意味着没有加速度，符合题意。

选项B“direction”（方向），方向与加速度并无直接关系，这里强调的是速度不变，B错误。

选项C“distance”（距离），距离与加速度没有直接关联，C错误。

选项D“time”（时间），时间不是描述汽车是否加速的因素，D错误。

3. A ball is thrown vertically upwards. At the highest point, its _____ is zero.A. velocityB. accelerationC. forceD. mass答案解析：A。

(完整版)kmust期末理论力学试卷及答案1

α 昆明理工大学理论力学测验试卷理论力学B(1) 日期：年月日专业：学号：姓名：一、是非题（每题2分。

正确用√，错误用×，填入括号内。

）1、在平面任意力系中，若其力多边形自行闭合，则力系平衡。

（ × ）2、若将某力沿两不相互垂直的轴分解，则其分力的大小一定不等于该力在这两个轴上的投影的大小。

（ ∨ ）3、不平衡的任意力偶系总可以合成为一个合力偶，合力偶矩等于各分力偶矩的代数和。

（ × ）4、若平面力系对一点的主矩为零，则此力系不可能合成为一个力偶。

（ ∨ ）5、若点的法向加速为零，则该点轨迹的曲率必为零。

（ × ）6、定轴转动刚体上点的速度可以用矢积表示为r v，其中是刚体的角速度矢量，r是从定轴上任一点引出的矢径。

（ ∨ ）二、选择题（每题3分。

请将答案的序号填入划线内。

）1、平面任意力系向作用平面内任意一点简化，其主矢与简化中心 ② ，主矩与简化中心 ① 。

①有关； ② 无关。

2、若斜面倾角为α，物体与斜面间的摩擦系数为f ，欲使物体能静止在斜面上，则必须满足的条件是 ③ 。

① tg f ≤α； ② tg f > α ； ③ tg α≤ f. ； ④ tg α> f 。

3、作用在一个刚体上的两个力F A 、F B ，满足F A ＝－F B 的条件，则该二力可能是 ② 。

① 作用力和反作用力或一对平衡的力； ② 一对平衡的力或一个力偶； ③ 一对平衡的力或一个力和一个力偶； ④ 作用力和反作用力或一个力偶。

4、汇交于O 点的平面汇交力系，其平衡方程式可表示为二力矩形式。

即0)(i AF m， 0)(i B F m ，但必须 ② 。

① A 、B 两点中有一点与O 点重合； ② 点O 不在A 、B 两点的连线上； ③点O 应在A 、B 两点的连线上。

5、半径为r 的车轮沿固定弧面作纯滚动，若某瞬时轮子的角速度为ω，角加速度为α，则轮心O 的切向加速度和法向加速度的大小分别为 ④ 。

Physics课程试卷(a)

Physcis （A ）Department Grade Specialty Number Name Score一．Fill in the blanks (40)1. For the situation of figure, Barbara’s velocity relative to Alex is a constant v BA =52km/h and car P is moving in the negative direction of the x axis. If Alex measures a constant velocity v PA =-78km/h for car P, the velocity v PB measured by Barbara is ; If car P brakes to a stop relative to Alex (and thus the ground) in time t=10s at constant acceleration, the acceleration a PA relative to Alex is ; the acceleration a PB of car P relative to Barbara during the braking is .2. In figure, three blocks are connected and pulled to the right on a horizontal frictionless table by a force with a magnitude of T 3=65.0N. If m1=12.0kg, m2=24.0kg, and m3=31.0kg, calculate (a) the acceleration of the system is ;(b) the tensions in the interconnecting cords T 1= ,T 2= .3．As shown in figure, a 1.34kg ball is connected by means of two massless strings to a vertical, rotating rod. The strings are tied to the rod and are taut. The tension in the upper string is 35N. (a) The tension in the lower string is ;(b) the net force on the ball is ;(c) the speed of the ball is .oO’PT1T2T34. A 1.50kg snowball is fired from a cliff 12.5m high with an initial velocity of 14.0m/s, directed 41.0° above the horizontal. (a)Work done on the snowball by the gravitational force during its flight to the flat ground below the cliff is ;(b) The change in the gravitational potential energy of the snowball-Earth system during the flight is ;(c) If the gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground .5. In figure, two particles, each with mass m, are fastened to each other, and to a rotation axis at o, by two thin rods, each with length d and mass M. The combination rotates around the rotation axis with angular velocity ω. In terms of these symbols, and measured about o , (a) the combination’s rotational inertia is ;(b) the combination’s kinetic energy is .6. A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertial of the first, is suddenly coupled to the same shaft. (a) The angular speed of the resultant combination of the shaft and two wheels is ;(b) What fraction of the original rotational kinetic energy is lost .7. In figure, water flows through a horizontal pipe, and then out into the atmosphere at a speed of 15m/s. The diameters of the left and right sections of the pipe are 5.0cm and 3.0cm, respectively. (a) The volume of water flows into the atmosphere during aoωd MMd1.70m10min period is ;(b) in the left section of the pipe, the speed v2 is ;(c) the gauge pressure is .V2 V1d2 d18. A wave traveling along a string is described by y(x,t)=0.00327sin(72.1x-2.72t), in which the numerical constants are in SI units (0.00327 m, 72.1 rad/m, and 2.72 rad/s).(a) The amplitude of this wave is ;(b) The wavelength of this wave is , the period is , the frequency is ;(c) the velocity of this wave is ;(d) the displacement y at x=22.5 cm and t=18.9 s is .9. A cylinder contains 12 L of oxygen at 20°C and 15 atm. The temperature is raised to 35°C, and the volume is reduced to 8.5 L. The final pressure of the gas in atmospheres is . (Assume that the gas is ideal.)10. Imagine a Carnot（卡诺）engine that operates between the temperatures T H=850 K and T L=300K. The engine performs 1200J of work each cycle, which takes 0.25s.(a) The efficiency of this engine is ;(b) The energy |Q H| extracted as heat from the high-temperature reservoir every cycle is ;(c) The energy |Q L| delivered as heat to the low-temperature reservoir every cycle is (d) What is the entropy change of the working substance for the energy transfer to it from the high-temperature reservoir? From it to the low-temperature reservoir? , .二．Problems (60)1. A railroad flatcar of weight W can roll without friction along a straight horizontal track. Initially, a man of weight w is standing on the car, which is moving to the right with speed v0. What is the change in velocity of the car if the man runs to the left (in the figure) so that his speed relative to the car is v rel?W2. A block of mass m1=2.0kg slides along a frictionless table with a speed of 10m/s. Directly in front of it, and moving in the same direction, is a block of mass m2=5.0kg moving at3.0m/s. A massless spring with spring constant k=1120N/m is attached to the near side of m2, as shown in figure. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic at this point.)m1m2 v1iv2i3. In figure, one block has mass M=500g, the other has mass m=460g, and the pulley （滑轮）, which is mounted in horizontal frictionless bearing, has a radius of 5.00cm. When released from rest, the heavier block falls 75.0cm in 5.00s (without the cord slipping on the pulley). (a) What is the magnitude of the blocks’ acceleration? What is the tension in the part of the cord that supports (b) the heavier block and (c) the lighter block? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia?mM4. In figure, a block weighing 14.0N. which slides without friction on a 40.0° incline, is connected to the top of unstretched length 0.450m and spring constant 120N/m. (a) How far from the top of the incline does the block stop? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations?5. Figure shows a hypothetical（假设的）speed distribution for a sample of N gas particles (note that P(v)=0 for v>2v0). (a) Express a in terms of N and v0. (b) How many of the particles have speeds between 1.5v0 and 2.0v0? (c) Express the average speed of the particles in terms of v0. (d) Find v rms.P(v)aV02V06. One mole of an ideal monatomic（单原子的）gas is taken through the cycle shown in figure. Assume that p=2p0, V=2V0, p0=1.01*105Pa, and V0=0.0225m3. Calculate (a) the work done during the cycle, (b) the energy added as heat during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of a. Carnot engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)?bPressure c V, pa dV0, p0V olumePhyscis （A ）Answer一．Fill in the blanks1. -130km/h;2.2m/s 2; 2.2m/ s 2 2.123Tm m m ++; 1123m T m m m ++; 12123()m m T m m m +++3. 8.74N; 37.9N; 6.45m/s4. 184J; -184J; -184J5. 5md 2+8/3Md 2; (5/2m+4/3M)d 2ω26. 267rev/min; 2/37. 6.4m 3; 5.4m/s 9.8*104Pa8. 3.27mm; 8.71cm, 2.31s, 0.433Hz; 3.77cm/s; 1.92mm 9. 22atm10. 65%; 1855J; 655J; 2.18J/K, -2.18J/K 二．Problems 1. 略 2．25cm3. 6.00cm/s 2;4.87N; 4.54N; 1.20rad/s2; 0.0138kgm2 4. 2/3v 0; N/3; 122v 0; 1.31v 05. 略6． 2270J; 14800J; 15.4%; 75.0%。

