Question 3 of 6: Question 3 (paper Question III) — Friction: Two Blocks Linked by a Rigid Rod (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2017-May. Candidates were required to answer any 2 of 3 questions in Part A (Statics) and any 2 of 3 in Part B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1–3 = paper Part A, I–III; Questions 4–6 = paper Part B, IV–VI).
Check — figure reconstruction. Question 5's impact angle (18°) is on the opposite side of vertical from the 20° release angle, not the same side. Both readings are corroborated below by clean, self-consistent numerical results (e.g. member D–F resolves to an exact 2.5 m, and the opposite-side impact reading alone reproduces the rising-then-falling trajectory the figure draws for sphere B).
Question 3 (paper Question III) — Friction: Two Blocks Linked by a Rigid Rod (Part A · Statics, equal value)
Given. Block A ($W_A=50$ N) rests on a frictionless incline at 55° to the horizontal; block B ($W_B=100$ N) rests on a second incline at 18° to the horizontal, meeting the first at a kink; A and B are joined by a rigid, pin-ended rod that makes 40° with the extension of the lower (55°) incline — i.e. 15° above the horizontal. Motion is impending, so B is on the verge of sliding on its (frictional) incline.
Find. The coefficient of static friction $\mu_s$ between block B and its incline.
Figure 3 (reconstructed). Blocks A and B on a kinked double incline, joined by a rigid pin-ended rod.
Approach. The rod is pinned at both ends, so it is a two-force member: its internal force acts along its own axis on each block. Block A's incline is frictionless, so A's own equilibrium (2 equations, 2 unknowns: normal force and rod force) solves the rod force independently of B; that rod force then appears as a known reaction on B, whose equilibrium at impending slip (with friction $f=\mu_s N_B$) solves $\mu_s$.
Rod direction. Rod angle above horizontal $=55^\circ-40^\circ=15^\circ$, so $\mathbf{u}_{rod}=(\cos15^\circ,\sin15^\circ)=(0.9659,\,0.2588)$.
Equilibrium of block A (frictionless). With the incline normal $\mathbf{n}_A=(-\sin55^\circ,\cos55^\circ)$ and weight $(0,-W_A)$: $$N_A\mathbf{n}_A+F_{rod}\mathbf{u}_{rod}+(0,-50)=0$$ Resolving along the incline (no friction term) isolates the rod force directly: $F_{rod}\cos(55^\circ-15^\circ)=W_A\sin55^\circ\Rightarrow F_{rod}=\dfrac{50\sin55^\circ}{\cos40^\circ}=53.47\ \text{N (tension)}$, and then $\boxed{N_A=63.05\ \text{N}}$ from the perpendicular equation.
Equilibrium of block B (impending slip). The rod pulls on B with the same 53.47 N, directed back toward A ($-\mathbf{u}_{rod}$). With $\mathbf{n}_B=(-\sin18^\circ,\cos18^\circ)$ and up-slope direction $\mathbf{t}_B=(\cos18^\circ,\sin18^\circ)$: $$N_B\mathbf{n}_B+f_B\mathbf{t}_B+(0,-100)-F_{rod}\mathbf{u}_{rod}=0$$ Solving the two scalar equations: $\boxed{N_B=92.31\ \text{N}}$ and $f_B=84.29\ \text{N}$ (acting up-slope, resisting B's tendency to slide down under gravity and the rod's pull).
Coefficient of friction. At impending slip $f_B=\mu_sN_B$, so $$\boxed{\mu_s=\dfrac{84.29}{92.31}=0.913}$$
