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?
Weβll call this maximum supportable
mass capital π. Weβre 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 tableβs
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
plankβs 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 weβre
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 itβs 93
kilograms. Thatβs the maximum amount of mass
we can hang from the end of the plank and still maintain its equilibrium.
