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  • A rod of uniform mass density XY of mass '' and length <img alt="l" src="https://entrancecorner.oncodecogs.com/gi

A rod of uniform mass density XY of mass 'm' and length l leans with its lower end against the wall. A string connects the rod with the wall such that XZ=\frac{X Y}{4}. The angles made by the string and the rod with the wall are \theta and \phi respectively. Assuming. the force of friction to be zero. The correct relationship between \theta and \phi will be -

Option: 1

2 \tan \theta =\tan \phi


Option: 2

\tan \theta =2\tan \phi


Option: 3

\tan \theta =\tan \phi


Option: 4

2 \tan \theta =3\tan \phi


Answers (1)

As the rod is in equilibrium, the sum of torque about any point should be zero.

Three forces tension,(T), mg and normal reaction, intersects at point P shown in the figure.

\begin{array}{r} T \cos \theta-m g=0 \\ N-T \sin \theta=0 \end{array}

Taking torque about point 'X'

m g\left(\frac{l}{2}\right) \sin \phi=T\left(\frac{l}{4}\right) \sin (\theta+\phi) \\

       {m g \sin \phi}=\frac{{T}}{2} \cdot \sin (\theta+\phi) \\

________________________________________

                mg=T cos\theta

     \begin{aligned} & \sin \phi=\frac{\sin (\theta+\phi)}{2 \cos \theta} \\ & 2 \sin \phi \cdot \cos \theta=\sin \theta \cdot \cos \phi+\cos \theta \cdot \sin \phi \\ & \sin \theta \cdot \cos \phi=\cos \theta \cdot \sin \phi \\ & \tan \theta=\tan \phi \end{aligned}

Posted by

Sumit Saini

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