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  • Three massless rods of length '  ' each are hinged at points X and Y 

Three massless rods of length ' 1 \mathrm{~m} ' each are hinged at points X and Y lying on the same horizontal plane. These rods are joined through hinges at points P and Q respectively. If the distance between points X and Y is 2 \mathrm{~m}, and a mass of1 \mathrm{~kg}' is suspended at hinge P. If the minimum force is being applied to hinge Q, such that middle rod is kept horizontal. Then find the magnitude of such force. [Given,g=10 \mathrm{~m} / \mathrm{s}^2 ]

Option: 1

5 N


Option: 2

10N


Option: 3

\frac{5}{2} N


Option: 4

\frac{5}{\sqrt{2}} N


Answers (1)

Assuming,masses of the hinges as M,

for equilibrium condition at p-

\begin{aligned} & T \cos 30^{\circ}=(M+1) g \cos 60^{\circ} \\ & T=(M+1) g \frac{(1 / 2)}{(\sqrt{3} / 2)}=\frac{(M+1)}{\sqrt{3}} g \end{aligned}

for the minimum force at hinge "Q", force should be

applied perpenticular to the rod.

\begin{aligned} T \cos 30^{\circ} & =F_{\text {min }}+M g \sin 30^{\circ} \\ T\left(\frac{\sqrt{3}}{2}\right) & =F_{\text {min }}+\frac{M g}{2} \\ \end{aligned}

As

,\begin{aligned} & \frac{(M+1)}{\sqrt{\3}} g \times \frac{\sqrt{\3}}{2}=F_{\text {min }}+\frac{M g}{2} \\ & F_{\text {min }}=\frac{M g}{R}+\frac{g}{2}-\frac{M g}{2}=\frac{g}{2} \end{aligned}

 

 

Posted by

Kshitij

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