Discussion
A horizontal effort P is applied to raise a block of weight W on a rough surface inclined at an angle θ with the horizontal. If \(\mu_{s}\) is the coefficient of static friction between the block and the surface, the minimum effort P required to impend the upward motion of the block along the surface is
Answer: B
The minimum effort is \(W\left(\frac{\mu_{s} + \tan\theta}{1 - \mu_{s}\tan\theta}\right)\).
When upward motion impends, friction \(\mu_{s}N\) acts down the slope. Resolve along and normal to the incline for the horizontal force P:
\[ \begin{aligned} P\cos\theta &= W\sin\theta + \mu_{s}N \\ N &= W\cos\theta + P\sin\theta \end{aligned} \]
Substituting and dividing by \(\cos\theta\) gives \(P(1 - \mu_{s}\tan\theta) = W(\tan\theta + \mu_{s})\), the keyed expression (equal to \(W\tan(\theta + \phi)\) with \(\tan\phi = \mu_{s}\)).
The expressions with \(\mu_{s} - \tan\theta\) belong to a block about to slide down, and the one with \(1 + \mu_{s}\tan\theta\) in the denominator gets the sign of the normal-force contribution of P wrong.
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