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Find the force required to move a load of 100 N up a rough inclined plane, the force being applied parallel to the plane. The inclination of the plane is such that when the same body is kept on a perfectly smooth plane inclined at that angle, a force of 24 N applied at an inclination of 30 degree to the plane keeps the same in equilibrium. Assume coefficient of friction between the rough plane and load equal to 0.3
Asked by donehacking97 | 21 Dec, 2020, 10:09: AM
Figure shows the free body diagram for the forces acting on an object placed on inclined plane.
Weight mg that is acting downward is resolved into two components.
(i) one component mgcosθ normal to inclined surface and (ii) other component mgsinθ parallel to inclined surface

Applied force F is at an angle 30º to the incline plane . This applied force is also resolved into two components
(i) one component ( F cos30 ) parallel to inclined surface  and (ii) other component ( F sin30 ) normal to inclined sutface

we get normal reaction force R from contact surface between object and inclined plane surface .

At equilibrium this normal force R is given as .   R = ( mg cosθ - F sin30 )

Hence we get friction force = μ R = μ ( mg cosθ - F sin30 )

To keep the object at equilibrium on a smooth surface ( μ = 0 ) required force = mg sinθ

hence if object of weight 100 N is kept in equilibrium on inclined surface using 24 N newton force that is inclined at 30o ,

then we have,  mg sinθ = 24 co30    or sinθ = ( 24 cos30 )/ ( m g ) =  20.785 / 100 = 0.208  or  θ = sin-1(0.208) = 12 o

To move the load of 100N upward on rough inclined force , minimum force required in the direction parallel to inclined surface is given as

F =  mgsinθ + μ mg cosθ = ( mg ) [ sinθ + μ cosθ ] = 100 × [ sin12 + 0.3 cos12 ] = 50 N

Answered by Thiyagarajan K | 21 Dec, 2020, 12:13: PM

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