How do you analyse a body on a rough inclined plane for impending motion up or down?

In one line

Resolve forces along and perpendicular to the incline and put friction at \( \mu N \) acting opposite to the motion that is about to occur.

Explained simply

Pushing a trolley up a ramp needs extra effort for both gravity and friction. Holding it from rolling down is easier, because friction now helps you.

Technically, on an incline at \( \alpha \), weight W has components \( W\sin\alpha \) down the plane and \( W\cos\alpha \) into the plane. With P parallel to the plane: \( N = W\cos\alpha \).

How it works

1. Take axes along and perpendicular to the plane.
2. Perpendicular: \( N = W\cos\alpha \) (plus any perpendicular component of P).
3. Impending motion up: friction acts down the plane: \( P = W\sin\alpha + \mu W\cos\alpha \).
4. Impending motion down (body would slide if \( \alpha > \phi \)): friction acts up the plane: \( P = W\sin\alpha - \mu W\cos\alpha \).
5. If \( \alpha \le \phi \), the body rests on its own, no P needed to hold it.
6. For a horizontal P, resolve P too; N then changes.

Key formula

Up: \( P = W(\sin\alpha + \mu\cos\alpha) \); down (to hold): \( P = W(\sin\alpha - \mu\cos\alpha) \). W, P in N, \( \alpha \) incline angle.

Quick example

W = 1000 N, \( \alpha = 30^\circ \), \( \mu = 0.25 \). Up: \( 1000(0.5 + 0.2165) = 716.5 \) N. To prevent sliding down: \( 1000(0.5 - 0.2165) = 283.5 \) N. Since \( \tan30^\circ = 0.577 > 0.25 \), the block would slide without P.

Common mistakes

• Using \( W\cos\alpha \) along the plane and \( W\sin\alpha \) for the normal.
• Keeping friction in the same direction for both up and down cases.
• Ignoring the component of an inclined or horizontal P in the normal reaction.

How MU asks it

Very common 6 to 8-mark numericals: a block on an incline, two blocks connected by a string over a pulley on two inclines, or wedge-like arrangements. Draw a separate FBD for each block and state the direction of impending motion before writing equations.