This is a 2 step problem and i have the answer to the first part of the problem

Question

This is a 2 step problem and i have the answer to the first part of the problem i just need help with the second part.

first part:
A block of mass 2.69 kg is placed on top of a
second block of mass 5.13 kg. The coefficient
of kinetic friction between the 5.13 kg block
and the surface is 0.29. A horizontal force F
is applied to the 5.13 kg block.Calculate the magnitude of the force neces-
sary to pull both blocks to the right with an
acceleration of 1.73 m/s2. Assume no slipping
between the two blocks. The acceleration of
gravity is 9.8 m/s2 . Answer in units of N

Answer: 35.75 N

Part 2: (the part i need help with)
Find the minimum coefficient of static friction
µt between the blocks such that the 2.69 kg
block does not slip under an acceleration of
1.73 m/s2.

Explanation / Answer

we have, ma = f 2.69 x 1.73 = u x 2.69 x 9.8 coefficient = u = 0.1765

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