A rod of length L and mass M is acted on by t | vedclass

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Question

Net force on the rod $=F_{1}-F_{2} \quad\left(\because F_{1}>F_{2}\right)$

As mass of the rod is $M,$ hence acceleration of the rod is:

$a=\f

Vedclass · Accepted Answer

${F_1}\left( {1 - \frac{y}{L}} \right) + {F_2}\left( {\frac{y}{L}} \right)$

Vedclass · Answer

Net force on the rod $=F_{1}-F_{2} \quad\left(\because F_{1}>F_{2}ight)$

As mass of the rod is $M,$ hence acceleration of the rod is:

$a=\frac{\left(F_{1}-F_{2}ight)}{M}$         $...(1)$

If we now consider the motion of part $A B$ of the rod [whose mass is equal to $(M/L)y$ ],then

$F_{1}-T=\frac{M}{L} y 	imes a$

Where $T$ is the tension in the rod at the point $B$ Now, $\quad F_{1}-T=\frac{M}{L} y 	imes\left(\frac{F_{1}-F_{2}}{M}ight)$

or $\quad T=F_{1}\left(1-\frac{y}{L}ight)+F_{2}\left(\frac{y}{L}ight)$

Alternative Method: Considering motion of the other part $B C$ of the rod also, we can calculate tension at the point $B$. In this case, $T-F_{2}=\frac{M}{L}(L-y) 	imes a$

or $\quad T=F_{2}+\frac{M}{L}(L-y) 	imes \frac{\left(F_{1}-F_{2}ight)}{M}$

$=F_{1}\left(1-\frac{y}{L}ight)+F_{2}\left(\frac{y}{L}ight)$

A rod of length $L$ and mass $M$ is acted on by two unequal forces $F_1$ and $F_2 (< F_1 )$ as shown in the following figure.The tension in the rod at a distance $y$ from the end $A$ is given by

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