Consider a rope that is connected to the top of a 40 kg crate, which is sitting

Question

Consider a rope that is connected to the top of a 40 kg crate, which is sitting on the ground. The other end of the rope is passed over a frictionless, massless pulley that is mounted to the ceiling. A child, 32 kg in mass, climbs up the free end of the rope accelerating upwards at a rate of 0.4 m/s 2 . If the crate leaves the ground, assume it does so instantaneously.

a) What is the magnitude and direction of the crate’s acceleration? (Hint: Consider carefully the conditions necessary to raise the crate off the ground and whether those are met)

b) What if the child begins to lose her grip and accelerates downward at 2.2 m/s, 2, what then will be the crate’s acceleration? (Hint: Again, you need to know whether the crate has left the ground from part a)).

c) What is the maximum acceleration the child can have (upwards) without the crate leaving the ground?

Explanation / Answer

A) Let the tension be T, Writing force equation on the child

T- mg = ma

T = 32*9.8 + 32*0.4 = 326.4 N

Acceleration of crare a= (T-Mg) /M

= (326.4 - 40*9.8)/40 = - 1.64

Acceleration can't be negative, hence crate is still not moving.

acceleration of crate = 0 m/s^2

B) In this case T = mg +ma = 32*(9.8-2) = 249.6 N

This tension is even more smaller, so crate will not move.

Acceleration of crate = 0 m/s^2

C) Tension to lift crate = Mg = 40*9.8 = 392 N

Now Net force on child = T-mg

ma = T-mg

a = (392 - 32*9.8)/32

= 2.45 m/s^2 Answer

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