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
[tex]= \frac{11 }{25} \times 5, \ \ \frac{14}{25} \times 5\\\\= (2.2, \ 2.8)[/tex]
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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