Question 3 of 6: Question 3 (paper Question III) — Range of Weight A for Equilibrium (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 2016 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Separately, Question 1's bracket figure (Figure 1) gives only one anchor point's coordinates (E) plus two angle pairs for the other two cables; every printed dimension (50 mm×2, 100 mm×2, 75 mm, the two 45° arcs at C, the 30°/60° pair at D) was used, but the exact 3-D position of points B and C required one reasonable reconstruction, disclosed at first use below. All other figures (2–6) reconstruct cleanly and unambiguously from the printed dimensions, confirmed by internal checks that return clean values (member lengths BD = 5 ft and DE = 10 ft in Question 6) — strong evidence the readings are correct.
Question 3 (paper Question III) — Range of Weight A for Equilibrium (Part A · Statics, equal value)
Given. Block B: mass 100 kg, resting on a 60° incline, coefficient of static friction $\mu=0.6$ between B and the incline. A frictionless pulley redirects the (massless, inextensible) cable from hanging weight A to block B, so the cable tension equals the weight of A throughout.
Given data
Quantity
Value
$m_B$
100 kg
Incline angle $\theta$
60°
$\mu$ (B–incline)
0.6
Pulley
Frictionless; cable massless, inextensible
Find. The range of the weight (and equivalent mass) of A for which the system remains in equilibrium.
Figure 3 — weight A hangs from a frictionless pulley; the cable runs down the 60° incline to block B ($\mu=0.6$).
Approach. Since the pulley is frictionless and the cable massless, the tension $T$ is the same throughout and equals the weight of A. Block B is on the verge of sliding either up or down the incline at the two limits of equilibrium; apply $\sum F=0$ along the incline at each limit with friction at its maximum value $\mu N$, opposing the impending motion.
Normal force on B. Perpendicular to the incline: $N=m_Bg\cos\theta = 100(9.81)\cos60^{\circ}=490.5\ \text{N}$, so $f_{\max}=\mu N = 0.6(490.5)=294.3\ \text{N}$.
Lower limit (B about to slide down; friction acts up-slope, helping T).
$$T_{\min}+f_{\max}=m_Bg\sin\theta\ \Rightarrow\ T_{\min}=981\sin60^{\circ}-0.6(981)\cos60^{\circ}=\boxed{555.3\ \text{N}}$$
Upper limit (B about to slide up; friction acts down-slope, opposing T).
$$T_{\max}=m_Bg\sin\theta+f_{\max}=981\sin60^{\circ}+0.6(981)\cos60^{\circ}=\boxed{1143.9\ \text{N}}$$
Convert to the weight/mass of A. Since $T=W_A=m_Ag$:
$$m_{A,\min}=\frac{555.3}{9.81}=\boxed{56.6\ \text{kg}},\qquad m_{A,\max}=\frac{1143.9}{9.81}=\boxed{116.6\ \text{kg}}$$
