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A steel ball ...

Question

A steel ball is suspended in an accelerating cabin by two wires $A$ and $B$ . If the acceleration of the frame is $a = \frac{g}{3 \sqrt{3}}$ , then the tension in $A$ is $x$ times the tension in $B$ . Find $x$ .

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Solution

Given, acceleration of cabin, $a = \frac{g}{3 \sqrt{3}}$
and $\frac{T_{A}}{T_{B}} = \frac{x}{1}$
Hence, $T_{A} = x T_{B}$ ... (1)
FBD of the steel ball:

Vertically, the ball is in equilibrium.
i.e $T_{B} cos 30^{\circ} + T_{A} cos 30^{\circ} = m g$ .. (2)
Horizontally, the ball is moving with cabin having acceleration $a$ .
i.e $T_{A} sin 30^{\circ} - T_{B} sin 30^{\circ} = m a$ ..... (3)

From eqn. (1) and (2), we get
$T_{B} (\frac{\sqrt{3}}{2}) + x T_{B} (\frac{\sqrt{3}}{2}) = m g$
$\Rightarrow \sqrt{3} T_{B} (1 + x) = 2 m g$ ..... (4)
Again, from eqn. (1) and (3), we get
$x T_{B} (\frac{1}{2}) - T_{B} (\frac{1}{2}) = m \frac{g}{3 \sqrt{3}}$
$\Rightarrow T_{B} (x - 1) = \frac{2 m g}{3 \sqrt{3}}$ ..... (5)
By dividing (4) and (5),
$\frac{\sqrt{3} (1 + x)}{x - 1} = 3 \sqrt{3}$
$\Rightarrow 1 + x = 3 x - 3$
$\Rightarrow 2 x = 4$
$\Rightarrow x = 2$

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Free Body Diagrams

Standard XII Physics

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A steel ball is suspended in an accelerating cabin by two wires A and B. If the acceleration of the frame is a=g3√3, then the tension in A is x times the tension in B. Find x.

A steel ball is suspended in an accelerating cabin by two wires $A$ and $B$ . If the acceleration of the frame is $a = \frac{g}{3 \sqrt{3}}$ , then the tension in $A$ is $x$ times the tension in $B$ . Find $x$ .