The board sandwiched between two other boards weighs 92.6 N. If the coefficient of friction between the boards is 0.587, what must be the magnitude of the horizontal forces acting on both sides of the center board to keep it from slipping downward? Answer in units of N.
mg=force friction=2*mu(forcehorizontal)
solve for force horizontal
I get mg/2mu
yay thank you!!
To find the magnitude of the horizontal forces acting on both sides of the center board, we need to consider the forces involved in this scenario.
Let's denote the weight of the center board as "W" and the magnitudes of the horizontal forces on both sides as "F."
The weight of the center board is given as 92.6 N. This weight exerts a normal force on the boards above and below it. According to Newton's second law, the normal force is equal in magnitude and opposite in direction to the weight.
Therefore, the normal force acting between each board is also 92.6 N.
The frictional force is proportional to the normal force and is given by the equation: frictional force = coefficient of friction × normal force.
In this case, the coefficient of friction is 0.587, and the normal force between the center board and the other two boards is 92.6 N. Thus, the frictional force on each side is given by: frictional force = 0.587 × 92.6 N.
To prevent the center board from slipping downward, the frictional forces on both sides should balance the weight of the center board. Since the weight is acting downward, the frictional forces should act upward.
So, the magnitude of the horizontal forces on both sides of the center board should be equal to the frictional force on each side.
Therefore, the magnitude of the horizontal forces is: F = 0.587 × 92.6 N.
Calculating this expression gives: F ≈ 54.302 N.
Hence, the magnitude of the horizontal forces acting on both sides of the center board to keep it from slipping downward is approximately 54.302 N.