Question 3 of 6: Question 3 (paper Question III): Friction – linked block on an incline

Question 3 (paper Question III): Friction – linked block on an incline (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Block A (weight 100 N) rests on top of block B (weight 350 N) along an interface parallel to the 40° incline; block A is restrained by a horizontal pin-pin link to a fixed wall. Block B rests directly on the 40° incline. A horizontal force P is applied to block B.

Find. (A) the range of P for which block B remains in equilibrium; (B) the corresponding range of the force carried by the link.

Approach. Because the link fixes block A in space, block B cannot move without sliding relative to both A and the incline at once, in the same sense. The two extremes of the equilibrium range are therefore found by setting friction to its limiting value at both interfaces simultaneously: impending motion up-slope (friction acts down-slope at both interfaces, giving $P_{\max}$) and impending motion down-slope (friction acts up-slope at both interfaces, giving $P_{\min}$). Block A's own equilibrium then gives the link force at each extreme.

1. Resolve along/normal to the incline. Let $\hat{x}'$ be up-slope and $\hat{y}'$ the outward normal, both at $\theta=40^{\circ}$ to the horizontal. Block A carries $W_A$, the A–B contact ($N_{AB}$, friction $\pm\mu_{AB}N_{AB}$), and the horizontal link force $F_{\text{link}}$: $$N_{AB}\hat{y}'\mp\mu_{AB}N_{AB}\hat{x}'+F_{\text{link}}\hat{\imath}-W_A\hat{\jmath}=0$$ (upper sign = impending up-slope, lower = impending down-slope).
2. Solve block A's $2\times2$ system for $N_{AB}$ and $F_{\text{link}}$ at each extreme: $$\text{Impending up: } N_{AB}=149.34\text{ N},\ F_{\text{link}}=113.15\text{ N}\qquad\text{Impending down: } N_{AB}=115.95\text{ N},\ F_{\text{link}}=61.21\text{ N}$$
3. Block B's equilibrium carries $W_B$, the reaction from A ($-N_{AB}\hat{y}'\pm\mu_{AB}N_{AB}\hat{x}'$, Newton's third law), the incline reaction ($N_B\hat{y}'\mp\mu_B N_B\hat{x}'$), and P: $$N_B\hat{y}'\mp\mu_B N_B\hat{x}'+P\hat{\imath}-W_B\hat{\jmath}-N_{AB}\hat{y}'\pm\mu_{AB}N_{AB}\hat{x}'=0$$
4. Solve for $N_B$ and P at each extreme using the $N_{AB}$ values from Step 2: $$\boxed{P_{\max}=199.0\text{ N}}\ (N_B=392.2\text{ N}),\qquad \boxed{P_{\min}=75.6\text{ N}}\ (N_B=279.5\text{ N})$$

Block B therefore remains in equilibrium for $\boxed{75.6\text{ N}\le P\le199.0\text{ N}}$, and the link force ranges over $\boxed{61.2\text{ N}\le F_{\text{link}}\le113.2\text{ N}}$ (compression throughout, i.e. the link pushes block A away from the wall).