中国石油大学工程力学答案合集精品PPT课件

Statics-Problems-11
(b)
+ C R M A
T2 sin (1.5 1.2 1.2) T2 cos (0.9 0.5 0.8)
142.5 1.5 50 (1.5 1.2)
0(N.m)
2021/1/26
Kylinsoft
Statics-Problems-12
静力学第2章作业/P64-2.32: Solution: Original forces and couple can be reduced to a forceR and a coupCleR acting at point O:
Rx 20cos30 10 3(N )
Ry 20sin 30 60 50(kN )
Fx 0 Px 80N Fy 0 Py 60N
Fz 0 Pz 90N
P 80i 60 j 90k
2021/1/26
Kylinsoft
Statics-Problems-14
静力学第3章作业/P85-3.2*: Solution: From the FBD of the beam in Fig:
静力学第2章作业/P60-2.14*:
Solution:
按定义有:
MO
(F
)
r1
F
F
r1
r2 F F r2
So, expressions (a), (d), (e) are valid.
2021/1/26
Kylinsoft
Statics-Problems-7
静力学第2章作业/P61-2.20:
Ry
5
d CR R
0.441(m)
(或d x
CR Ry
0.467(m)
,或 d y
CR Rx
1.348(m)
)
2021/1/26
Kylinsoft
Statics-Problems-13
静力学第2章作业/P64-2.34*:
Solution:
+ C R M 0
M 0 90 0.3 60 0.2 39N.m
由 M 0 RL Ry L
x L M 0 36 2.25(m) Ry 16
2021/1/26
Kylinsoft
Statics-Problems-10
静力学第2章作业/P63-2.30*:
Solution:
（a) T1与水平面夹角为，T2与水平面夹角为
tan 0.8 8
1.5 15
2021/1/26
Kylinsoft
Statics-Problems-8
(b)
rz 0.6i 0.4 j 0.16k (m)
rz
F 12i 8 j 6k (N )
z k
0.6 0.4 0.16
M z (F ) rz F z 12 8 6 0(N m)
00 1
In vector form:
静力学第3章作业/P86-3.11: Solution:
Studying person and ladder, draw FBD (a) x=1.5m
M A 0 +
GP x GL 2 RB 2 0
RB
GP
x 2
GL
76g
744.8(N)
Fx 0 +
RB
3 5
RAx
0
3 RAx RB 5 446.88(N)
sin 8
17
cos 15
17
tan 0.9 3 1.2 4
sin
3 5
cos 4
5
+
FX T1 cos T2 cos 0
+
Fy T1 sin T2 sin 142.5 50 0
T1 170(N ) T2 187.5(N )
2021/1/26
Kylinsoft
Statics-Problems-3
静力学第1章作业/P27-1.3(c):
2021/1/26
Kylinsoft
Statics-Problems-4
静力学第1章作业/P27-1.3(e)*:
2021/1/26
KFra Baidu biblioteklinsoft
Statics-Problems-5
静力学第2章作业/P58-2.3*: Solution:
静力学第1章作业/P23-1.1(l)*:
2021/1/26
Kylinsoft
Statics-Problems-1
静力学第1章作业/P23-1.1(g)*:
2021/1/26
Kylinsoft
Statics-Problems-2
静力学第1章作业/P24-1.2(k):
2021/1/26
Kylinsoft
Rx P cos 40cos60 30 90(kN )
Ry P sin 40sin 60 0(kN )
由此得到如下方程组：
P cos 100 P sin 20 3
解得：
P 40 7 105.8(kN )
tan1 3 19.1()
5
2021/1/26
Kylinsoft
Statics-Problems-6
Fx 0 M B 0
Fy 0
+
+ RBx 0(kN ) + RAy 12 2 9 12 3 2 0
RAy 0(kN ) RAy 2 3 RBy 0
RBy 1(kN )
2021/1/26
RAy
Kylinsoft
3kN
2m
RB
x
RBy
Statics-Problems-15
M z (F ) 0(N m)
2021/1/26
Kylinsoft
Statics-Problems-9
+
静力学第2章作业/P63-2.28*:
Solution:
Rx Fx 0 Ry Fy 20 12 16 16kN
+ M0 M 20 2 125 48 16 7 36kN.m
Solution:
(a)
ra 0.48i 0.4 j 0.16k (m)
F 12i 8 j 6k (N )
a j
0.48 0.4
M a (F ) ra F a 12 8
01
In vector form:
Ma (F ) 0.96 j(N m)
ra
0.16 6 0.96(N m) 0
CR 60 0.5 20cos30 0.5 20sin 30 1 12
dx
O Rx
d
R
Ry
dy
32 5 3 23.3(N m)
于是得到合力的大小、方向和作用线位置（如图所示）：
R Rx2 Ry2 20 7 52.9(N )
tan1 Rx tan1 3 19.1()