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Homework Statement
1. An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall. If the ladder has a mass of 20 kg and it makes an angle of 60° with the ground, find the force of friction of the ground on the ladder.
Homework Equations
[itex]T[/itex] = Torque
∑[itex]T[/itex] = [itex]T[/itex][itex]_{man}[/itex] + [itex]T[/itex][itex]_{ladder}[/itex] + [itex]T[/itex][itex]_{wall}[/itex]
The Attempt at a Solution
Well I tried to find the Force of Friction F[itex]_{f}[/itex] = μn; F[itex]_{f}[/itex] = μmgcosθ:
mg = μmgcosθ
(20 kg)(-10 m/s^2) = μ(20 kg)(-10 m/s^2)cosθ
μ = 1.05
and
[itex]T[/itex] = F*r*sinθ
[itex]T[/itex] = (800 N)(10 m)(sin 60°)
[itex]T[/itex] = -2438.5 N
I have no idea which of that is right. I only got 2 out of like 10 points for this Free Response question and I need to fix it, but I'm horrible at Physics and really need some help figuring it out. Thank you!