A block of mass $$5$$ kg is suspended by a massless rope of length $$2$$ m from the ceiling. A force of $$50$$ N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is $$(Take$$ g $$=$$ $$10$$ $$ms-2)m$$

Question

A block of mass $$5$$ kg is suspended by a massless rope of length $$2$$ m from the ceiling. A force of $$50$$ N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is $$(Take$$ g $$=$$ $$10$$ $$ms-2)m$$

Infinity Learn · Accepted Answer

Let θ be the angle made by the rope with the vertical in equilibrium.The free body diagram of $$5$$ kg block is as shown in fie.(b)In$$ $$equilibriumT2$$ $$=$$ $$5$$ g $$=$$ $$5$$ x $$l0$$ $$=$$ $$50$$ NThe free body diagram of the point P is as shown in Fig. $$(c).In $$equilibriumT1$$ $$sin$$ θ $$=$$ $$50$$ $$N--------(i)T1$$ $$cos$$ θ $$=$$ $$T2$$ $$=$$ $$50$$ $$N------(ii)Dividing$$ (i) by (ii), we gettan θ $$=$$ $$5050$$ $$=$$ $$1$$θ $$=$$ $$tan$$$$-1(1)$$ $$=$$ $$450$$

A block of mass 5 kg is suspended by a massless rope of length 2 m from the ceiling. A force of 50 N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is (Take g = 10 ${ms}^{- 2}$ )m

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A block of mass 5 kg is suspended by a massless rope of length 2 m from the ceiling. A force of 50 N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is (Take g = 10 ms-2)m

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A block of mass 5 kg is suspended by a massless rope of length 2 m from the ceiling. A force of 50 N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is (Take g = 10 ${ms}^{- 2}$ )m