Consider the contraption below. Let the coefficient of static friction between t

Question

Consider the contraption below. Let the coefficient of static friction between the blocK and sled be mu s and the coefficient of kinetic friction between the block and the sled be There is no friction between the sled and the ground. What is the maximum force one can pull on the rope such that the block does not slide on the sled? What is the acceleration of the system given the force in part (a)? What is the non-zero force applied to the rope such that the block has zero acceleration? What is the acceleration of he sled given the force in part (c)?

Explanation / Answer

a)

mb and ms are mass of block and sled resp.
acceleration a= T/(mb + ms)
from FBD of block when its not moving relative 2 sled:
T=s*mb*g - mb*a

So T = s*mb*g - mb*T/(mb + ms)

Solving we hav, T=s*mb*g /(1+(mb/(mb+ms)))

b)

a= T/(mb + ms) = s*mb*g /((1+(mb/(mb+ms))*(mb+ms))

c)

for a=0, frm FBD we get

T=k*mb*g

d)

from FBD of sled:

acceleration = (2T+k*mb*g )/ms = 3k*mb*g /ms

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