Find the tension $T_{2}$ in the string for the figure shown, if the surface is frictionless. $(g = 10 m / s^{2})$

12 N

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24 N

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36 N

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60 N

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Solution

The correct option is A $12 N$
Considering $(2 k g + 4 k g + 6 k g)$ as a system, the FBD of the $12 k g$ body is

As $12 g sin 37^{\circ} (= 72 N) > 24 N$ , the system will accelerate down the plane.
Hence net acceleration using Newton's second law, $a = \frac{72 - 24}{12} = 4 m / s^{2}$
Considering $(2 k g + 4 k g)$ as a system,

$6 g sin 37^{\circ} - T_{2} = m a = 6 \times 4$
$T_{2} = (6 \times 10 \times \frac{3}{5}) - 24 = 12 N$

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Similar questions

g = 10 m / s^{2}

Calculate the tension in the string shown in figure. The pulley and the string are light and all surfaces are frictionless. Take g = 10 $m / s^{2}$ .

T_{1}

g = 10 {m/s}^{2}

70 N

4 k g

2 k g

1 k g

T_{1} = 60 N

T_{2}

= 10 {m / s}^{2}

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