Forces, Density and Pressure: Question 8

Worked solution

Setting up the problem

Take A as the origin, with the beam lying horizontally towards B. The forces on the beam are: its own weight $120\text{ N}$, acting at its midpoint, $1.5\text{ m}$ from A; the lamp’s weight $60\text{ N}$, acting at B, $3.0\text{ m}$ from A; the cable’s tension $T$, acting at B at $40^\circ$ to the beam, with vertical component $T\sin40^\circ$ and horizontal component $T\cos40^\circ$ pulling B towards the wall; and the hinge force at A, with horizontal component $H$ and vertical component $V$.

Part (a): Tension in the cable

Take moments about A. The hinge force acts at A, so it has zero moment about A. The horizontal component of the tension, $T\cos40^\circ$, acts along the same horizontal line as the beam, so its perpendicular distance from A is also zero. Only the vertical component $T\sin40^\circ$, acting at B ($3.0\text{ m}$ from A), produces a moment about A.

The beam’s weight and the lamp’s weight both act downward, tending to rotate the beam so B dips down (clockwise); the vertical component of the tension acts upward at B, opposing this (anticlockwise). By the principle of moments: $T\sin40^\circ \times 3.0 = (120\times1.5) + (60\times3.0)$ $T\sin40^\circ \times 3.0 = 180 + 180 = 360$ $T\sin40^\circ = 120$ $T = \frac{120}{\sin40^\circ} = \frac{120}{0.6428}$ $T \approx 186.7\text{ N} \approx 187\text{ N (3 s.f.)}$

Part (b): Components of the hinge force

Vertically ($\sum F_y = 0$): the vertical component of the tension (upward) must balance the beam’s weight, the lamp’s weight, and the hinge’s vertical component: $V + T\sin40^\circ = 120 + 60$ $V + 120 = 180$ $V = 60\text{ N (upward)}$

Horizontally ($\sum F_x = 0$): the only other horizontal force is the horizontal component of the tension, which pulls the beam towards the wall. The hinge’s horizontal component must balance this, pushing the beam away from the wall: $H = T\cos40^\circ = 186.7\times\cos40^\circ = 186.7\times0.7660$ $H \approx 143.0\text{ N} \approx 143\text{ N (3 s.f.), directed away from the wall}$

Part (c): Resultant force at the hinge

Combine the perpendicular components with Pythagoras’ theorem: $R = \sqrt{H^2+V^2} = \sqrt{143.0^2+60.0^2} = \sqrt{20{,}449+3{,}600} = \sqrt{24{,}049}$ $R \approx 155.1\text{ N} \approx 155\text{ N (3 s.f.)}$

Final answers

• (a) $T \approx \boxed{187}\text{ N}$
• (b) Horizontal component $\approx \boxed{143}\text{ N}$ (away from the wall); vertical component $= \boxed{60}\text{ N}$ (upward)
• (c) Resultant hinge force $\approx \boxed{155}\text{ N}$