See the figure given below. A mass of 6 ; kg | vedclass

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(B) Consider the equilibrium of the weight $W$ suspended at the end of the rope.Clearly,the tension $T_{2}$ in the lower part of the rope must balance

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Consider the equilibrium of the weight $W$ suspended at the end of the rope.Clearly,the tension $T_{2}$ in the lower part of the rope must balance the weight of the mass.$T_{2} = m 	imes g = 6 \; kg 	imes 10 \; m s^{-2} = 60 \; N$.Now,consider the equilibrium of the point $P$ under the action of three forces: the tension $T_{1}$ in the upper part of the rope,the tension $T_{2}$ in the lower part of the rope,and the horizontal force of $50 \; N$.The horizontal and vertical components of the resultant force at point $P$ must vanish separately for equilibrium.For the vertical components: $T_{1} \cos 	heta = T_{2} = 60 \; N$.For the horizontal components: $T_{1} \sin 	heta = 50 \; N$.Dividing the two equations,we get:$\frac{T_{1} \sin 	heta}{T_{1} \cos 	heta} = \frac{50}{60} \implies 	an 	heta = \frac{5}{6}$.Therefore,$	heta = 	an^{-1}\left(\frac{5}{6}ight) \approx 39.8^{\circ}$,which is approximately $40^{\circ}$.

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