Question

In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.​

Answer 1

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

• Vertical slope, = 2
• Horizontal slope = 3
• Clockwise moment = 330 ft-lb
• Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

• the A coordinate = (0, 5)
• the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

$= \frac{11 }{25} \times 5, \ \ \frac{14}{25} \times 5\\\\= (2.2, \ 2.8)$

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2,  2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

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