Question

A steel ball is suspended from the ceiling of an accelerating carriage by means of two cords $\text{A}$ and $\text{B}$. Determine the acceleration $a$ ( in ${\text{m/s}}^2$) of the carriage which will cause the tension in A to be twice that in B

• A. $g\sqrt{3}$
• B. $\frac{g}{\sqrt{3}}$
• C. $3\sqrt{3}g$
• D. $\frac{g}{3\sqrt{3}}$

Answer 1

The correct option is D $\frac{g}{3\sqrt{3}}$


Given Condition: $\begin{array}{}{T}_A=2T_B& \text{(i)}\end{array}$
For vertical direction
$(T_A+T_B)\sin {60}^0=mg$
By using equaion $\text{(i)}$
$\begin{array}{}3T_B\sin {60}^0=mg& \text{(ii)}\end{array}$
In horizontal direction
$(T_A-T_B)\cos {60}^0=ma$
By using equaion $\text{(i)}$
$\begin{array}{}{T}_B\cos {60}^0=ma& \text{(iii)}\end{array}$
Dividing $\left(\text{ii}\right)$ by $\left(\text{iii}\right)$
$3\sqrt{3}=\frac{g}{a}$
$\Rightarrow a=\frac{g}{3\sqrt{3}}{\text{m/s}}^2$