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?
2 answers:
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
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
The compression forces acting on both sides of the center board = <u>70.338 N</u>
<h3>Further explanation </h3>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
- 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
<h3>Learn more </h3>
law of motion
displacement of a skateboarder
The distance of the elevator
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Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s
Explanation:
Let's start out with finding the force acting downwards because of the mass of 'The Rock':
Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N
Now the problem also states that the kinetic friction of the desk in this problem is 370 N
Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N
Now lets use F = ma to calculate for the acceleration of the desk:
787.58 = 63 x acceleration
acceleration = 12.501 m/s
Finally, we can use the motion equation:
here u = 0 m/s (since initial speed of the desk is 0)
a = 12.501 m/s
and s = 10 m
Solving this we get:
Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.