Question

A rotational axis is directed perpendicular to the plane of a square and is located as shown in the figure. Two forces, 1 and 2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the figure. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of 2 is 2.0 times that of 1. Find the distances (a) a and (b) b that locate the axis.

Answer 1

Moment = Force * lever arm

In part a, F1 has a lever arm of b and F2 has a lever arm of a.

Sum of the moments about the axis in part a) = 0 (CCW is +)

0 = F1*b - F2*a

With F2 = 2*F1

0 = F1*b - 2*F1*a

0 = b - 2*a

b = 2*a

In part b, F1 has a lever arm of (1-a) because the side is 1 m long and F2 has a lever arm of b.

Sum of the moments about the axis in part b) = 0 (CCW is +)

0 = F1 * (1-a) - F2*b

Sub in F2 = 2*F1

0 = F1 - F1*a - 2*F1*b

0 = 1 - a - 2*b

Sub in b = 2*a

0 = 1 - a - 2*2*a

0 = 1 - a - 4*a

0 = 1 - 5*a

5*a = 1

a = 0.20m

Use that in the b = 2*a equation

b = 2*0.2= 0.4 m

So, a = 0.20 m.....answer.

b = 0.40 m...... answer.

Hope this helps!

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