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

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SUMMARY

The discussion focuses on calculating the horizontal forces required to prevent a board weighing 86.5N from slipping, given a coefficient of friction of 0.544. The normal force is determined by multiplying the weight of the board (86.5N) by the acceleration due to gravity (9.8 m/s²), resulting in a normal force of 847.7N. The horizontal forces can then be calculated by applying the formula for frictional force, which is the coefficient of friction multiplied by the normal force.

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Momentum09
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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.
 
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