First slide

Different types of Forces

Question

A rod of length L and mass M is acted on by two unequal force F1 and F2 (< F1) as shown in fig.
The tension in the rod at a distance x from end B will be

Easy

A

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

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

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

none of these

Solution

Net force on the rod = F₁ - F₂
$∴ Acceleration of the rod = (F_{1} - F_{2}) / M$
Now consider t}re motion of part BC. Mass of this part = (M / L) x. Then
$F_{1} - T = [(\frac{M}{L}) x] a$
where I is the tension at point C. So
$F_{1} - T = [(\frac{M}{L}) x] [\frac{F_{1} - F_{2}}{M}] = \frac{(F_{1} - F_{2}) x}{L}$
$or T = F_{1} - \frac{(F_{1} - F_{2}) x}{L}$
$or T = F_{1} (1 - \frac{x}{L}) + F_{2} (\frac{x}{L})$ .

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