A 1,080-N crate is being pushed across a level floor at a constant speed by a fo

Question

A 1,080-N crate is being pushed across a level floor at a constant speed by a force F of 220 N at an angle of 20.0 degree below the horizontal, as shown in the figure (a) below. What is the coefficient of kinetic friction between the crate and the floor? If the 220-N force is Instead pulling the block at an angle of 20.0degree above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).

Explanation / Answer

Let W be the weight of the crate. and R the frictional force

(a)
You can separate the force F into a horizontal and vertical component:
F_h = F·cos(20°)
F_v = -F·sin(20°)

The change of momentum in horizontal direction is equal
to the sum of the forces acting in horizontal direction:
dp/dt = F_h - R
<=>
m·a = F·cos(20°) - µ·N
N is the normal force exerted by the plane, which is equal to the sum of the vertical forces acting downwards on the crate:
N = W + F·sin(20°)
=>
m·a = F·cos(20°) - µ·(W + F·sin(20°))

Because the crate moves at constant velocity, the acceleration is zero. Hence:
F·cos(20°) - µ·(W + F·sin(20°)) = 0
<=>
µ = F·cos(20°) / (W + F·sin(20°))
= 200N · cos(20°) / (1080N + 300N · sin(20°))
= 0.158 N

(b)
In this case F acts partially upward:
F_h = F·cos(20°)
F_v = F·sin(20°)
This changes the normal force to:
N = W - F·sin(20°)

The overall momentum balance becomes to:
m·a = F·cos(20°) - µ·(W - F·sin(20°))
<=>
a = [ F·cos(20°) - µ·(W - F·sin(20°)) ] / m

The mass of the crate is given by
W = m·g
<=>
m = W /g

Therefore the acceleration of the crate is:
a = g · [ (F/W)·cos(20°) - µ·(1 - (F/W)·sin(20°)) ]
= 9.81m/s² · [ 0.20 · cos(20°) - µ·(1 - 0.20·sin(20°)) ]
= 0.399m/s²

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