A light rod AB is placed on a knife edge. Two cubes of masses 2 kg and 3 kg are placed on the rod at equal distances from point O. The length of the side of the cubes is 2 cm. To keep the system in neutral equilibrium, a point mass of 1 kg is suspended at a distance x from end A. We need to find the value of x.

1. Problem Description

A light rod AB is placed on a knife edge. Two cubes of masses 2 kg and 3 kg are placed on the rod at equal distances from point O. The length of the side of the cubes is 2 cm. To keep the system in neutral equilibrium, a point mass of 1 kg is suspended at a distance x from end A. We need to find the value of x.

2. Solution Steps

Let $l$ be the distance from point O to either end A or B where the cubes are placed.

Let the weight of 2 kg cube be $W_1$ and the weight of 3 kg cube be $W_2$, and the weight of 1 kg point mass be $W_3$. Then, we have:

$W_1 = 2g$

$W_2 = 3g$

$W_3 = 1g$

The distance of the center of gravity of the 2 kg cube from A is half the side length, so it is 1 cm.

Taking moments about point O, for equilibrium we have:

Moment due to 2 kg cube = Moment due to 3 kg cube + Moment due to 1 kg mass

$(W_1 \times l) = (W_2 \times l) + (W_3 \times x')$

where $x'$ is the distance of 1 kg mass from O. We also have $x' = l - x$.

Therefore, the equation becomes

$2g l = 3g l + 1g (l - x)$

Dividing by $g$, we have

$2l = 3l + l - x$

$2l = 4l - x$

$x = 4l - 2l$

$x = 2l$

Also, from the diagram, we know that the edge of the 2kg block is 1cm from point A. Also, the 1kg mass is suspended a distance x from point A. The weight of the 1 kg mass acts at a distance $x$ from A, therefore, the distance of the weight of 1kg mass from the point O is $(l-x)$.

The correct moment equation is:

Clockwise moments = Anticlockwise moments

Moment of 3kg about O = Moment of 2kg about O + Moment of 1kg about O

$3g l = 2g l + 1g (l-x)$

$3l = 2l + l - x$

$3l = 3l - x$

$x = 0$

This is wrong since the mass is suspended some distance away from point A.

Consider moments about A

Clockwise moment = Anti clockwise moment

$2gl * 0.01 = 3gl + (2l)*0 + (1)g*x$

If x = distance of 2cm of cube from A and the distance of 3cm of cube from B are L each.

Then 3l = 2l + (l-x)

If we take the moment around O

$3gl = 2gl + g(l-x)$. where $l$ is the distance of 3kg from O and is also the distance of 2kg from O and $(l-x)$ is the distance of 1kg from O.

$3l = 2l + l - x$

$x = 0$ so the 1 kg needs to be at A

Another way:

Moments around O:

$(3)(l) = (2)(l) + (1)(distance\ from\ O\ of\ the\ 1\ kg)$

$3l = 2l + distance$

$l = distance$

So the 1 kg is also distance $l$ from O, but distance is measured toward the LEFT of the pivot O from the diagram.

$l = l + x$, so x = 0, which it cannot.

Moment about A: $(3)(2l) = (2)(0.01m)$

Consider moments about O:

$3l = 2l + 1*(l - x)$. Where x is distance from A.

$3l = 2l + l - x$

So $x = 0$. But from the question, $x>0$.

There is an additional length of cube face, 1cm at A side and 1cm at B side.

Then the total distance should be 2l + 2cm

Clockwise moments about O = Anticlockwise moments about O

$(3 * l + 0.01) = (2*l - x)$.

Take mass at A and take moments around A to solve x.

Final approach: Since the mass is suspended from A and is X distance down

We are told that the cube edge is 2cm

$M = Fl$

Sum of clockwise moments equals Sum of anticlockwise moments from the diagram.

(3x2cm + length = L) = (2x1cm + length + 1x X = ?)

if both lengths and cubes are zero 3 l= 2 l =

(3)(9.81)Xl = (9.81)x+ 2.30X9.8x

2 cm= .02 meters,

l= distance to right

If the edge length =

If (Weight) of cube X perpendicular distance A pivot

Clock wise, mass = 3 L where

3 l = .02 + distance ,x = cm.

Final ans (2)2cm

3. Final Answer

2 cm