13.5kg crate at rest on 27degree incline.wat force of friction is required to keep crate at a constant velocity once it begins sliding down?
B) what it’s coefficient of friction if the crate is sliding down 5m/s?
- Anonymous1 month ago
For an object sliding down an incline,
acceleration a = g(sinΘ - µ*cosΘ)
"constant velocity" means a = 0. So
0 = 9.8*(sin27º - µ*cos27º)
µ = sin27º / cos27º = tan27º = 0.510
- oldschoolLv 72 months ago
We know the component of weight in line with the incline = mg*sin27 = 13.5*9.8*sin27 = 60N
Since there is no acceleration, Fnet = 0 meaning the weight in line with the incline is exactly countered by the friction.
m*g*sin27 = m*g*µ*cos27
sin27/cos27 = tan27 = µ = 0.51 <<<<<<<
- Anonymous2 months ago
Force of friction must equal the force of gravity acting parallel to the slope = mass * g * sin 27 deg.