If W is weight of a body, α is angle of an inclined plane and ϕ is angle of friction, then the force required to drag the body when it is just impending to move the plane, is_____

Explanation:

• The screw is considered as an inclined plane with inclination α.
• When a load is being raised, the following force acts on it at a point on the inclined plane.
• Load (W): It always acts in a vertically downward direction, Normal Reaction (N): It acts perpendicular to the inclined plane.
• Frictional force (μN): Frictional force acts opposite to the motion. Since the load is moving up the inclined plane, the frictional force acts along the inclined plane in the downward direction.
• Effort (P): Effort P acts perpendicular to the load W. It acts towards right for lifting the load and towards left for lowering the load.

From FBD of the figure.

ΣF_x = 0

P × sin α + W × cos α = N.

ΣF_y = 0

W × sin α + μN = P × cos α

W × sin α  + μ(P × sin α + W × cos α) = P × cos α 

W × sin α + μW × cos α = P × cos α - μP × sin α 

\(∴ P = \frac{{W\left( {\sin α + \mu \cos α } \right)}}{{\cos α - \mu \sin α }}\)

\( \Rightarrow P = W\left( {\frac{{\tan α + \mu }}{{1 - \mu \tan α }}} \right)\)

\( \Rightarrow P = W\left( {\frac{{\tan α + \tan ϕ }}{{1 - \tan α \tan ϕ }}} \right)\;\;\;\;\left[ {\because\mu = \tan ϕ } \right]\)

∴ P = W × tan (α + ϕ).