理论力学教案2

2.5 Moment of a Force about a Point

1. Definition ：The moment of the force F about point O is defined as: 力F 对点A 的矩定义为

Explanation:

F : an arbitrary force (任一个力)

O : an arbitrary point (任一点)

A : any point on the line of action

of F (力F 作用线上任一点)

r : the vector from point O to point A

(从点O 到点A 的矢量)

(a) Dimension: [F L ];

(b) Direction: perpendicular to both r and F .

2. Geometric interpretation ：

The magnitude of M O is given by: M O = | M O | = |r × F| = rF sin θ

Note that d = r sin θ, so

M O = Fd

Vector direction: by the right-hand rule. 矢量指向: 右手定则

3. Principle of moments ：

The moment of a force about a point is equal to the sum of the moments of its components about the point. 力对一点的矩等于该力的各个分量对该点矩的和。

推论：(合力矩定理)：平面汇交力系对于平面内任一点的矩等于所有各力对于该点矩的代数和。

4. Vector and scalar methods

(a). Vector method

· Write F in vector form;

· Write r in vector form;

· Use the determinant form of r ×F ;

(b). Scalar method

M O = Fd The scalar method is convenient only when the moment arm d can be easily determined.

F

r )F (M O ?==?=F r )F (M

O

Sample Problem 2.3

Determine (1) the moment of the force F about point C ; and (2) the perpendicular distance between C and the line of action of F.

Solution: (1)

·

F

· Chose and write an r in vector form: r CA = – 2i m

· Use the determinant form of r CA × F:

m

(2) The magnitude of M C is:

d = M C

/ F = 928.6 / 500 = 1.857m

2.6 Moment of a Force about an Axis

Explanation:

F : an arbitrary force

AB : an axis, the moment axis

O : any point on the axis

M O : the moment of force F about point O

M AB : the moment of force F about the axis AB

1. Definition ：The moment of the force F about

the axis AB is the orthogonal component of M O

along the axis AB, where O is any point on AB.

(力F 对轴AB 的矩等于其对点O 的矩M O 沿轴AB

的正交分量，其中O 为轴AB 上的任一点。)

The above definition gives the moment of F about

the axis AB as:

Let λ be a u nit vector directed from A toward B , so M O · λ = M O cos α

So, M AB can be expressed as the following form: M AB = M O · λ = r × F · λ The vector form of M AB :

M AB = M AB λ = (r × F · λ)λ

Rectangular components of M O :

=?=F r M CA C αcos O AB M M =

M x = M O· i M y = M O· j M z = M O· k

Conclusion: The rectangular components of the

moment of a force about the origin O are equal

to the moments of the force about the coordinates

axes.

Or,M O = M x i + M y j + M z k

? Special case: moment axis is perpendicular to F.

In this case, it can be drawn that M O= M AB

2. Geometric interpretation：

Explanation:

F: an arbitrary force

AB: an arbitrary axis

P: a plane perpendicular to the axis AB

O, C: the points where the axis and the line

of action of the force intersect P, respectively

M AB = F2d

Conclusion:

) has

· A force parallel to the moment axis(such as F

1

no moment about that axis.

· If the line of action of a force intersects the moment axis(d=0), the force has no moment about that axis.

· The moment of a force is proportional to its component that is perpendicular to the moment axis (such as F

), and the moment arm (d) of that component.

2

·The sense of the moment is consistent with the direction where the force would tend to rotate a body.

3. Vector and scalar methods

· Vector method:

M AB = M O·λ = r ×F · λ

where x , y and z are the rectangular components of r. · Scalar method: M AB = F 2d

Sample Problem 2.5

The force F of magnitude 195kN acts along the line AB.

(1) Determine the moments of F about each of the coordinates axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that M O = M x i + M y j + M z k.

Solution: (1) Calculate the rectangular

F = F λ= 45i + 180j – 60k kN

· Moment about the x -axis

F = 45i + 180j – 60k kN

M x = – F y · 4 = – 720 kN · m

M y =

F x · 4 = 180 kN · m

M z = 0 kN · m

(2) The moment of F about point O :

Chose and write an r in vector form: r OA = 4k m So, the moment of F about point O is:

= – 720i + 180j N·m

So, M O = M x i + M y j + M z k

Homework: P52: 2.31, 2.39; P63: 2.50, 2.59 =?=F r M OA O