Question

The figure below shows an initially stationary block of mass m on a floor

The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F = 0.550mg is then applied at upward angle θ = 24°.  

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(a) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are μs = 0.610 and μk = 0.505?  

(b) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are μs = 0.395 and μk = 0.295?

 What is the vertical component and horizontal component of the applied force? What is the gravitational force? What is the vertical acceleration? From Newton's second law for vertical motion, what is the normal force? What is fs,max? How does that value compare with the horizontal component of the applied force?


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Answer #1

FN fs 8

(a) The magnitude of acceleration of the block across the floor which is given as :

from a free body diagram, we have

on the y-direction :     Fapp sin \theta + FN = Fg

Fapp sin 240 + FN = mg

FN = mg - (0.55 mg) sin 240

FN = mg [1 - (0.55) sin 240]                                                          { eq.1 }

on the x-direction :        Fapp cos \theta - Ff = Fx

Fapp cos 240 - \muk FN = m a                                                          { eq.2 }

(0.55 mg) cos 240 - \muk mg [1 - (0.55) sin 240] = m a

using, \muk = 0.505

(because, the block is initially stationary and required a force greater than the maximum force of kinetic friction)

a = (0.55 g) cos 240 - (0.505) g [1 - (0.55) sin 240]

a = (0.55) (9.8 m/s2) (0.9135) - (0.505) (9.8 m/s2) [1 - (0.55) (0.4067)]

a = (4.92 m/s2) - (3.84 m/s2)

a = 1.08 m/s2

(b) The magnitude of acceleration of the block across the floor which is given as :

Fapp cos 240 - \mus FN = m a

(0.55 mg) cos 240 - \mus mg [1 - (0.55) sin 240] = m a

using, \mus = 0.395

(because, the block is initially stationary and required a force greater than the maximum force of static friction)

a = (0.55 g) cos 240 - (0.395) g [1 - (0.55) sin 240]

a = (0.55) (9.8 m/s2) (0.9135) - (0.395) (9.8 m/s2) [1 - (0.55) (0.4067)]

a = (4.92 m/s2) - (3 m/s2)

a = 1.92 m/s2

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