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
Hope this Helps!!!
You might be interested in
Answer:
d = 90 ft
Explanation:
Here in each swing the distance sweeps by the swing is half of the initial distance that it will move
So here we can say that total distance in whole motion is given as
since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2
so sum of such GP is given by the formula