An object is resting at the bottom of two strings which are inclined at an angle of $120^{\circ}$ with each other. Each string can withstand a tension of $20 N$ . The maximum weight of the object that can be sustained without breaking the strings is

10 N

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20 N

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20 \sqrt{2} N

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40 N

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Solution

The correct option is B $20 N$

Considering the forces acting in the vertically upward direction, $2 T cos 60^{o} = W$
$W_{m a x} = T_{m a x} = 20 N$
In the horizontal direction, the component of the forces cancel out.

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An object is resting at the bottom of two strings which are inclined at an angle of 120∘ with each other. Each string can withstand a tension of 20 N. The maximum weight of the object that can be sustained without breaking the strings is

The correct option is B 20 N Considering the forces acting in the vertically upward direction, 2Tcos60o=W Wmax=Tmax=20 N In the horizontal direction, the component of the forces cancel out.

An object is resting at the bottom of two strings which are inclined at an angle of $120^{\circ}$ with each other. Each string can withstand a tension of $20 N$ . The maximum weight of the object that can be sustained without breaking the strings is

The correct option is B $20 N$

Considering the forces acting in the vertically upward direction, $2 T cos 60^{o} = W$
$W_{m a x} = T_{m a x} = 20 N$
In the horizontal direction, the component of the forces cancel out.