NLM BYU 30

The block $P$ is described as “light,” which implies its mass is negligible ($m_P \approx 0$). For a massless block to exert a normal force on the table (and thus experience friction), the tension forces acting on it must have a downward vertical component. This implies the pulleys are slightly lower than the attachment points on block P, creating a small angle $\theta$ with the horizontal.

Let $F$ be the force applied by the hand and $T = mg$ be the tension due to the load. The normal reaction $N$ on block P is due to the vertical components of these tensions: $$N = F \sin\theta + mg \sin\theta = (F + mg)\sin\theta$$ The limiting friction force is $f = \mu N = \mu (F + mg)\sin\theta$.

Case 1: Minimum Force ($F_{\min}$)
The hand pulls weakly, so the block tends to slide towards the load $m$. Friction acts towards the hand (left). $$mg \cos\theta = F_{\min} \cos\theta + f$$ $$mg \cos\theta = F_{\min} \cos\theta + \mu (F_{\min} + mg)\sin\theta$$ Rearranging to isolate $F_{\min}$: $$F_{\min}(\cos\theta + \mu\sin\theta) = mg(\cos\theta – \mu\sin\theta)$$ $$F_{\min} = mg \left( \frac{\cos\theta – \mu\sin\theta}{\cos\theta + \mu\sin\theta} \right)$$

Case 2: Maximum Force ($F_{\max}$)
The hand pulls strongly, so the block tends to slide towards the hand. Friction acts towards the load (right). $$F_{\max} \cos\theta = mg \cos\theta + f$$ $$F_{\max} \cos\theta = mg \cos\theta + \mu (F_{\max} + mg)\sin\theta$$ Rearranging to isolate $F_{\max}$: $$F_{\max}(\cos\theta – \mu\sin\theta) = mg(\cos\theta + \mu\sin\theta)$$ $$F_{\max} = mg \left( \frac{\cos\theta + \mu\sin\theta}{\cos\theta – \mu\sin\theta} \right)$$

Notice that the expression for $F_{\max}$ is the reciprocal of the term in brackets for $F_{\min}$, scaled by $(mg)^2$. Let’s multiply the two equations: $$F_{\min} \cdot F_{\max} = \left[ mg \left( \frac{\cos\theta – \mu\sin\theta}{\cos\theta + \mu\sin\theta} \right) \right] \cdot \left[ mg \left( \frac{\cos\theta + \mu\sin\theta}{\cos\theta – \mu\sin\theta} \right) \right]$$ The bracketed terms cancel out perfectly: $$F_{\min} F_{\max} = (mg)^2$$ $$m = \frac{\sqrt{F_{\min} F_{\max}}}{g}$$