Calculating Forces of Friction with Coefficient of 0.544 and Weight of 86.5N

[Momentum09]

The board which weighs 86.5N is sandwiched between two other boards. If the coefficient of friction between the boards is 0.544, what must be the magnitude of the horizontal forces acting on both sides of the center board to keep it from slipping downward?



2. Force of friction = coefficient x Normal force
Normal force = mg



3. I first found out what the normal force is by multiplying 86.5 by 9.8 = 847.7. Do I then find the horizontal forces by multiplying that value by the coefficient?

Thank you so much.

 

[malawi_glenn]

it would be eaiser if you supplied a picture.

 

[m2003]

I would like to clarify that the normal force is not equal to the weight of the object. The normal force is the force exerted by a surface on an object that is in contact with it, and it is equal to the weight of the object only when the object is placed on a horizontal surface. In this case, the normal force would be equal to the weight of the center board, which is 86.5N.

To find the magnitude of the horizontal forces acting on both sides of the center board, we can use the equation: F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force. Plugging in the values, we get F = 0.544 x 86.5 = 47.08N. Therefore, the magnitude of the horizontal forces acting on both sides of the center board to keep it from slipping downward is approximately 47.08N. It is important to note that this is the maximum force of friction that can be applied before the board starts to slip. The actual force required to keep the board in place may be lower, depending on other factors such as the roughness of the surfaces and the angle at which the boards are placed.