Aperson is pulling on a block of mass m with a force equal to its weight directed 30 above the horizontal plane across a rough surface, generating a friction 'f' on the block. if the person is now pushing downward on the block with the same force 30 degrees above the horizontal plane across the same rough surface, what is the friction on the block?

a) f

b) 1.5f

c) 2f

d) 3f

The friction on the block is 3f. Answer D.

Explanation:

Hi there!

Please, see the attached figure for a better description of the problem.

Let´s write the sum of all forces in the  vertical direction for the first case in which the person pulls on the block:

∑Fy = 0 (because the block is not accelerated in the vertical direction).

N + Fy - W = 0   (see figure, the forces in the upward direction are positive and those in the downward direction are negative)

By trigonometry: Fy = F · sin (30°) = W · sin (30°). Then:

N + W · sin (30°) - W = 0

Solving for N:

N = W (1 - sin (30°))

Then the friction force can be written as follows:

f = μ · N (where  μ is the coefficient of friction)

f =  μ · W (1 - sin (30°))  = μ· 0.5 · W

In the case in which the person is pushing, the sum of vertical forces will be expressed as follows:

N - W - Fy = 0

N - W - W · sin (30°) = 0

N = W(1 + sin (30°))

Then Fr will be:

Fr = μ · N

Fr = μ · W (1 + sin (30°))

Fr = μ · 1.5 W

Now if we divide FR / f:

Fr/f = μ · 1.5 · W /  μ· 0.5 · W

Fr/f = 3

Fr = 3f

The friction on the block is 3f. Answer D.

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