RRB Junior Engineer / DMS / CMA 2019 · Question 3482 of 3719
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 = W × tan (α + ϕ) .
Source: RRB JE 2019 (CBT 2) (ME) Previous Year Paper (31 Aug 2019) — prepp.in solved-paper PDF (answer key with explanations) · reliable-secondary
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