A mass m is suspended by two strings T_1 and | vedclass

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(B) From the geometry of the figure,the angle between $T_1$ and the vertical is $60^{\circ}$ (since the angle with the wall is $30^{\circ}$),and the a

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$mg, \frac{\sqrt{3}}{2}mg$

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(B) From the geometry of the figure,the angle between $T_1$ and the vertical is $60^{\circ}$ (since the angle with the wall is $30^{\circ}$),and the angle between $T_2$ and the vertical is $60^{\circ}$ (since the angle with the horizontal is $30^{\circ}$).Let the vertical axis be the $y$-axis and the horizontal axis be the $x$-axis.Resolving forces in the horizontal direction ($x$-axis):$T_2 \sin 60^{\circ} - T_1 \sin 60^{\circ} = 0 \implies T_1 = T_2$.Resolving forces in the vertical direction ($y$-axis):$T_1 \cos 60^{\circ} + T_2 \cos 60^{\circ} = mg$.Substituting $T_1 = T_2$:$2 T_1 \cos 60^{\circ} = mg$$2 T_1 (1/2) = mg$$T_1 = mg$.Since $T_1 = T_2$,we have $T_2 = mg$.

$A$ mass $m$ is suspended by two strings $T_1$ and $T_2$ as shown in the figure. Calculate the tensions $T_1$ and $T_2$.

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