posted by Donna
A 24.5-kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 77.0 N and is directed at an angle of 30.0° above the horizontal. Determine the coefficient of kinetic friction.
ok, first find the vertical and horizontal components.
for the vertical part, it reduces weight.
upward force= 77Sin30
Now for friction force, that with no acceleration, horizontal component=24.5*cos30=fricton force
put those in the equation above, solve for mu.
ok according to you the answer must be 1.5 but that is wrong :(
x: Fcosα-μN =0,
y: mg-N-Fsinα=0 =>N= mg-Fsinα,
μ= Fcosα/N = Fcosα/(mg-Fsinα)=