Applying Newton's Laws — Question 7

Question 7

Two unequal support angles require unequal tensions

Take $g=10$ m/s$^2$. A 10 kg lamp is at rest. Its left support cord makes $30^\circ$ above horizontal; the right makes $60^\circ$ above horizontal. Both cords are light and taut.

1. Write horizontal and vertical balance equations.

2. Find both tensions.

3. Check the force sum and explain why splitting weight equally is invalid.

Question 7 - Solution

Two unequal support angles require unequal tensions

(a) Resolve each cord in common axes.

$$-T_L\cos30^\circ+T_R\cos60^\circ=0,\quad T_L\sin30^\circ+T_R\sin60^\circ=100$$ $-T_L\cos30^\circ+T_R\cos60^\circ=0$ and $T_L\sin30^\circ+T_R\sin60^\circ=100$ N.

(b) Eliminate one tension.

$$T_R=\sqrt3T_L,\quad2T_L=100\ \Rightarrow\ \boxed{T_L=50\,\mathrm N,\quad T_R=50\sqrt3\,\mathrm N}$$ Horizontal balance gives $T_R=\sqrt3T_L$. The vertical equation becomes $2T_L=100$, so $T_L=50$ N and $T_R=50\sqrt3\approx86.603$ N.

(c) Verify both components.

$$\sum F_x=-25\sqrt3+25\sqrt3=0,\quad\sum F_y=25+75-100=0\quad\mathrm N$$ The horizontal components are $\mp25\sqrt3$ N. The upward components are 25 and 75 N, summing to 100 N. Unequal geometry gives unequal vertical shares; tension itself is also not identical to a vertical share.