A vice pushes on a system of three boards, each oriented vertically and held up by horizontal forces. The outer boards weigh 90
N and the inner board weighs 118 N. The coefficient of friction between the inner and the outer boards is 0.67. What is the magnitude of one of the compression forces acting on either side of the inner board?
1 answer:
Answer:
88.1 N
Explanation:
As shown in the free body diagram attached to the question, the only forces acting on the inner block include in the required vertical direction.
- Its Weight.
- Frictional forces as a result of the Two compression forces.
If the block is not to slip off, the weights must match the two frictional forces
Let the compressive forces be F and frictional force be Fr
Fr = μ F
Sum of force acting on the inner block in the y-direction
Fr + Fr - W = 0
μ F + μ F = mg
2 μF = 118
2(0.67) F = 118
F = 118/1.34
F = 88.1 N
Each of the compression forces is 88.1 N
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Answer:
η = 0.667 = 66.7%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the box = Wh
input = work done by man = Fd
Therefore,
where,
W = weight of box = 200 N
h = height gained by box = 1 m
F = force exerted by man = 60 N
d = length of ramp = 5 m
Therefore,
<u>η = 0.667 = 66.7%</u>