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A crate of potatoes of mass 9.0 kg is on a ramp with angle of incline 30° to the horizontal. The coefficients of friction are μ...Asked by nadine
A crate of potatoes of mass 12.0 kg is on a ramp with angle of incline 30° to the horizontal. The coefficients of friction are μs = 0.69 and μk = 0.43. Find the frictional force (magnitude and direction) on the crate if the crate is sliding down the ramp.
Answers
Answered by
Kouder Dakhlallah
First, we'll find the force vector of the crate parallel to the slope on which it rests.
Force = (mass)(gravity)cos(30deg)
N = (12.0kg)(9.8m/s^2)cos(30deg)
N = 101.84
Becasue your crate is sliding down the ramp, your direction will be up the ramp.
ìs= Static Friction
ìk= Kinetic Friction
Because you said the crate is moving, we will only be using ìk. (so this means that ìs is negligible information in this problem).
So, (101.84N)(uk) = (101.84)(0.43) = 43.79N
So, the answer to (a) magnitude is 43.79N and (b) the direction is up the ramp.
Hope this helps!
Force = (mass)(gravity)cos(30deg)
N = (12.0kg)(9.8m/s^2)cos(30deg)
N = 101.84
Becasue your crate is sliding down the ramp, your direction will be up the ramp.
ìs= Static Friction
ìk= Kinetic Friction
Because you said the crate is moving, we will only be using ìk. (so this means that ìs is negligible information in this problem).
So, (101.84N)(uk) = (101.84)(0.43) = 43.79N
So, the answer to (a) magnitude is 43.79N and (b) the direction is up the ramp.
Hope this helps!