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!!!
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f is the focal length
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p = 13 cm (object distance)
f = 40 cm (focal length, positive for a converging lens)
So the image distance is
The negative sign means that the image is virtual.
B)
In order to calculate the image height, we use the magnification equation:
where
y' is the image height
y is the object height
In this problem, we have:
y = 1.0 cm (object height)
p = 13 cm
q = -19.3 cm
Therefore, the image heigth is
And the positive sign means the image is upright.