理论力学英文版试卷A(附参考答案)

XXX工业大学20XX ～20XX 学年第XX学期期末考试试卷学院_________________ 班级__________ 姓名__________ 学号___________ 《理论力学》课程试卷B（附参考答案）Instructor: Yijian(Time: 2 Hours) Course Code:题号 1 2 3 4 5 6 总得分题分20 15 15 15 15 20得分1．Choose the correct answers with proper justification (10×2=20, 20%)1 2 3 4 5 6 7 8 9 10D D C A B B A C B C(1) For vector addition you have to use law.A) Newton’s Second B) the arithmetic C) Pascal’s D) the parallelogram(2) A particle moves along a horizontal path with its velocity varying with time as shown in theright figure. The average acceleration of the particle is _________.A) 0.4 m/s2 → B) 0.4 m/s2 ←C) 1.6 m/s2 → D) 1.6 m/s2 ←(3) If the position of a particle is defined by r = [(1.5t2 + 1) i + (4t – 1) j] (m), its speed at t=1 sis .A) 2 m/s B) 3 m/s C) 5 m/s D) 7 m/s(4) As shown in the right figure, the speed of block B is .A) 1 m/s B) 2 m/sC)4 m/s D) None of the above.(5) If the path function of a particle is s = 10 sin 2θ, the acceleration, a,is . (Noting that θ=ωt, ω is the angular velocity at time t).A) 20 a sin2θB) 20 a cos2θ-40ω2sin2θC) 20 a cos2θD) -40 a sin2θ(6) As shown in the right figure, the velocity of plane A with respect to plane B is .A) (400 i+ 520 j) km/hrB) (1220 i - 300 j) km/hr3C) (-181 i- 300 j) km/hrD) (-1220 i + 300 j) km/hr(7) If the pendulum is released from the horizontal position, the velocity of its bob in thevertical position is .A) 3.8 m/sB) 6.9 m/sC) 14.7 m/sD) 21 m/s(8) If a slender bar rotates about end A, its angular momentum with respect to A is .A) (1/12) m l2 ωB) (1/6) m l2 ωC) (1/3) m l2 ωD) m l2 ω(9) The tangential acceleration of an object .A) represents the rate of change of the velocity vector’s direction.B) represents the rate of change in the magnitude of the velocity.C) is a function of the radius of curvature.D) Both B and C.(10) Point A on the rod has a velocity of 8 m/s to the right. Where is the instantaneous center (IC)for the rod? .A) Point A. B) Point B. C) Point C. D) Point D.Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axisSolution:Using the vector analysis, we have the x - and y -components of forces 600N and 800N as ()()()o 1o 122600N cos303003519.6N 600N sin 30300N 0800N x y x y F F F F ====-=-== (2%)Hence, the force vectors are obtained as111222519.6300800x y x y F F F F =+=-=+=F i j i j F i j j(4%)According to the vector addition rule, the resultant force is expressed in the vector form ()()121212519.6500R x x y y Rx Ry F F F F F F =+=+++=+=+F F F i j i j i j(5%)So, the magnitude and direction of the resultant force are obtained as ()()()()2222o519.6500721.10N519.6arccos arccos 43.90721.1R Rx RyRx R F F F F F θ=++=⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭(4%)The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles α, β, γ of the resultant force. Take x =20 m, y =15 m.Solution:Going from D to A , one must travel {-24k } m, then {20i } m and finally {15j } m. Thus, the unit vector for rope DA is (2%) ()222201524m 20152434.6634.6634.66201524m A DA Ar +-===+-++-i j k r u i j k (){}201524400N 230.81173.11276.98N 34.6634.6634.66A A DA F ⎛⎫==+-=+- ⎪⎝⎭F u i j k i j kGoing from D to B , one must travel {-24k } m, then {-6i } m and finally {4j } m. Thus, the unit vector for rope DB is (2%) ()()2226424m 642425.0625.0625.066424m BDB Br -+-===-+--++-i j k r u i j k (){}6424800N 191.54127.69766.16N 25.0625.0625.06B B DB F ⎧⎫==-+-=-+-⎨⎬⎩⎭F u i j k i j kGoing from D to C one must travel {-24k } m, then {-18j } m and finally {16i }m. Thus, the unit vector for rope DC is (2%) ()()2228912171717161824mC DC C r ===--++-r u i j k(){}8912600N 282.35317.65423.53N 171717C C DC F ⎧⎫==--=--⎨⎬⎩⎭F u i j k i j kHence, the resultant force is (5%){}321.6216.851466.67N R A B C =++=--F F F F i j kThe magnitude and the coordinate direction angles of R F are (4%)oo1501.7N321.62arccos arccos 77.631501.790.64arccos