A 5-m uniform ladder of mass 15 kg is held stationary against a frictionless wall. If the ladder is in a state of impending motion when it makes an angle of 65° with respect to the floor, calculate the force exerted by the wall onto the ladder.
A 5-m uniform ladder of mass 15 kg is held stationary against a frictionless wall. If the ladder is in a state of impending motion when it makes an angle of 65° with respect to the floor, calculate the force exerted by the wall onto the ladder.
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter12: Static Equilibrium And Elasticity
Section: Chapter Questions
Problem 34P: In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the...
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Torque And Static Equilibrium
4. A 5-m uniform ladder of mass 15 kg is held stationary against a frictionless wall.
If the ladder is in a state of impending motion when it makes an angle of 65° with
respect to the floor, calculate the force exerted by the wall onto the ladder.
Solution and Answer:
Since the ladder is uniform, then the center of gravity of
the ladder is at its geometric center. We use the first and
second conditions for equilibrium in order to tackle this
problem. The chosen axis of rotation is at the point
where the ladder makes contact with the ground. Let
Fwall and f be the forces due to the wall and due to
friction, respectively. From the first conditions of
equilibrium, we get:
force exerted by wall
weighe
ermal force
ction
=f+(-Fwatt) = 0
B = n + (-W) = 0
The second condition for equilibrium is: Twall + (-Triction) + Tn + (-Tweight) = 0
The frictional force and the normal force are located at the pivot point. This
further simplifies to:
Twall + (-Tweight) = 0
(Fwatl)(5 m)(sin65°) = (W)(2.5 m)(sin25°)
Now, compute for the value of the force exerted by the wall on the ladder, Fwall-
Hint: Compute for weight, W = mladder9 first before evaluating the value for Fwall-
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