SATHEE: Rotational Motion Question 82

Rotational Motion Question 82

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

Options:

A) $\frac{w x}{d}$

B) $\frac{w d}{x}$

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

D) $\frac{w (d - x)}{d}$

Show Answer

Answer:

Correct Answer: C

Solution:

As the weight w balances the normal reactions. So, $w = N_{1} + N_{2}$ (i) Now balancing torque about the COM, i.e.

anti-clockwise momentum = clockwise momentum

$\Rightarrow$ $N_{1} x = N_{2} (d - x)$ Putting the value of $N_{2}$ from Eq.

(i), we get $N_{1} x = (w - N_{1}) (d - x)$

$\Rightarrow$ $N_{1} x = w d - w x - N_{1} d + N_{1} x$

$\Rightarrow$ $N_{1} d = w (d - x)$

$\Rightarrow$ $N_{1} = \frac{w (d - x)}{d}$

Rotational Motion Question 82

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

Options:

Answer:

Solution:

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Rotational Motion Question 82

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

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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