arccos 90.641501.7167.6arccos arccos 167.61501.7R Rx R Ry R RzRF F F F F F F αβγ==⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭oDetermine the horizontal and vertical components of reaction at the pin A and the reaction on the beam at C .Solution:(1) Free-body diagram (5%)We draw a free-body diagram for the overhanging beam as below:(2) Equations of equilibrium (5%)Consider the counterclockwise moment of the force positive. According to the free-body diagram, we have x- and y-component equations and moment about point A of equilibrium as follows()()()()o o o 0,cos 4500,sin 454kN 00,sin 45 1.5m 4kN 3m 0x Ax C y Ay C A C F F F F F F M F =+==+-==-=∑∑∑ After solving, we have (5%)11.31kN,8kN,4kN Ax Ay C F F F ==-=-The negetive sign indicates that the direction sence is oppisite to that shown in the free-body diagram.A projectile is launched from point A with the initial conditions shown in the figure. Determine the slant distance s which locates the pointB of impact. Calculate the time of flight t .Solution:Establish a fixed x,y coordinate system (in the solution here, the origin of the coordinate system is placed at A ). Apply the kinematic relations in x and y-directions. Motion in x-direction: (4%)()o 00ocos400120m/s cos40B A x AB A AB B ABx x v t x v t x t =+=+=+Motion in y-direction: (5%)()()()2o 20o 2211120m/s sin 402210120m/s sin 409.81kg/s 2B A y AB AB A AB ABB AB ABy y v t gt y t gt y t t =+-=+-=+-Noting that (2%)()o 800m tan 20B B y x =--, o /sin 20B s y =After solving, we obtain the slant distance s and the flight time t as follows (4%) 279.8m, 3.04AB s t t s ===The rod AB has a mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when θ= 0°. Neglect the mass of the pistons. Determine the angular velocity of rod AB at θ= 0° if the rod is released from rest when θ= 30°.Solution:Initial position (2%)Final position (2%)(1) Potential Energy:(5%)Let’s put the datum in line with the rod when θ= 0°. Then, the gravitational potential energy and the elastic potential energy will be zero at position 2. => V2 = 0Gravitational potential energy at position 1: - (10)( 9.81) ½ (0.4 sin 30)Elastic potential energy at position 1: ½ (800) (0.4 sin 30)2So, V1 = - 9.81 J + 16.0 J =6.19 J(2) Kinetic Energy:(5%)The rod is released from rest from position 1 (so v G1 = 0, ω1 = 0). Therefore, T1 = 0. At position 2, the angular velocity is ω2 and the velocity at the center of mass is v G2.Therefore,T2 = ½ (10) (v G2)2 + ½ (1/12)(10)(0.42) (ω2)2At position 2, point A is the instantaneous center of rotation. Hence, v G2 = r ω = 0.2 ω2 . Then, T2 = 0.2 ω22 + 0.067 ω22 = 0.267 ω22(3) conservation of energy equation (6%)Now apply the conservation of energy equation and solve for the unknown angular velocity, ω2. T1 + V1 = T2 +V20 + 6.19 = 0.267ω22 + 0 => ω2 = 4.82 rad/s。

高二英语物理概念单选题50题

高二英语物理概念单选题50题1. A force that causes an object to move in a circular path is called _.A. centripetal forceB. centrifugal forceC. frictional forceD. gravitational force答案：A。

解析：向心力（centripetal force）是使物体做圆周运动的力。

离心力 centrifugal force）是一种虚拟力，不是使物体做圆周运动的实际力，B选项错误。

摩擦力（frictional force）是阻碍物体相对运动的力，与圆周运动的产生原因无关，C选项错误。

重力gravitational force）是物体由于地球的吸引而受到的力，和圆周运动没有直接关系，D选项错误。

2. Which of the following is a vector quantity related to mechanics?A. MassB. TemperatureC. VelocityD. Energy答案：C。

解析：速度 velocity）是矢量，既有大小又有方向，在力学中是很重要的矢量。

质量（mass）是标量，只有大小没有方向，A选项错误。

温度（temperature）是与热学相关的物理量，不属于力学范畴，且为标量，B选项错误。

能量 energy）是标量，D选项错误。

3. The force exerted on an object due to gravity is _.A. weightB. massC. inertiaD. momentum答案：A。

解析：物体由于地球吸引而受到的力叫重力，重力的大小叫重量（weight）。

质量（mass）是物体所含物质的多少，不是重力，B选项错误。

惯性（inertia）是物体保持原有运动状态的性质，C 选项错误。

动量 momentum）是质量和速度的乘积，与重力概念不同，D选项错误。

初一英语物理原理单选题40题

初一英语物理原理单选题40题1. The ball rolls on the ground because of _____.A.gravityB.frictionC.magnetismD.electricity答案：B。

