Chapter 4 Force System Resultant, Exercises of Advanced Physics

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Chapter 4

Force System Resultant

4.1 MOMENT OF A FORCE-SCALAR FORMULATION, The moment of a force about a point provides a measure of thetendency for rotation (sometimes called a torque).

MOMENT OF A FORCE - SCALAR FORMULATION

For example, M

O

= F d and the

direction is counter-clockwise.

Often it is easier to determine M

O

by using the components of

F

as shown. Using this approach, M

O

= (F

Y

a) – (F

X

b). Note the different

signs on the terms! The typical sign convention for a moment in2-D is that counter-clockwise is considered positive.

We can

determine the direction of rotation by imagining the body pinnedat O and deciding which way the body would rotate because ofthe force.

F

a

b

d O

a

b

O

F F

x

F

y

4.2 CROSS PRODUCT

In general, the cross product of two vectors

A

and

B

results in

another vector

C

, i.e.,

C = A

×

B

. The magnitude and

direction of the resulting vector can be written as

C

A

×

B

= A B sin

θ

U

C

Here

U

C

is the unit vector perpendicular to both A and B

vectors as shown (or to the plane containing theA and B vectors).

CROSS PRODUCT

Of even more utility, the cross product can be written as Each component can be determined using 2

×

2 determinants.

4.3 MOMENT OF A FORCE – VECTOR FORMULATION

Moments in 3-D can be calculated using scalar (2-D) approach butit can be difficult and time consuming.

Thus, it is often easier to

use a mathematical approach called the vector cross product.Using the vector cross product,

M

O

= r

×

F

.

Here

r

is the position vector from point O to any point on the line

of action of

F

Given:

A 400 N force is

applied to the frameand

θ

Find:

The moment of theforce at A.

Plan: 1) Resolve the force along x and y axes.2) Determine M

A

using scalar analysis.

EXAMPLE

EXAMPLE

(continued)

Solution +

F

y

= -400 cos 20° N

F

x

= -400 sin 20° N

+ M

A

= {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m= 1160 N·m

PROBLEM # 1

Given:

A 40 N force is

applied to the wrench.

Find:

The moment of theforce at O.

Plan:

1. Resolve the forcealong x and y axes.2) Determine M

O

using

scalar analysis.

Solution

F

y

= - 40 cos 20° N

F

x

= - 40 sin 20° N

M

O

= {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm= -7107 N·mm = - 7

11 N·m

PROBLEM SOLVING

Given

: a = 3 in , b = 6 in and c = 2 in

Find

Moment of F about point P

Plan

: 1) Find

r

PA

.

2. Determine

M

P

r

PA

x

F

.

Solution

r

PA

i

j

k

} in

i

j

k

3

6 -

3

2 -

M

P

=

{

i

j

k

}

lb · in