How do the readings of spring and compression balances change with upthrust?

[Johnny Leong]

I would like to discuss the following problem which is about upthrust. Please give me corrections if there is anything I did wrongly.

Question:
The solid metal ball is suspended from the spring balnace S as shown so that it is just above the surface of the water in the displacement can. The spout of the can is just above an empty beaker which rests on a compression balance C as shown in the figure. Assume that the weight of the beaker on the compression balance C is negligible. For the figure, please visit here.

Describe how the readings of both balances will change as
(a) the ball is lowered slowly until it is just below the surface of the water, and
(b) the ball is now lowered slowly until it rests on the botton of the displacement can.

Draw a diagram of the ball showing the forces acting on the ball when it is at rest just below the surface of the water. What is the relationship between these forces?
State and explain how (if at all) the readings of S and C would vary if the displacement can contained a liquie whose density was less than that of water.

My answers:
(a) The reading of spring balance S decreases because there is upthrust acting on the ball. But the reading of the compression balance C increases because the water displaced from the displacement can is dripped on the beaker that is above the balance C. The reading of S will be equal its original reading minus the reading of C.

(b) The reading of the spring balance S will become zero and the reading of the balance C will become the initial reading of the balance S.

The forces acting on the ball when it is at rest just below the surface of the water:
1. Upthrust produced by water (upwards), namely U
2. Weight of the ball (downwards). namely W
3. Tension of the string of the balance S (upwards), namely T
The relationship between them: T = W - U

If the density of the liquid in the displacement can is less than that of water, then the upthrust on the ball should be smaller than that produced by water before. Therefore, the reading of S will be greater and the reading of C will be smaller than in part (a).

 

[Doc Al]

Almost perfect.

Originally posted by Johnny Leong
My answers:
(a) The reading of spring balance S decreases because there is upthrust acting on the ball. But the reading of the compression balance C increases because the water displaced from the displacement can is dripped on the beaker that is above the balance C. The reading of S will be equal its original reading minus the reading of C.

Exactly right.

(b) The reading of the spring balance S will become zero

Right.

and the reading of the balance C will become the initial reading of the balance S.

Why? Once the ball is under the surface, no additional liquid goes into the beaker.

The forces acting on the ball when it is at rest just below the surface of the water:
1. Upthrust produced by water (upwards), namely U
2. Weight of the ball (downwards). namely W
3. Tension of the string of the balance S (upwards), namely T
The relationship between them: T = W - U

If the density of the liquid in the displacement can is less than that of water, then the upthrust on the ball should be smaller than that produced by water before. Therefore, the reading of S will be greater and the reading of C will be smaller than in part (a).

Exactly right.

 

[M98Ranger]

This is because the buoyant force (upthrust) is directly proportional to the density of the liquid displaced. So, with a lower density liquid, there will be less upthrust acting on the ball, resulting in a smaller difference between the readings of S and C.