A mass of 10 kg is suspended by a rope of length 4 m , from the ceiling. A force $F$ is applied hori

Question

A mass of 10 kg is suspended by a rope of length 4 m , from the ceiling. A force $F$ is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45^{\circ}$ with the vertical. Then $F$ equals (Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and the rope to be massless)\

Professor PC Thomas Classes · Answer

(b) : $m=10 \mathrm{~kg}$, $l=4 \mathrm{~m}$\ $g=10 \mathrm{~m} / \mathrm{s}^{2}$\ Let $T$ be the tension in the string.\ $A B=2 \mathrm{~m}, B C=2 \mathrm{~m}$\ Using components of force\ we can write that,\ $$\begin{equation*} T \cos 45^{\circ}=m g 	ag{i} \end{equation*}$$ $$\begin{equation*} T \sin 45^{\circ}=F 	ag{ii} \end{equation*}$$ Dividing equation (ii) by (i),\ we get, $\frac{F}{m g}=	an 45^{\circ} ; F=m g=10 	imes 10=100 \mathrm{~N}$\