本题主要考查对不同物理现象原因的英语表达的理解。

A 选项“gravity”是重力，球在地面滚动不是因为重力。

C 选项“magnetism”是磁力，与球在地面滚动无关。

D 选项“electricity”是电，也不符合球在地面滚动的原因。

B 选项“friction”是摩擦力，球在地面滚动是因为摩擦力的作用。

2. When we push a box, we need to overcome _____.A.forceB.pressureC.frictionD.tension答案：C。

A 选项“force”是力的总称，比较宽泛。

B 选项“pressure”是压力。

D 选项“tension”是张力。

我们推箱子需要克服的是摩擦力，所以选C。

3. The book stays on the table due to _____.A.gravityB.magnetismC.electricityD.friction答案：A。

书能在桌子上静止是因为重力的作用。

B 选项磁力、C 选项电、D 选项摩擦力都不符合书在桌子上静止的原因。

4. A car can move forward because of the force provided by _____.A.the engineB.the wheelsC.gravityD.friction答案：A。

汽车能前进是因为发动机提供的动力。

B 选项轮子只是传递动力的部分。

C 选项重力不是汽车前进的动力。

D 选项摩擦力在某些情况下会阻碍汽车前进。

5. When we stretch a spring, we are applying _____.A.forceB.pressureC.frictionD.tension答案：A。

理论力学试卷及答案-A

专业年级理论力学试题考试类型：闭卷试卷类型：A卷考试时量：120分钟一、判断题：（10分，每题1分，共10题）1、作用在刚体上两个力平衡的充要条件是：两个力的大小相等，方向相反，作用在同一条直线上。

（）2、力的可传性原理既适用于刚体，也适用于弹性体。

（）3、两个力系等效的条件是主矢相等，主矩可以不等。

（）4、力偶只能产生转动效应，不能产生移动效应，因此不能用一个力来平衡。

（）5、在两个力作用下保持平衡的构件称为二力构件。

（）6、求解物体系平衡的受力问题时，当系统中的未知量数目等于独立平衡方程的数目时，未知数都能由平衡方程求出，这种问题称为静定问题。

（）7、空间任意力系平衡的充要条件是：力系的主矢和对于任一点的主矩都等于0。

（）8、点作匀速圆周运动时，其在某一瞬时的切向加速度不一定等于0，而其法向加速度一定等于0。

（）9、刚体作平移时，其上各点的轨迹形状相同，在每一瞬时，各点的速度相同，但加速度不一定相同。

（）10、在某一瞬时，平面图形内速度等于0的点称为瞬时速度中心，所以不同时刻速度瞬心的位置可能不同。

（）二、填空题：（15分，每空1分，共7题）1、在外力作用下形状和大小都保持不变的物体称为。

2、平面力系向作用面内任一点简化，一般情形下，可以得到一个和。

3、力对物体的作用效应取决于三个要素，力的、和。

4、固定铰链支座只能限制构件的，不能限制。

5、平面运动中，平移的和与基点的选择有关，而平面图形绕基点的角速度和角加速度与基点的选择。

6、刚体的简单运动包括和。

7、点的速度矢对时间的变化率称为。

三、选择题：（20分，每题2分，共10题）1、刚体受处于同一平面内不平行的三力作用而保持平衡状态，则此三力的作用线（）（A）汇交于一点（B）互相平行（C）都为零（D）其中两个力的作用线垂直2、关于平面力系的主矢和主矩，以下表述中错误的是（）（A）主矢的大小、方向与简化中心的选择无关（B）主矩的大小、转向一般情况下与简化中心的选择有关（C）当平面力系对某点的主矩为零时，该力系向任何一点简化结果为一合力（D）当平面力系对某点的主矩不为零时，该力系向某一点简化的结果可能为一合力3、下列属于二力构件的是( )（A）杆件AC（B）杆件BC（C）杆件DE（D）杆件AC、BC和DE均是二力构件4、图示三铰刚架上作用一力偶矩为m的力偶，则支座B的约束反力方向应为（）（A）沿BC连线（B）沿AB连线（C）平行于AC连线（D）垂直于AC连线5、下列说法中错误的是（）（A）力偶对其作用平面内任一点的矩恒等于力偶矩，而与矩心位置无关（B）平面汇交力系的合力对平面内任一点的力矩等于力系中各力对该点的力矩的代数和（C）力偶不能与一个力等效也不能与一个力平衡（D）力使物体绕矩心逆时针旋转为负6、下列哪项不属于静力学公理( )（A）二力平衡公理（B）力的可传性原理（C）胡克定律（D）三力平衡汇交定理7、若点作匀变速曲线运动，其中正确的是（）（A）点的加速度大小a=常量（B）点的加速度a=常矢量a=常量（C）点的切向加速度大小ta=常量（D）点的法向加速度大小n8、作用在刚体上的力偶转动效应的大小与下列哪个要素无关（）（A）力偶矩的大小（B）矩心的位置（C）力偶矩的转向（D）力偶的作用面9、如图所示，点M沿螺线自内向外运动，它走过的弧长与时间的一次方成正比，则点的加速度越来越，点M越跑越。

