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59. Problem 13.56P (HRW) A rigid body can turn around a vertical axis until it presses
against identical rubber stoppers A and B attached to rigid walls at distances
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and
from
the axle. Initially the stoppers touch the walls, without being compressed. Then
force F is applied perpendicular to the rod at a distance R from the axle. We
have to find expressions for the forces compressing (a) stopper A and
(b) stopper B. 
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Solution: Click For PDF Version We draw free-body diagram of the rod. When the rod presses against the
stopper A it will experience force
For static equilibrium the vector sum of forces acting on the rod has to be zero. That is
And, the condition of rotational equilibrium is that the vector sum of torques about the axle has to be zero. That is
Solving these equations for
and
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acting on
it in the anti-clockwise direction and force
acting on
it in the clockwise direction as it compresses the stopper B. According
to Newton’s third law of motion the magnitude of forces compressing stoppers
A and B will be
.
.
,
.