First slide

Connected bodies

Question

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

Moderate

A

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

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

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

none of these

Solution

$\begin{array}{l} a = \frac{F_{2} - F_{1}}{M} \\ m = y \frac{M}{L} \\ T - F_{1} = ma = \frac{(F_{2} - F_{1})}{M} \frac{yM}{L} \\ T = F_{1} + (F_{2} - F_{1}) \frac{y}{L} \\ ∴ T = F_{1} (1 - \frac{y}{L}) + F_{2} \frac{y}{L} \end{array}$

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