# Question Video: Static Equilibrium

A uniform plank of length 8.0 m rests on a level surface with 1.2 m of its lengths overhanging the end of the surface, as shown in the diagram. The plank has a mass of 40.0 kg. What is the maximum mass that can be supported at the overhanging end of the plank that maintains the equilibrium of the plank?

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### Video Transcript

A uniform plank of length 8.0 meters rests on a level surface with 1.2 meters of its lengths overhanging the end of the surface, as shown in the diagram. The plank has a mass of 40.0 kilograms. What is the maximum mass that can be supported at the overhanging end of the plank that maintains the equilibrium of the plank?

Weβll call this maximum supportable mass capital π. Weβre told in this statement that the overall mass of the plank is 40.0 kilograms. We can depict this scenario as hanging a mass of value capital π from the end of the overhanging plank. If this mass was heavy enough to cause the plank to rotate, that rotation would happen at the point where the tableβs edge is. So, that is effectively our axis of rotation. And we want to calculate the torques acting on the plank with respect to that point.

In addition to the torque created by our hanging mass, there is also a torque created by the mass of the plank itself. That torque is equal to the force of gravity acting on the plank equal to π sub π times π multiplied by the distance between the center of mass of the plank and our fulcrum point. We know that the condition of our mass capital π is that it be the largest mass possible without upsetting the plankβs equilibrium. This means we can set the magnitude of the torque created by the mass capital π equal to the magnitude of the torque created by the plank under its own weight.

Recalling that torque magnitude is equal to force magnitude times the distance magnitude times the sine of the angle between these vectors. And recognizing that in our case, π is always 90 degrees. Since the forces being exerted are always perpendicular to the lever arm directions, we can write that capital π times π times 1.2 meters, the distance from the axis of rotation to the point where the force of capital π is applied, is equal to the mass of the plank times π times π.

And we see that the acceleration due to gravity cancels from both sides. We can solve for π by knowing the overall length of the plank as well as the length of the plank on the surface. Based on our diagram, π is equal to 6.8 meters minus 40.0 meters, or 2.8 meters. Now that we know π and since weβre given the mass of the plank π sub π, we can rearrange and solve for capital π.

Capital π is equal to π sub π times 2.8 meters divided by 1.2 meters. Plugging in for π sub π 40.0 kilograms, when we calculate capital π to two significant figures, we find itβs 93 kilograms. Thatβs the maximum amount of mass we can hang from the end of the plank and still maintain its equilibrium.