A 1.7 kg block is placed on top of a 7.29 kg block. The coefficient of kinetic friction between the 7.29 kg?

Question

block and the surface is .26. A horizontal force F is applied to the 7.29 kg block. The acceleration of gravity is 9.8 m/s/s. Calculate the magnitude of the forces necessary to pull both blocks to the right with an acceleration of 1.71 m/s/s. Assume no slipping between the two blocks. Answer in units of N.

block and the surface is .26. A horizontal force F is applied to the 7.29 kg block. The acceleration of gravity is 9.8 m/s/s. Calculate the magnitude of the forces necessary to pull both blocks to the right with an acceleration of 1.71 m/s/s. Assume no slipping between the two blocks. Answer in units of N.
Part 2... Find the minimum coefficient of static friction mu(sub t) between the blocks such that the 1.7 kg block does not slip under an acceleration of 1.71m/s/s.

Answer 1

First, let's find the force necessary for the acceleration

F=m*a, looking at a FBD
F-f=(1.7+7.29)*1.71
where f is the frictional force
f=(1.7+7.29)*9.8*0.26
so F=(1.7+7.29)*(1.71+9.8*0.26)
=38.28 N

To find the minimum coefficient of static friction for the top block to not slip, again use F=m*a and look at a FBD of the small block
f=1.7*1.71
where f is the force of friction and
1.7*9.8*u=f
solve for u
u=1.71/9.8
=0.174

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