Some Particular Forces — Question 9

Question 9

Two inclined cords share a load through their vertical components

A 6 kg lamp hangs at rest from two identical light cords, each at $30^\circ$ above horizontal and directed upward to opposite sides. Take $g=10$ m/s$^2$.

1. Find the tension in each cord.

2. Find tension if each angle becomes $60^\circ$.

3. Explain what happens to the required tension as the cords approach horizontal.

Question 9 - Solution

Two inclined cords share a load through their vertical components

(a) Cancel horizontal components and support weight.

$$2T\sin\theta-mg=0\ \Rightarrow\ \boxed{T_{30}=\frac{60}{2\sin30^\circ}=60\,\mathrm N}$$ $2T\sin30^\circ=60$ N, so each tension is 60 N.

(b) Increase the vertical share.

$$T_{60}=\frac{60}{2\sin60^\circ}=20\sqrt3=34.641\,\mathrm N$$ $T=60/(2\sin60^\circ)=20\sqrt3\approx34.641$ N.

(c) Examine the small-angle limit.

$$T=\frac{mg}{2\sin\theta},\quad\lim_{\theta\to0^+}T=+\infty$$ $T=mg/(2\sin\theta)$ grows without bound as $\theta\to0^+$. Exactly horizontal cords supply no upward component at finite tension, so they cannot support this nonzero weight in the ideal geometry.