See Figure given below. A mass of $6 \;kg$ is suspended by a rope of length $2 \;m$ from the ceiling. A force of $50\; N$ in the horizontal direction is applied at the midpoint $P$ of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take $g = 10 \;m s^{-2}$). Neglect the mass of the rope.
See Figure given below. A mass of $6 \;kg$ is suspended by a rope of length $2 \;m$ from the ceiling. A force of $50\; N$ in the horizontal direction is applied at the midpoint $P$ of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take $g = 10 \;m s^{-2}$). Neglect the mass of the rope.
Consider the equilibrium of the weight $W$
$\text { Clearly, } T_{2}=6 \times 10=60 \,N$
Consider the equilibrium of the point P under the action of three forces - the tensions $T_{1}$ and $T_{2},$ and the horizontal force $50 N$. The horizontal and vertical components of the resultant force must vanish separately
$T_{1} \cos \theta=T_{2}=60 \,N$
$T_{1} \sin \theta=50 \,N$which gives that
$\tan \theta=\frac{5}{6} \text { or } \theta=\tan ^{-1}\left(\frac{5}{6}\right)=40^{\circ}$
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