静力学第二章力系的简化
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M LL = r × F ⋅ λ = r × (F1 + F2 ) ⋅ λ = r × F1 ⋅ λ + r × F2 ⋅ λ
S
Substitution of r×F2 · λ=F2d, where d is the perpendicular distance from O to the line of action of F2, yields
mBC(F) = mB(F)⋅λBC = rBA×F⋅λBC rBA=4i ft F=FλAD F − 2i + 3 j − 3k )= = F( (−2i + 3 j − 3k ) 2 2 2 22 2 +3 +3 4 j − 2k 1 λ BC = = ( 4 j − 2k ) 20 42 + 22
2.4 Moment of a Force About a Axis
力对轴之矩是一个力使一 个物体绕轴转动趋势的度量
a. Physical characteristic (l) Py is the only component that will rotate the door; (2) the door will be easier to rotate if the magnitude of Py is increased; (3) the door will be easier to open if the distance from the hinge axis to the doorknob is increased; (4) Py will cause the door to rotate in the direction as shown.
R = R + R = 27.5
2 x 2 y
θ = tg
−1
Ry Rx
= 24.4°
Method 2 Vector
2.3 Moment of a Force About a Point
1.Definition The moment of a force F about a point O, called the moment center, is defined as A vector Mo= r×F where: r is the vector from the point O to any point on the line of action of F. Unit: N ⋅m. The magnitude of Mo is
The equivalent scalar equations are:
Rx = ∑ Fx R y = ∑ Fy Rz = ∑ Fz
If the original forces lie in xy-plain, then ∑Fz=0, thus:
Rx = ∑ Fx R y = ∑ Fy
Example 2-1 Determine the resultant force for the forces shown in Fig.. Solution: Method 1 Scalar
推论:
1. Changing the magnitude F of each force and the perpendicular distance d, while keeping the product Fd constant
2. Rotating the couple in its plane
Therefore, if the moment of a force about an axis is to correctly describe the tendency of the force to rotate a body about that axis, we anticipate that its definition will include the following physical characteristics. 1. A force that is parallel to an axis has no moment about that axis. 2. A force that intersects an axis has no moment about that axis. 3. The moment of a force about a given axis depends on the magnitude of the force component that is perpendicular to the axis and on the distance of this component from the axis. 4. The direction of the moment of the force represents the direction of the rotational tendency of the force.
For MLL=Mo·λ ∴ Mx=Mo·i; My=Mo·j; ∴ Mo= Mxi+ Myj+ Mzk Mz=Mo·k
Sample problem 2.3
Solution: From principle of moments:
mx(P) = mx(Pz)+mx(Px) = 0.4×120=48 Nm + mx(Q) = -0.4 × 50 = -20 Nm ∴ + mx(P+Q) = 48-20 = 28 Nm +
vector calculation:
M
c. Characteristics: (1)a couple has no resultant force; (2)moment of a couple is the same about every point; (3)two couples that have the same moment are said to be equivalent.
where d is the perpendicular distance from O to the line of action of F, called the moment arm of F. Then: Mo=Fd
2.Principle of Moments The moment of a force about a given point is equal to theБайду номын сангаасsum of the moments of its components about the point.
M O = i (ry Fz − rz Fy ) − j(rx Fz − rz Fx ) + k (rx Fy − ry Fx )
Sample Problem 2.2 Determine the moment of the force F in Fig. about point A. 1.vector solution Writing vector F as following
MLL=F2d
C. Rectangular components Writing r and F in rectangular representations: r=rxi+ryj+rzk F= Fxi+Fyj+Fzk rx ry rz M LL = r × F ⋅ λ = Fx Fy Fz λx λ y λz
3.Rectangular components Writing r and F in scalar form: r=rxi+ryj+rzk F=Fxi+Fyj+Fzk
i M O = r × F = rx Fx j ry Fy k rz Fz
Expanding the determinant gives
MLL=MO cosα
L
where α is the angle between MO and λ.
