A block with a mass of 2.00 kg is initially held at rest against the ceiling by

Question

A block with a mass of 2.00 kg is initially held at rest against the ceiling by a force P of magnitude 80.0 N, acting vertically upwards. The coefficients of static and kinetic friction between the block and the ceiling are 0.600 and 0.200, respectively. Suddenly, P changes its direction (without changing its magnitude), so that it points at an angle of 70.0 degrees above the horizontal, as shown in the figure below.

(a) Does the block slide?

(b) What is the force of friction on the block? (Answer: 27.4 N)

Explanation / Answer

a)

force , P = 80 N

mass , m = 2 Kg

let N is the normal force on the ceiling

a) for theta = 70 degree

N = P*sin(theta) - m* g

N = 80 * sin(70) - 2 * 9.8

N = 55.6 N

Now , for the block

horizontal force = 80 * cos(70) = 27.3 N

maximum friction , f = 0.6 * 55.6 N

f = 33.36 N

as the maximum friction force is more than the force for moving the block

the block will not slide.

b)

the force of friction on the block = force applied in horizontal force

the force of friction on the block = 27.3 N

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