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A rod of leng...

Question

A rod of length $L$ and mass $M$ is acted on by two unequal to force $F_{1}$ 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 :

F_{1} (1 - \frac{y}{L}) + F_{2} (\frac{Y}{L})

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F_{2} (1 - \frac{y}{L}) + F_{1} (\frac{Y}{L})

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(F_{1} - F_{2}) \frac{y}{L}

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None of these

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Solution

The correct option is A $F_{1} (1 - \frac{y}{L}) + F_{2} (\frac{Y}{L})$
Let the acceleration is $a$ and the tension is $T$ .

The acceleration is given as,

$a = \frac{F_{1} - F_{2}}{M}$

The tension is given as,

$F_{1} - T = m a$

$F_{1} - T = \frac{M}{L} \times y \times \frac{F_{1} - F_{2}}{M}$

$T = F_{1} (1 - \frac{y}{L}) + F_{2} \frac{y}{L}$

Thus, the tension is $F_{1} (1 - \frac{y}{L}) + F_{2} \frac{y}{L}$ .

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