# Where Will the Center of Gravity of the Wooden Plank Be?

• zorro
In summary: The Attempt at a SolutionIn summary, the centre of mass of the plank is the mid point of the plank and the torque on both sides is equal.
zorro

## Homework Statement

A thin wooden plank 1m long is kept on the protruding stone with some of its part immersed in water.
Will the centre of gravity of the wooden plank be inside the water or outside?

## The Attempt at a Solution

For the wooden plank to be in equilibrium, moments of buoyancy force and weight about the point of contact of stone should be equal.
Buoyancy force acts at the midpoint of the immersed part.
But what will be the position of the centre of mass (gravity) of the plank?

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How long is the rod?

1 metre

The position of the centre of mass (gravity) of the plank is the mid point of the plank.

rl.bhat said:
The position of the centre of mass (gravity) of the plank is the mid point of the plank.

then does it mean the torque on both side is equal?

rl.bhat said:
The position of the centre of mass (gravity) of the plank is the mid point of the plank.

What I asked was whether it will should lie inside the water or outside to get balanced?

Assume that x-axis is oriented along the thin plank and area of cross section of the plank 1 cm^2.
Density of wood = 0.7 g/cm^3 and that of water = 1 g/cm^3
Let x cm be the portion of the plank in the water.
Length of the plank is 100 cm. So the mass of the plank = 70 g.
Mass of the displaced water = x g. Then buoyancy force is proportional to x.

If R is the reaction of the stone, R = (70 - x)
Take the moment about the end of the plank out side the water.

R*10 + x(100 - x/2) = 70*50

Solve the quadratic. If x is more than 50, C.G. is inside the water. Otherwise it is outside the water.

How did you get R = 70-x ?

I have assumed that the component of the weight of the plank perpendicular to x-axis is

L*d*g*cosθ. Length of the plank is 100 cm, density 0.7 g/cm^3

Component of the weight of the displaced liquid = x*1*g*cosθ

For equilibrium, total downward with respect x-axis must be equal to the total upward force. Weight is in the downward direction and buoyancy and the reaction are in the upward direction.
Hence R*g*cosθ = 70*g*cosθ - x*g*cosθ

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Thanks!

## 1. What is buoyancy?

Buoyancy is the upward force exerted by a fluid on an object immersed in it. It is caused by the difference in pressure between the top and bottom of the object.

## 2. How does an object's shape affect its buoyancy?

An object's shape affects its buoyancy because it determines the amount of fluid it displaces. Objects that displace more fluid will experience a greater buoyant force.

## 3. What is the center of gravity?

The center of gravity is the point at which an object's weight is evenly distributed in all directions. It is the point where an object will balance perfectly if suspended.

## 4. How does an object's center of gravity affect its stability?

An object's center of gravity plays a crucial role in its stability. If the center of gravity is located above the object's base, it will be stable, but if it is outside the base, the object will be prone to tipping over.

## 5. How are buoyancy and center of gravity related?

Buoyancy and center of gravity are closely related because the position of an object's center of gravity affects its buoyancy. If the center of gravity is above the object's center of buoyancy, it will be stable and remain upright. However, if the center of gravity is below the center of buoyancy, the object will be top-heavy and may tip over.

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