A 0.2 kg ball on a thread is held 37° from the vertical by a horizontal force. Find the force and the tension — Physics problems | PhysLabo
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Equilibrium of a ball on a thread under a horizontal force

Medium 10 grade Mechanics / Statics
Problem statement

A ball of mass $m = 0.2$ kg hangs on a light thread. Under a horizontal force $F$ the ball is in equilibrium and the thread is deflected from the vertical by an angle $\alpha = 37°$ (see the figure). Find the force $F$ and the tension $T$ in the thread. Take $g = 10$ m/s², $\sin 37° = 0.6$, $\cos 37° = 0.8$.

A ball on a thread deflected from the vertical by angle α under a horizontal force F; mg, T and F are shown
Given
  • $m = 0.2$ kg
  • $\alpha = 37°$
  • $\sin 37° = 0.6$, $\cos 37° = 0.8$
  • $g = 10$ m/s²
  • $F = ?$
  • $T = ?$
Hints
Hint 1 / 2
Three forces balance each other. Project them onto the vertical and horizontal axes.
Hint 2 / 2
The vertical component of the tension balances $mg$: $T\cos\alpha = mg$. The horizontal component balances $F$: $T\sin\alpha = F$.
Solution

The ball is in equilibrium, so the net force is zero. The forces are gravity $m\vec g$ (downward), the horizontal force $\vec F$ and the tension $\vec T$ (along the thread toward the suspension point). Project them onto the axes.

Force triangle: the three vectors add up to zero
Force triangle: the three vectors add up to zero

1. Vertically (the vertical component of the tension supports the ball):

$$T\cos\alpha - mg = 0 \;\Rightarrow\; T = \frac{mg}{\cos\alpha} = \frac{2}{0.8} = 2.5 \text{ N}$$

2. Horizontally, the horizontal component of the tension balances $F$:

$$F - T\sin\alpha = 0 \;\Rightarrow\; F = T\sin\alpha = 2.5 \cdot 0.6 = 1.5 \text{ N}$$

The same result: $F = mg\tan\alpha = 2 \cdot \frac{0.6}{0.8} = 1.5$ N.

Answer

$F = 1.5$ N, $T = 2.5$ N

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