Chapter 9, Problem 32A

Rank the scale readings from greatest to least. (Ignore friction.)

To determine

To rank: The scale readings from greatest to least.

Answer to Problem 32A

The rank of the scale readings from greatest to least is $T_{\text{A}}>T_{\text{B}}=T_C$ .

Explanation of Solution

Given:

The given figure is shown below.

  

Also, friction is considered negligible.

Formula used:

The expression for the weight of block is,

  $W=mg$

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

Calculation:

Sketch the free body diagram for each case as shown below.

  

For case A: The mass is supported by a string.

Find the tension on the string. For Equilibrium of forces,

  $T=mg$

For case B: The mass is supported by two strings.

Find the tension on the string. For Equilibrium of forces,

  $\begin{array}{l}T+T=mg\\ 2T=mg\\ T=\frac{mg}{2}\end{array}$

For case C: The mass is supported by two strings.

Find the tension on the string. For Equilibrium of forces,

  $\begin{array}{l}T+T=mg\\ 2T=mg\\ T=\frac{mg}{2}\end{array}$

By comparing the values of tension on string $T_{\text{A}}>T_{\text{B}}=T_C$ .

Conclusion:

Thus, the rank of the scale readings from greatest to least is $T_{\text{A}}>T_{\text{B}}=T_C$ .