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A rod of leng...

Question

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 :

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

$Now \begin{matrix} F_{1} - T & = (\frac{M}{L}) y \times a \end{matrix}$ $(∵ (\frac{M}{L}) y is mass of A B)$

$\begin{matrix} F_{1} - T = (\frac{μ_{1}}{L}) y \times (\frac{F_{1} - F_{2}}{m/}) T = F_{1} (1 - \frac{y}{L}) + F_{2} (\frac{y}{L}) \end{matrix}$

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A rod of length 'L' and mass 'M' is acted on by two unequal forces F1 and F2 (<F1) 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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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 :