Answer:
The friction on the block is 3f. Answer D.
Step-by-step 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.