双语物理试卷091001a

2009─2010学年第一学期《力学》(双语)课程考试试卷(A 卷)专业：物理学年级：08级考试方式：闭卷学分：4 考试时间:120分钟引力常量 G=6.67×10-11N 2·m 2·kg -2 重力加速度 g=9.8m/s -2 真空中光速 c=3.00×10m/s一Choose the Correct Answer (3’ ⨯10=30’)Note: Please fill answers in the following table1. A particle moves in a circle with radius R at variable-speed, the magnitude of the acceleration is (v is the speed at any one time) d v/d t .(B) v 2/R .(C) d v/d t + v 2/R .(D) [(d v/d t )2+(v 4/R 2)]1/2.2. A projectile initial velocity v 0 and projectile angle θ , the centripital acceleration and the tangential acceleration at highest point , respectively are:(A) g cos θ , 0 , (B) g cos θ , g sin θ , (C) g sin θ , 0,(D) g , g , 3. Shown in Figure 1, a cylindrical cage with radius R, rotates along the central axis OO ' . A block A is stick to the cylinder block wall with the friction coefficient μ between the block A and cylinder.To make block A do not fall, the angular velocity ω of cylinder should be at least(A) R g μ. (B) gμ.(C)()R g μ.Fig 1(D)R g .4. The mass of particle B is four times of that of Particle. At the beginning the speed of particle A and B is respectively (3i +4 j), (2i-7j). Due to the interaction between the two particles, the speed of A becomes (7i-4j). At this time the speed of particle B is(A) 2i －7j . (B) i －5j . (C) 0.(D) 5i －3j .5. Shown in Figure 2, a smooth, thin rod, rotating along the OO ′at constant angular velocity,is fixed at point O, with a small ring set in the upper end of rod.. At the beginning the pole in a cone movement, and then the small ring along the rod fell from the stationary. In the course of ring drop,the small ring, rod and the Earth are composed of a systemwhose mechanical energy and angular momentum are(A) Mechanical energy, angular momentum are notconserved(B) Mechanical energy is conserved, while angular momentum is not. (C) angular momentum is conserved, while mechanical energy is not. (D) Mechanical energy, angular momentum are conserved6. There are two same radius fine quality ring A with uniform mass distribution and B of the unevenmass distribution. Their rotational inertia along the axis which is perpendicular to ring plane, are respectively J A and J B , then the(A) J A >J B . (B) J A <J B . (C) J A =J B . (D) Not sure.7. In figure 3, the masses of Pulley, rope are negligible. Ignoring all the friction, if the m 1 of the object A, is greater than the m 2 of object B, during the movement, the spring balance reading is(A) (m 1+m 2 )g . (B) (m 1－m 2)g .(C) 2m 1m 2g/(m 1+m 2). (D) 4m 1m 2g/(m 1+m 2).Fig 2Fig 38. A simple harmonic wave propogating in elastic medium, in one instant, an element is in a position of equilibrium, this time its energy is(A) Zero kinetic energy, maximum potential energy. (B) Zero kinetic energy, Zero potential energy.(C) Maximum kinetic energy, maximum potential energy. (D) Maximum kinetic energy, Zero potential energy. 9. A locomotive is running away from a observer at rest at a speedof 25ms -1, with the whistle frequency of 600Hz , then the observer will receive the voice with the frequency of (assuming the speed of voice in the air is 340 m/s)(A) 558Hz . (B) 646 Hz . (C) 555 Hz . (D) 649 Hz .10. Two incidents happen at the same time while two different places inan inertial reference frame, then will they happen at a same time in another inertial reference frame? Which of the following is correct?(A) They must happen at different time. (B) They must happen at same time.(C) They might happen at a same time or different time. (D) It’s uncentain二. Fill in the Blanks (2’ ⨯10=20’)1.An object osciooate in thevertical direction isconnected to a spring. The vibration equation is y=A sin ω t , in which A 、ω are constant, then(1)The speed of an object relationship with a function of time is ;(2) The speed of an object relationship with a function of coordinate is .2. A particle subject to two constant force has a displacement of∆r =3i+8j (SI), and its kinetic energy increases by 24J, if one of the constantforce F 1=12i －3j (SI), then the work done by the other force is3. A particle oscillates simple harmonic motion, with period of T. Along the positive x-axis, it moving from the equilibrium position to one-half the maximum displacement will cost time .4. The two basic assumptions of special relativity are and5. Altair (牛郎星) is 16 light years away from the Earth, if a spacecraft travelwith a velocity of , it will take the spacecraft 4 years (time shown in the spacecraft ) to arrive Altair.6. The so-called vacuum degree of fluid, means the difference with fluid inside pressure and the atmospheric pressure ， as the mercury vacuum gauge shown inthe figure, let h= 50cm, then the vacuum degree inside the container B isthe N/m 2？7. The homogeneous pole, with length L, make it stand upright to smooth the desktop, and then release hands, as the pole can not be locatedabsolute along the vertical direction, so then it fall. Let the pole move in the o-xy plane.Atthe beginning, the pole coincides with the axis, then the center of mass moves along the horizontal direction. Then the pole trajectory equation for the endpoint is三. (15 points)There is a mass of m 1, length l uniform thin rods, located onthe desktop with coefficient of sliding friction μ. It canrotates in the desk plane, about the fixed axis at O. Another block of the mass m 2 collides with the end of rod the other end. Before and after the collision the speed of block are respectively v 1 and v 2, such as the as shown in Figure 5. If the collision time is very short, (1) what is the rotational inertia of the rod about point A?(2)how long have process from thin rod starting rotate to stop after the collision?P 2 (Fig 5四. (15 points)A skierfree falls from point A, when he arrives at platform C across a trench of width d via point B, his velocity v c is just in the horizontal direction. The vertical distance between point A and B is40 m. The tangential direction of the slope at point B is 30º to the horizon, the friction is neglected. (1) What ’s the skier ’s velocity v B leaving point B? (2) What ’s vertical distance between B and C? What about the trench width d ? (3) What ’s the skier ’s velocity v c arriving at C?五. (10 points)shows The waveform of simple harmonic wave at t = 0moment is shown in figure 7(1)What is the wave equation?(2) What is the vibration equation at point P.六. (10 points) A railway bridge of length L , the length of a train at rest is l . When the train is passing the bridge with high speed of 0.6c , what will be the bridge ’s lengthmeasured by the observer on the train? How long will it take him to measure the time that the whole train through the bridge?－Fig 72009─2010学年第一学期《力学》课程考试试卷(A 卷) 参考答案及评分标准一Choose the Correct Answer. (3’ ⨯10=30’)二. Fill in the Blanks. (2’ ⨯10=20’)1. v=A ωcos ωt , v=22y -A ω2. 12J.3. T /12 .4. relativity, c=constant.5. c c 97.017/16=.6. 426.66410/N m ⨯7. no ，4x 2+y 2=L 2三. (15points)Solution : (1) the rotational inertial of the rod about point A is J =()⎰lx l m x 012d /=m 1l 2/3Or J =()⎰212d /2lx l m x + m 1l 2/4=m 1l 2/3 (5 points)（2）Due to a very short time collision , M μ<<M F , The angular momentum of the thin rod and slider is approximate conserved. So we can getm 2v 1l=- m 2v 2l+ (m 1l 2/3)ωm 1ωl 2/3= m 2 ( v 1+v 2) l （4 points ） ,yielding M μ=()⎰lx l /m g 01d μ=μ m 1gl/2After the impact, the Impulse theorem reduces⎰=-tt M 0d μ-μm 1glt/2=0-m 1ωl 2/3=-m 2( v 1+v 2)l （4 points ）So we can obtaint=2m 2( v 1+v 2)/(μ m 1g ) （2 points ）四. (15points)解：Solution :⑴ The mechanical energy of the skier is conserved, taking point B aszero potential energy.∵mgH = mv B 2/2∴28.00/B v m s == （4 points ）⑶The skier dose projectile motion from B to C, and C is known as the maximum point. Sov c = v B cos30º = 19.80m/s （4 points ）⑵∵mv B 2/2 = m v c 2/2+mgh ∴h = (v B 2-v c 2)/2g = 20.00m ；（4 points ） The transit time ：∵0 = v B sin30º-gt ∴t = v B /2g =1.43sThe transit distance d = v B cos30ºt = 28.29m. （3 points ）五. (10 points)解：(1)When wave is at t =0 y 0=A cos ϕ0=0 v 0=-A ωsin ϕ0＞0 so we can get ϕ0=-π/2.（2 points ）and we know T=λ/v=0.40/0.08=5(s) （1points ）So the wave equation isy=0.04cos[2π( t/5-x/0.4)-π/2] (SI) （3 points ）(2) The vibration equation at point P is y P =0.04cos[2π( t/5-0.2/0.4)-π/2] = 0.04cos(0.4π t -3π/2)(SI)（4points ）六. (10 points)Solution : L '=L (1-v 2/c 2)1/2 =0.6L （5分） ∆t '=( L '+l )/v =(0.6 L +l )/0.8c （5分）－。

