# Rotational Equilibrium: Seesaw with Three Bodies

On the seesaw shown to the right (see attachment), mass A is 60 kg. mass B is 30 kg, and mass C is 10 kg. Mass A is 1.0m from the pivot. Mass C is 3.0m from the pivot. The see saw is at an angle of 30 degrees from the horizontal. Calculate the torques of mass A and mass C about the pivot. Where you should place mass B in relation to the pivot for the system to be in rotational equilibrium. (I.e. there is no net torque on the system)?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attached.

The torque is the rotational effect of a force and is given by the product of force and the perpendicular distance of the line of action of the force from the axis of rotation called lever arm.

As the distance of mass A is 1 m from the pivot but this distance makes angle 300 with horizontal hence the horizontal distance (perpendicular to weight which is vertical) from pivot is the component of it and given by d*cos .

Hence torque due to weight of mass A about pivot will be (clockwise positive)

Similarly torque due to weight of mass C about pivot will be (anticlockwise negative)

Now torque due to weight of mass B about pivot will be (anticlockwise negative)

Now for the equilibrium net torque about pivot must be zero and hence we get

Or 509.22 - 254.61 - 254.61 dB = 0

Gives dB = 254.61/254.61 = 1 m

Thus the distance of mass B from the pivot in equilibrium position is 1 m.

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