M LL = M O ⋅λ = r × F ⋅λ
具体计算时可利用:
(a) Principle of moment The moment of a force about a given axis is equal to the sum of the moments of its components about that axis. (b) Geometric interpretation
b.Definition The moment of F about the axis LL is the rectangular component of MO along the axis LL, where O is any point on LL. Letting λ be a unit vector parallel to LL, this definition gives
in vector form: mx(P+Q)= 28 i Nm
Sample problem 2.4 Determine the magnitude of the force F given that its moment about an axis directed from point B toward point C is 137.3 lb·ft. Solution:
4 0 0 F 1 F = 4 × [3 × (−2) − (−3) × 4] × ∴ mBC ( F ) = − 2 3 − 3 22 20 440 0 4 −2 = 24 F = 137.3 lb ⋅ ft 440
x
∴ F = 120.0 lb
2.5 Couples
a. Definition A couple is defined to consist of two forces that are equal in magnitude, opposite in direction, and have parallel noncollinear lines of action. d: perpendicular distance between the lines of action of two forces ---couple arm b.Moment of a couple about a point scalar calculation:
Chapter 2 Resultants of force systems
2.1 Introduction 2.2 Reduction of Concurrent Force Systems 1.Definition:各力作用线汇交于一点的力系---concurrent forces . 2.Reduction----replacing a system of concurrent forces with a single equivalent force.
3. Moving the couple to a parallel position in its plane
4. Moving the couple to a parallel plane
以上四种变化均不会改变力偶对物体的作用效果
因此, 对图(a)原力偶的表示可简化为(b)or(c).
A couple is a free vector. d. The addition and resolution of couples
3.Resultant force----the vector sum of all forces acting on the body. The line of action of R must pass through original point of concurrency, “O”.
R = ∑ F = F1 + F2 + F3
4 3 F = −( )200i + ( )200 j 5 5 = −160i + 120 j Ib
2.Scalar solution: component of F at point B in vector form:
3. Scalar solution: component of F at point C
S
Substitution of r×F2 · λ=F2d, where d is the perpendicular distance from O to the line of action of F2, yields
mBC(F) = mB(F)⋅λBC = rBA×F⋅λBC rBA=4i ft F=FλAD F − 2i + 3 j − 3k )= = F( (−2i + 3 j − 3k ) 2 2 2 22 2 +3 +3 4 j − 2k 1 λ BC = = ( 4 j − 2k ) 20 42 + 22
2.4 Moment of a Force About a Axis
力对轴之矩是一个力使一 个物体绕轴转动趋势的度量
a. Physical characteristic (l) Py is the only component that will rotate the door; (2) the door will be easier to rotate if the magnitude of Py is increased; (3) the door will be easier to open if the distance from the hinge axis to the doorknob is increased; (4) Py will cause the door to rotate in the direction as shown.
R = R + R = 27.5
2 x 2 y
θ = tg
−1
Ry Rx
= 24.4°
Method 2 Vector
2.3 Moment of a Force About a Point
1.Definition The moment of a force F about a point O, called the moment center, is defined as A vector Mo= r×F where: r is the vector from the point O to any point on the line of action of F. Unit: N ⋅m. The magnitude of Mo is
The equivalent scalar equations are:
Rx = ∑ Fx R y = ∑ Fy Rz = ∑ Fz
If the original forces lie in xy-plain, then ∑Fz=0, thus:
Rx = ∑ Fx R y = ∑ Fy
Example 2-1 Determine the resultant force for the forces shown in Fig.. Solution: Method 1 Scalar
推论:
1. Changing the magnitude F of each force and the perpendicular distance d, while keeping the product Fd constant
2. Rotating the couple in its plane
Therefore, if the moment of a force about an axis is to correctly describe the tendency of the force to rotate a body about that axis, we anticipate that its definition will include the following physical characteristics. 1. A force that is parallel to an axis has no moment about that axis. 2. A force that intersects an axis has no moment about that axis. 3. The moment of a force about a given axis depends on the magnitude of the force component that is perpendicular to the axis and on the distance of this component from the axis. 4. The direction of the moment of the force represents the direction of the rotational tendency of the force.
