First slide

mechanical equilibrium

Question

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

Moderate

A

$\frac{W (d - x)}{x}$
B

$\frac{W (d - x)}{d}$
C

$\frac{W x}{d}$
D

$\frac{W d}{x}$

Solution

Given situation is shown in figure.
$\begin{array}{l} N_{1} = Normal reaction on A \\ N_{2} = Normal reaction on B \\ W = Weight of the rod \end{array}$

$I n v e r t i c a l e q u i l i b r i u m, N_{1} + N_{2} = W$ . . . .(i)
Torque balance about centre of mass of the rod,
$N_{1} x = N_{2} (d - x)$
Putting value of $N_{2}$ from equation (i)
$N_{1} x = (W - N_{1}) (d - x)$
$\begin{array}{l} \Rightarrow N_{1} x = W d - W x - N_{1} d + N_{1} x \\ \Rightarrow N_{1} d = W (d - x) \end{array}$
$∴ N_{1} = \frac{W (d - x)}{d}$

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