A body of mass 60,kg is suspended by means of | vedclass

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Question

(d)

Horizontal component $= T _1 \cos 30^{\circ}- T =0$

Vertical component $=T_1 \sin 30^{\circ}- mg =0$

$	herefore \frac{ T _1 \cos 30^{\ci

Vedclass · Accepted Answer

$103.9$

Vedclass · Answer

(d)

Horizontal component $= T _1 \cos 30^{\circ}- T =0$

Vertical component $=T_1 \sin 30^{\circ}- mg =0$

$	herefore \frac{ T _1 \cos 30^{\circ}}{ T _1 \sin 30^{\circ}}=\frac{ T }{ mg }$

$T = mg \cot 30^{\circ}$

$=60 	imes 10 	imes 1.732$

$=103.9\,gN$

A body of mass $60\,kg$ is suspended by means of three strings $P, Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is $..........g\,N$

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