理论力学和材料力学试题A

绝密★启用前《理论力学、材料力学》试卷A 卷一、填空题。

(每空1分，共10分)1、如下图所示，AB 杆自重不计，在五个力作用下处于平衡状态。

则作用于B 点的四个力的合力 R F = ，方向。

2、作用在刚体上的两个力偶的等效条件是。

3、由二力杆平衡条件可知，二力杆两端所受两个力大小、方向，作用线两个力的作用点的连线。

4、力偶由两个大小相等、且的力组成的力系组成。

5、材料力学主要研究任务中衡量构件承载能力指标是、和。

二、判断题。

（每小题1分，共10分）1、物体处于平衡状态一定是静止的。

（）2、约束是通过约束反力阻碍物体运动的。

（）3、因为构成力偶的两个力满足F =－F `，所以说此力偶的合力等于零。

（） 4、物体受大小相等、方向相反的两个平行力时一定平衡。

（） 5、由平面假设可知，受挤压的杆件，挤压面上的应力集中分布。

（） 6、一个力可以与一个力加一个力偶可等效。

（）7、力可以在平面内任意移动，不改变力对刚体的作用效果。

（） 8、轴向拉压杆就是承受拉力或者压力的杆件。

（） 9、轴力的正负定义为：拉伸为负，压缩为正。

（）10、若梁在某一段内无载荷作用，则该段内的弯矩图必定是一根斜直线段。

（）三、单项选择题。

（每小题1分，共40分）1、图示杆的重量为P ，放置在直角槽内。

杆与槽为光滑面接触，A 、B 、C 为三个接触点，则该杆的正确受力图是( )2、力对刚体作用效果，可使物体( )。

A.产生运动B.产生内力C.产生变形D.运动状态发生改变和产生变形 3、作用在刚体上的二力平衡条件是( )。

A.大小相等、方向相反、作用线相同、作用在两个相互作用物体上B.大小相等、方向相反、作用线相同、作用在同一刚体上C.大小相等、方向相同、作用线相同、作用在同一刚体上D.大小相等、方向相反、作用点相同4、刚体受三力作用而处于平衡状态,且两个力沿作用线可汇交与一点，则此三力的作用线( )。

理论力学期末考试A、B及其标答

华中农业大学本科课程考试试卷考试课程与试卷类型：理论力学A 姓名：学年学期：2007-2008-1 学号：考试时间：2008-1-23 班级：一、判断题（每题2分，共20分。

正确用√，错误用×，填入括号内。

）1、动平衡的刚体，一定是静平衡的；但静平衡的刚体，不一定是动平衡的。

（）2、质点系有几个虚位移就有几个自由度。

（）3、若在作平面运动的刚体上选择不同的点作为基点时，则刚体绕不同基点转动的角速度是不同的。

（）4、已知质点的质量和作用于质点的力，质点的运动规律就完全确定。

（）5、质点系中各质点都处于静止时，质点系的动量为零。

于是可知如果质点系的动量为零，则质点系中各质点必都静止。

（）6、凡是力偶都不能用一个力来平衡。

（）7、任意空间力系一定可以用一个力和一个力偶来与之等效。

（）8、两个作曲线运动的点，初速度相同，任意时刻的切向加速度大小也相同，则任意时刻这两点的速度大小相同。

（）9、对于做平面运动的平面图形，若其上有三点的速度方向相同，则此平面图形在该瞬时一定作平动或瞬时平动。

（）10、若质点的动量发生改变，则其动能也一定发生变化。

（）二、选择题（每题3分，共30分。

请将答案的序号填入划线内。

）1、曲柄OA 以匀角速度ω转动，当系统运动到图1所示位置（OA 平行O 1B 、AB 垂直OA ）时，有A V B V ，A a B a ，AB ω 0，AB α 0。

