One way to put a ladder in place on a wall is to \"walk\" the ladder to the wall
ID: 1696251 • Letter: O
Question
One way to put a ladder in place on a wall is to "walk" the ladder to the wall. One end of the ladder is held fixed (perhaps at the base of the wall); and a person slides his hands along the ladder as he walks toward the fixed end. Anyone who has raised a ladder in this way knows that the amount of force the person must apply grows larger as the person moves toward the end that is fixed. Assuming the force from the person is perpendicular to the ladder, find the force the person must apply to hold the ladder at (a) an angle of 10 degrees, and (b) an angle of 20 degrees?Explanation / Answer
To hold the ladder in place without moving means that the net force and net torque will have to equal 0. If you set the pivot point at the rigid end of the ladder where the ladder meets the wall the force of gravity and the force of the person holding the ladder will have to cancel out.
Gravity is going to create a counter-clockwise(CCW) torque on the ladder
The torque created by gravity will be Fg x lg [ lg in this case will be= 10m x cos(angle)]
Fgravity = 9.8(mass of ladder)
The person holding the ladder is going to create a clockwise(CW) torque on the ladder.
(bare with me this is kind of jumping around)
Since the force applied by the person is going to perpendicular to the ladder The torque the person will need to apply to the ladder will simply be the (Force of the person) x (distance of the ladder from the person to the wall)...e.g. the distance from the persons feet (if standing straight-up) divided by the cosine of the angle)
Force of the person=Fp
r=the length of the ladder from where the person applies the force to the end of the ladder that stays at rest
for net torque to equal zero CW + CCW = 0
so CCW = -CW
(Fg)(10m)(cos(angle)) = -(Fp)(r)
Rearranging this equation and you find that the Force needed to be applied by the person will have to equal
Fp= - [(Fg)(10)(cos(angle))] / r
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