Chapter 8, Problem 3TS

A 1-kg rock suspended above water weighs 10 N. When the rock is suspended beneath the surface of the water, the scale reads 8 N .

(a) What is the buoyant force on the rock?

(b) If the container of water weighs 10 N on the weighting scale, what is the scale reading when the rock is suspended beneath the surface of the water?

(c) What is the scale reading when the rock is released and rests at the bottom of the container?

To determine

The buoyant force on the rock.

Answer to Problem 3TS

The buoyant force on the rock is 2 N.

Explanation of Solution

Given Info: The weight of the rock is 10 N , and the apparent weight is 8 N .

Write the expression for the buoyant force.

$F=W-W_A$

Here,

W is the weight of the object

W_A is the apparent weight.

Substitute 10 N for W and 8 N for W_A in the above expression to get F.

$\begin{array}{c}F=10\text{N}-8\text{N}\\ F=2\text{N}\end{array}$

Conclusion:

Therefore, the buoyant force on the rock is 2 N .

To determine

The scale reading when the rock is suspended beneath the surface of the water.

Answer to Problem 3TS

The scale reading when the rock is suspended beneath the surface of the water is 12 N .

Explanation of Solution

Given Info: The weight of the rock is 10 N , and the apparent weight is 8 N .

The scale reading when the rock is suspended beneath the surface of the water can be evaluated adding the buoyant force and the actual scale reading.

Write the expression for the new scale reading.

$W^{\prime }=W_s+F$

Here,

W_s is the scale reading

Substitute 2 N for F and 10 N for W_s in the above expression to get W^'.

$\begin{array}{c}{W}^{\prime }=10\text{N}+2\text{N}\\ W^{\prime }=\text{12}\text{N}\end{array}$

Conclusion:

Therefore, The scale reading when the rock is suspended beneath the surface of the water is 12 N .

To determine

The scale reading when the rock is released and rest at the bottom.

Answer to Problem 3TS

The scale reading when the rock is released and rest at the bottom is 20 N .

Explanation of Solution

Given Info: The weight of the rock is 10 N , and the apparent weight is 8 N .

When the rock is released and it rest at the bottom of the container, the reading of the scale will be due to the weight of the rock and weight of the container and water. It can be then evaluated by adding both the weights.

Write the expression for the reading of the scale.

$W^{″}=W+W_s$

Substitute, 10 N for W_s and 10 N for W in the above expression to get ${W^{\prime }}^{\prime }$ .

$\begin{array}{l}{W}^{″}=10\text{N}+10\text{N}\\ W^{″}=20\text{N}\end{array}$

Conclusion:

Therefore, the scale reading when the rock is released and rest at the bottom is 20 N .