In one line
Write equilibrium along and perpendicular to the plane with friction at its limiting value \( \mu N \); remember an inclined pull changes the normal reaction.
Explained simply
Dragging a heavy suitcase by its handle at an angle feels easier than pushing it at an angle downward. Pulling slightly upward lifts some weight off the floor, which reduces the normal reaction and hence the friction.
Technically, for a pull P at angle \( \theta \) above the horizontal on a body of weight W, \( N = W - P\sin\theta \) and at impending motion \( P\cos\theta = \mu N \).
How it works
- Draw the FBD: W down, N up, P at \( \theta \), friction F opposite to the impending motion.
- Perpendicular to plane: \( N + P\sin\theta - W = 0 \).
- Along the plane: \( P\cos\theta - \mu N = 0 \).
- Solve: \( P = \dfrac{\mu W}{\cos\theta + \mu\sin\theta} \).
- For a push at \( \theta \) below the horizontal, N becomes \( W + P\sin\theta \), so \( P = \dfrac{\mu W}{\cos\theta - \mu\sin\theta} \), always larger.
- The pull is minimum when \( \theta = \phi \) (angle of friction), giving \( P_{min} = W\sin\phi \).
Key formula
Pull: \( P = \dfrac{\mu W}{\cos\theta + \mu\sin\theta} \); \( P_{min} = W\sin\phi \) at \( \theta = \phi \). W, P in N, \( \mu \) dimensionless.
Quick example
W = 500 N, \( \mu = 0.3 \), pull at 30°: \( P = 150/(0.866 + 0.15) = 147.6 \) N. With a horizontal pull, P = 150 N. Minimum pull: \( \phi = \tan^{-1}0.3 = 16.7^\circ \), \( P_{min} = 500\sin16.7^\circ = 143.7 \) N.
Common mistakes
- Taking N = W even when the force is inclined.
- Pointing friction the wrong way.
- Using \( F = \mu N \) when the question asks whether the body moves; first compare required friction with \( \mu N \).
How MU asks it
5 to 6-mark numericals: 'Find the force P required to just move the block'. Often combined with two blocks or with finding \( \mu \) from two experiments (pull and push). Show FBDs, the two equations and the final value.