The board sandwiched between two other boards shown below weighs 87.5 n. if the coefficient of friction between the boards is 0.622, what must be the magnitude of the compression forces (assume horizontal) acting on both sides of the center board to keep it from slipping?

The compression forces acting on both sides of the center board = 70.338 N

The force acting on a system with static equilibrium is 0

(forces acting as translational motion only, not including rotational forces)

For objects undergoing rotation, the equilibrium must be met

The force acting on the touchpad between the two fields is called the normal force  (N)

While the frictional force arises because of 2 objects that come into direct contact, especially when there is motion between two objects that are in direct contact.

There are 2 friction forces

The board sandwiched between two other boards will not fall if

a downward force = an upward force

or ΣFy = 0 (idle / equilibrium)

so that

A downward force = weight of the board sandwiched = 87.5 N

An upward force = arises from two static friction forces namely: us.N (N =  reaction force of the compression forces=F) ---> us.F

then the force that works can be stated:

EFy = 0

W = 2fs

W = 2 (us.F)

87.5 N = 2 (0.622. F)

F = 70,338 N

law of motion

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displacement of a skateboarder

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The distance of the elevator

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car a and b are moving in opposite directions

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