① 等于； ② 不等于。

图1 图22、边长为l 的均质正方形平板，位于铅垂平面内并置于光滑水平面上，如图2所示，若给平板一微小扰动，使其从图示位置开始倾倒，平板在倾倒过程中，其质心C 点的运动轨迹是。

① 半径为l /2的圆弧； ② 抛物线；③ 椭圆曲线； ④ 铅垂直线。

3、半径为R 的圆盘沿倾角为ϕ的斜面作纯滚动，在轮缘上绕以细绳并对轮作用水平拉力F （如图3所示）。

当轮心C 有位移r d 时，力F 的元功是____________。

专科《理论力学》_试卷_答案

实用文档专科【理论力学】一、〔共76题,共152分〕1. 考虑力对物体作用的两种效应，力是〔〕。

〔2分〕A.滑动矢量；B.自由矢量；C.定位矢量。

.标准答案：C2. 梁AB在D处受到的支座反力为( )。

〔2分〕A.５Ｐ/４B.ＰC.３Ｐ/４D.Ｐ/２.标准答案：A3. 图所示结构中，如果将作用于构件AC上的力偶m搬移到构件BC上，那么A、B、C三处反力的大小( )。

题3图〔2分〕A.都不变；B.A、B处反力不变，C 处反力改变；C.都改变；D.A、B处反力改变，C处反力不变。

.标准答案：C4. 在下述原理、法那么、定理中，只适用于刚体的有。

〔2分〕A.三力平衡定理；B.力的平行四边形法那么；C.加减平衡力系原理；D.力的可传性原理；E.作用与反作用定律。

.标准答案：D5. 三力平衡定理是。

〔2分〕A.共面不平行的三个力相互平衡必汇交于一点；B.共面三力假设平衡，必汇交于一点；C.三力汇交于一点，那么这三个力必互相平衡。

.标准答案：A6. 作用在一个刚体上的两个力F A、F B，满足F A=-F B的条件，那么该二力可能是。

〔2分〕A.作用力与反作用力或一对平衡力；B.一对平衡力或一个力偶；C.一对平衡力或一个力和一个力偶；D.作用力与反作用力或一个力偶。

.标准答案：A7. 点的运动轨迹由直线和弧线组成，点作匀加速运动，如果保证直线与圆弧相切，那么点的运动速度是函数，加速度是函数。

〔2分〕A.不连续，不连续B.连续，连续C.不连续，连续D.连续，不连续.标准答案：D8. 两动点在运动过程中加速度矢量始终相等，这两点的运动轨迹相同。

〔2分〕A.一定B.不一定C.一定不.标准答案：B9. 平面图形上各点的加速度的方向都指向同一点，那么此瞬时平面图形的等于零。

〔2分〕A.角速度B.角加速度C.角速度和角加速度.标准答案：B10. 重为G的钢锭，放在水平支承面上，钢锭对水平支承面的压力为F N，水平支承面对钢锭的约束力是F N。

理论力学期末考试试卷(含答案)

同济大学课程考核试卷（A 卷） 2006— 2007学年第一学期命题教师签名：审核教师签名：课号：课名：工程力学考试考查：此卷选为：期中考试( )、期终考试( )、重考( )试卷年级专业学号姓名得分题号一二三四五六总分题分 3010151515 15 100得分一、填空题（每题5分，共30分）1刚体绕O Z 轴转动，在垂直于转动轴的某平面上有A ，B 两点，已知O Z A =2O Z B ，某瞬时a A =10m/s 2，方向如图所示。

则此时B 点加速度的大小为__5m/s 2 ；（方向要在图上表示出来）。

与O z B 成60度角。

2刻有直槽OB 的正方形板OABC 在图示平面内绕O 轴转动，点M 以r =OM ＝50t 2（r 以mm 计）的规律在槽内运动，若t 2=ω（以rad/s 计），则当t =1s 时，点M 的相对加速度的大小为_0.1m/s 2_；牵连加速度的大小为__1.6248m/s 2__。

科氏加速度为_22.0m/s 2_，方向应在图中画出。

方向垂直OB ，指向左上方。

3质量分别为m 1=m ，m 2=2m 的两个小球M 1，M 2用长为L 而重量不计的刚杆相连。

现将M 1置于光滑水平面上，且M 1M 2与水平面成︒60角。

则当无初速释放，M 2球落地时，M 1球移动的水平距离为___（1）___。

（1）3L ；（2）4L；（3）6L；（4）0。

4已知OA =AB =L ，=常数，均质连杆AB 的质量为m ，曲柄OA ，滑块B 的质量不计。

则图示瞬时，相对于杆AB 的质心C 的动量矩的大小为__122ωmL L C =，（顺时针方向）___。

5均质细杆AB 重P ，长L ，置于水平位置，若在绳BC 突然剪断瞬时有角加速度，则杆上各点惯性力的合力的大小为_g PL 2α，（铅直向上）_，作用点的位置在离A 端_32L_处，并在图中画出该惯性力。

6铅垂悬挂的质量--弹簧系统，其质量为m ，弹簧刚度系数为k ，若坐标原点分别取在弹簧静伸长处和未伸长处，则质点的运动微分方程可分别写成_0=+kx xm _和_mg kx x m =+ _。

