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I’m self-studying for my A Level mechanics 2 exam (it’s in 4 days :S) and there’s a problem in my book that I just can’t solve. Any help would be really appreciated. :)

1. The problem statement, all variables and given/known data

Find the sum of the moments about the point P of the forces shown in the diagram:

2. Relevant equations

Moment about a point = Force x Perpendicular distance of the force from the point

3. The attempt at a solution

From the way the problem diagram is shown in my book, it seems as though APB is a rigid triangular frame.

Since we’re finding the sum of the moments about the point P, we won’t be considering the green 2N force and the blue 1N force because they both act along P. We won’t be considering the 5N force either because it doesn’t have any effect on the moments about P.

Moment of the blue 2N force about P

= 2N x PB

= 2N x 5m

= 10 Nm anticlockwise

Applying the sine rule:

(AP / sin 18 degrees) = (PB / sin 40 degrees)

Or, AP = sin 18 degrees x (5m / sin 40 degrees)

Hence, AP = 2.4037m

Moment of the 3N force about P

= 3N x 2.4037m

= 7.2111 Nm clockwise

Hence, moments about P

= 10 Nm – 7.2111Nm

= 2.7889Nm anticlockwise

= 2.79 Nm anticlockwise

My book’s answer is 2.07 Nm anticlockwise. Could someone please tell me where I’m going wrong?

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# Homework Help: Statics of rigid bodies – calculating the moments about a point. A Lvl exam in 4 days

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