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A thin plank of mass \mathrm{m} and length \mathrm{l} is pivoted at one end and it is held stationary in horizontal position by means of a light thread as shown in the figure then the force on the pivot is -


 

Option: 1

\mathrm{mg}
 


Option: 2

\mathrm{mg} / 2
 


Option: 3

2 \mathrm{mg}
 


Option: 4

3 \mathrm{Mg}


Answers (1)

best_answer

Free body diagram of the plank is shown in figure.

\therefore plank is in equilibrium condition -

So \mathrm{F_{\text {net }} \& \: \tau_{\text {net }}} on the plank is zero.

from, \mathrm{f_{\text {net }}=0}

         \mathrm{ F_{\text {netx }}=0 }

          \mathrm{ N_1=0 }

Now, \mathrm{\quad f_{\text {net\, y }}=0}

        \mathrm{ N_2+T=m g}..............(1)

from \mathrm{\tau_{\text {net }}=0}

\mathrm{\Rightarrow \tau_{net}} about point \mathrm{A} is zero

so, \mathrm{N_2 \cdot l=m g \cdot l / 2}

\mathrm{ N_2=\frac{m_g}{2} }

Hence option 2 is correct.
 

Posted by

HARSH KANKARIA

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