Question

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?

Answer 1

Refer to the diagram shown below.

W = 87.5 N, the weight of the sandwiched board.
μ = 0.622, the static coefficient of friction.

From the free body diagram of the sandwiched board, obtain
2μF = W
F = W/(2μ) = 87.5/(2*0.622) = 70.34 N

Answer: 70.34 N

Answer 2

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

Further explanation

The force acting on a system with static equilibrium is 0

$\large {\boxed {\bold {\sum F = 0}}$

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

$\displaystyle \sum F_x = 0 \\\\\ sum F_y = 0$

For objects undergoing rotation, the equilibrium must be met

$\large {\boxed {\bold {\sum \tau = 0}}$

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

• static friction: fs = us.N
• kinetic friction: fk = uk .N

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

Learn more

law of motion

brainly.com/question/75210

displacement of a skateboarder

brainly.com/question/1581159

The distance of the elevator

brainly.com/question/8729508