A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on top of the toboggan. The coefficient of static friction μs between the block and the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is 0.48. The block is pulled by a horizontal force of 30 N as shown.
a] Calculate the block’s acceleration.
b] Calculate the toboggan’s acceleration
c] If the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration?
2. Relevant equations
mBa = 30 – Ff
mTa = Ff
μs * mg = force of static friction
μk * mg = force of kinetic friction
3. The attempt at a solution
So for a) I get:
b) mTa = Ff (I’m still trying to figure out why friction reverses direction for the toboggan, but alas):
All frictions above were calculated using μk.
c) This one’s kinda tricky:
Have to find F when acceleration is equal. Since they accelerate the same, this means the block doesn’t move relative to the toboggan, so we’re using μs rather than μk.
m(B+T)a = F – μs*mg
m(B+T)a = μs*mg
Set them equal (masses of 6kg cancel out):
F – (0.58)*(2kg)(9.8m/s^2) = (0.58)*(2kg)(9.8m/s^2)
F-11.37 = 11.37
F = 22.74N
Please let me know where I made any logical errors, thanks!