What is the compressive force in the following case:Three boards are placed on a slope with the board sandwiched between two other boards weighing 96.5 N. What must be the magnitude of the...

What is the compressive force in the following case:

Three boards are placed on a slope with the board sandwiched between two other boards weighing 96.5 N. What must be the magnitude of the compression forces (assumed horizontal) acting on both sides of the center board to keep it from slipping if coefficient of friction between the boards is 0.57.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The angle made with the horizontal of the slope on which the boards lie is not given. It is assumed to be A.

The weight of the board that lies between the other boards is 96.5 N which is a force that acts vertically downwards. This can be divided into two components, one parallel to the other boards, which is 96.5*g*sin A and the other normal to the board that is below the board in the center equal to 96.5*g*cos A.

The compression forces are horizontal. Let them be equal to Cf. This force can be divided into two components, one along the boards equal to Cf*cos A and one normal to the boards equal to Cf*sin A.

The force of friction between the boards is (96.5*g*cos A+2*Cf*sin A)*0.57. To prevent slippage, the frictional force should be equal to the force acting downwards:

(96.5*g*cos A+2*Cf*sin A)*0.57 = 96.5*g*sin A + Cf*cos A

=> 96.5*g*cos A*0.57+2*Cf*sin A*0.57 = 96.5*g*sin A + Cf*cos A

=> 96.5*g*cos A*0.57 - 96.5*g*sin A = Cf*cos A - 2*Cf*sin A*0.57

=> Cf = (96.5*g*cos A*0.57 - 96.5*g*sin A)/(cos A - 2*sin A*0.57)

The force of compression required to prevent slippage isĀ (96.5*g*cos A*0.57 - 96.5*g*sin A)/(cos A - 2*sin A*0.57) if A is the angle made by the slope with the horizontal.

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