For MLL=Mo·λ ∴ Mx=Mo·i; My=Mo·j; ∴ Mo= Mxi+ Myj+ Mzk Mz=Mo·k
Sample problem 2.3
Solution: From principle of moments:
mx(P) = mx(Pz)+mx(Px) = 0.4×120=48 Nm + mx(Q) = -0.4 × 50 = -20 Nm ∴ + mx(P+Q) = 48-20 = 28 Nm +
vector calculation:
M
c. Characteristics: (1)a couple has no resultant force; (2)moment of a couple is the same about every point; (3)two couples that have the same moment are said to be equivalent.
where d is the perpendicular distance from O to the line of action of F, called the moment arm of F. Then: Mo=Fd
2.Principle of Moments The moment of a force about a given point is equal to theБайду номын сангаасsum of the moments of its components about the point.
M O = i (ry Fz − rz Fy ) − j(rx Fz − rz Fx ) + k (rx Fy − ry Fx )
Sample Problem 2.2 Determine the moment of the force F in Fig. about point A. 1.vector solution Writing vector F as following
MLL=F2d
C. Rectangular components Writing r and F in rectangular representations: r=rxi+ryj+rzk F= Fxi+Fyj+Fzk rx ry rz M LL = r × F ⋅ λ = Fx Fy Fz λx λ y λz
3.Rectangular components Writing r and F in scalar form: r=rxi+ryj+rzk F=Fxi+Fyj+Fzk
i M O = r × F = rx Fx j ry Fy k rz Fz
Expanding the determinant gives
MLL=MO cosα
L
where α is the angle between MO and λ.
M LL = M O ⋅λ = r × F ⋅λ
具体计算时可利用:
(a) Principle of moment The moment of a force about a given axis is equal to the sum of the moments of its components about that axis. (b) Geometric interpretation
b.Definition The moment of F about the axis LL is the rectangular component of MO along the axis LL, where O is any point on LL. Letting λ be a unit vector parallel to LL, this definition gives
in vector form: mx(P+Q)= 28 i Nm
Sample problem 2.4 Determine the magnitude of the force F given that its moment about an axis directed from point B toward point C is 137.3 lb·ft. Solution:
4 0 0 F 1 F = 4 × [3 × (−2) − (−3) × 4] × ∴ mBC ( F ) = − 2 3 − 3 22 20 440 0 4 −2 = 24 F = 137.3 lb ⋅ ft 440
x
∴ F = 120.0 lb
2.5 Couples
a. Definition A couple is defined to consist of two forces that are equal in magnitude, opposite in direction, and have parallel noncollinear lines of action. d: perpendicular distance between the lines of action of two forces ---couple arm b.Moment of a couple about a point scalar calculation:
Chapter 2 Resultants of force systems
2.1 Introduction 2.2 Reduction of Concurrent Force Systems 1.Definition:各力作用线汇交于一点的力系---concurrent forces . 2.Reduction----replacing a system of concurrent forces with a single equivalent force.
3. Moving the couple to a parallel position in its plane
4. Moving the couple to a parallel plane
以上四种变化均不会改变力偶对物体的作用效果
因此, 对图(a)原力偶的表示可简化为(b)or(c).
A couple is a free vector. d. The addition and resolution of couples
3.Resultant force----the vector sum of all forces acting on the body. The line of action of R must pass through original point of concurrency, “O”.
R = ∑ F = F1 + F2 + F3
4 3 F = −( )200i + ( )200 j 5 5 = −160i + 120 j Ib
2.Scalar solution: component of F at point B in vector form:
3. Scalar solution: component of F at point C