2015理论力学静力学期末英文试卷

第 1 页共 1 页Part I Theoretical Mechanics (100 points)1.(12 points)Hydraulic clamping refers to machine quick clamping device with hydraulic as power, which can be realized ona work-piece clamping action quick, saving a lot of manpower and time.It can be simplified as the following picture(题1图).We ’ve known that the structure remains equilibrium.And F is a given constant.Find the holding power acted by H to E.(B,C,E are movable hinges,D is hinge.)2.(12 points)Here is a still system consisting of two weightless beams connecting with each other at point C,Now external force F,uniformly distributed force Q,a couple are given,please find out the reactions at point A,B,C.(Everything shown on the graph is given)3.(10 points)An separated airfoil where a engine is set on it is shown as the following graph.There is a air force acting on it distributing in the shape of ladder-shaped,and q 1=60KN/m,q 2=40KN/m,the gravity of the airfoil P 1=45KN,engine weighs P 2=20KN,and a couple whose magnitude is 18KN`m.So please figure the forces acting on point O.题1图Cautions:①This test paper is about Theoretical Mechanics,and you have 75 mins to finish this part.②Reassure that each question and every sketch are answered in the right area. ③The unit of length in this test is always millimeter unless special announcement.AFBa2/a M2/a qCD题2图姓名︓学号︓班级︓题3图第 2 页共 2 页4.(11 points)Everything is frictionless,weightless,so please find the reactions about point A,D,B.(F is applied force.)5.(12 points)A truss is here,and please find the internal forces of beam 1,2,3.6.(13 points)Here is a ladder weighed 200N,and its length is l,the angle between the ground is 60°.The coefficient of still friction is0.25,now a man whose gravity is 650N go along the ladder,find the maximum distance between C where man can go the most high and A.7.(15 points)A truss is here,and please find the internal forces of beam 1,2,3. 8.Filling Blanks(15 points)[1]An conplanar for system is reduced to point A 、B 、C 、D ，only when it reduces to point c,itswhole moment is 0.Thus its resultant must pass through point .[2]Find the moment of F about y-axis=__________if the side length of the cubic is a.题4图题5图题6图题7图 OF xyz A B[3]_______forces won’t change the motion of particle,only_______forces do.(fill the kinds of force)第3页共3 页。

(完整word版)《理论力学》试卷集

中国矿业大学06-07学年第2学期《理论力学》试卷（A卷）考试时间：100分钟考试方式：闭卷学院班级姓名学号题号一二三四五总分得分阅卷人一．填空题（40分，每题4分。

请将简要答案填入划线内）1．已知A物块重500N，B物块重200N，A物与地面间摩擦系数为0。

20，A块与B块间的摩擦系数为0.25。

则拉动物块A的最小力P= ，A块与B块间的摩擦力大小= 。

2．两直角刚杆AC、CB支承如图，在铰C处受力F作用,则A处约束反力大小= .方向与x轴正向所成夹角= 。

3．平面桁架及载荷如图所示，则受力为零的杆的标号为 .4．在边长为a的正方体的顶角A处，作用一力F，则力F对O点之矩的大小为。

5．图示机构中,刚性板AMB与杆O1A、O2B铰接，若O1A=O2B=r，O1O2=AB=l,在图示瞬时，O1A的角速度为,角加速度为，则M点的速度大小= 。

6．图示机构中，已知O1A=O2B，当O1A//O2B时,O1A与O2B 杆转动的角速度、角加速度分别为1、1与2、2，则该瞬时相互间的大小关系为12；12。

（请填等于或不等于）7．质量为m，长度为l的均质细杆AB,绕其距端点A为l/3并垂直于杆的轴O以匀角速度转动,此时AB杆的动量大小= ，对轴O的动量矩大小8．如图所示，一质量为m，半径为r的钢管放在光滑的具有分枝的转轴上，为使钢管不滑动，则分枝转轴的角速度应满足 .9．质量为m、长为l的均质杆静止地放在光滑的水平面内，若某瞬时在杆的一端施加一冲量，大小为I，方向与杆垂直，则冲击结束时杆的角速度= 。

10．在图所示的平面机构中,试用OA杆的虚位移表达套筒B的虚位移y B= 。

二．计算题(本题15分）图示平面支架,直角弯杆BE与水平杆AD在C点铰接，AD杆的D端铰接一半径为r=0.3m的滑轮。

跨过滑轮的绳子，一端水平地系于弯杆的E点，另一端悬挂有重为Q=100kN的物块.设AB=AC=CD=1m，不计摩擦及其余各构件重量，试求系统平衡时，支座A和B处的约束力.三．计算题（本题15分）机构如图所示，已知：b，OA=e。

理论力学期末标准试卷及详解答案

2011～2012学年度第二学期《理论力学》试卷（A卷）一、填空题（每小题 4 分，共 28 分）1、如图所示结构,已知力F，AC＝BC＝AD＝a，则CD杆所受的力FCD＝（），A点约束反力FAx=（）。

2、如图所示结构，,不计各构件自重，已知力偶矩M，AC=CE=a，AB∥CD。

则B处的约束反力FB =（）；CD杆所受的力FCD=（）。

3、如图所示，已知杆OA L，以匀角速度ω绕O轴转动，如以滑块A 为动点，动系建立在BC杆上，当BO铅垂、BC杆处于水平位置时，滑块A的相对速度vr =（）；科氏加速度aC=（）。

4、平面机构在图位置时， AB 杆水平而OA 杆铅直，轮B 在水平面上作纯滚动，已知速度v B ，OA 杆、AB 杆、轮B 的质量均为m 。

则杆AB 的动能T AB =（），轮B 的动能T B =（）。

5、如图所示均质杆AB 长为L ,质量为m,其A 端用铰链支承,B 端用细绳悬挂。

当B 端细绳突然剪断瞬时, 杆AB 的角加速度 =（），当杆AB 转到与水平线成300角时，AB 杆的角速度的平方ω2=（）。

6、图所示机构中,当曲柄OA 铅直向上时,BC 杆也铅直向上,且点B 和点O 在同一水平线上；已知OA=0.3m,BC=1m ，AB=1.2m,当曲柄OA 具有角速度ω=10rad/s 时,则AB 杆的角速度ωAB =（）rad/s,BC 杆的角速度ωBC =（）rad/s 。

AB1.57、图所示结构由平板1、平板2及CD 杆、EF 杆在C 、D 、E 、F 处铰接而成，在力偶M 的作用下，在图上画出固定铰支座A 、B 的约束反力F A 、F B 的作用线方位和箭头指向为（）（要求保留作图过程）。

二、单项选择题（每小题 4 分，共28 分）1、如图所示，四本相同的书，每本重均为P，设书与书间的摩擦因数为，书与手间的摩擦因数为，欲将四本书一起抱起，则两侧手应加的压力至少大于（）。