Chapter 11, Problem 21A

In a physics lab you find four different vertically mounted cart wheels that are not free to rotate. Each has a block that hangs from a string wrapped around the wheel. Rank the torques these blocks produce about the wheel axes from greatest to least.

To determine

The rank of the torque of the blocks produce about the wheel axes from greatest to least.

Answer to Problem 21A

The rank of the torque of the blocks produce about the wheel axes from greatest to least is ${\tau }_{\text{C}}>{\tau }_{\text{D}}>{\tau }_{\text{A}}>{\tau }_{\text{B}}$ .

Explanation of Solution

Formula used:

The expression for the applied force as follows:

  $F=mg$

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

The expression for torque as follows:

  $\tau =r\times F$

Here, $r$ is the lever arm and $F$ is the applied force.

Calculation:

Consider the acceleration due to gravity is $g=10\text{m}/{\text{s}}^{\text{2}}$ .

The torque applied on the cart for case A as follows:

  $\begin{array}{l}\tau =r\times F\\ {\tau }_{\text{A}}=r_{\text{A}}{m}_{\text{A}}g\\ {\tau }_{\text{A}}=\left(10\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)\times \left(8\text{ g}\times \frac{1\text{ kg}}{1000\text{ g}}\right)\left(10\text{m}/{\text{s}}^{\text{2}}\right)\\ {\tau }_{\text{A}}=8\times {10}^{-3}\text{ N}\cdot \text{m}\end{array}$

The torque applied on the cart for case B as follows:

  $\begin{array}{l}\tau =r\times F\\ {\tau }_{\text{B}}=r_{\text{B}}{m}_{\text{B}}g\\ {\tau }_{\text{B}}=\left(\text{8 cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)\times \left(\text{9 g}\times \frac{1\text{ kg}}{1000\text{ g}}\right)\left(10\text{m}/{\text{s}}^{\text{2}}\right)\\ {\tau }_{\text{B}}=7.2\times {10}^{-3}\text{ N}\cdot \text{m}\end{array}$

The torque applied on the cart for case C as follows:

  $\begin{array}{l}\tau =r\times F\\ {\tau }_{\text{C}}=r_{\text{C}}{m}_{\text{C}}g\\ {\tau }_{\text{C}}=\left(\text{6 cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)\times \left(\text{14 g}\times \frac{1\text{ kg}}{1000\text{ g}}\right)\left(10\text{m}/{\text{s}}^{\text{2}}\right)\\ {\tau }_{\text{C}}=8.4\times {10}^{-3}\text{ N}\cdot \text{m}\end{array}$

The torque applied on the cart for case D as follows:

  $\begin{array}{l}\tau =r\times F\\ {\tau }_{\text{D}}=r_{\text{D}}{m}_{\text{D}}g\\ {\tau }_{\text{D}}=\left(\text{9 cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)\times \left(\text{9 g}\times \frac{1\text{ kg}}{1000\text{ g}}\right)\left(10\text{m}/{\text{s}}^{\text{2}}\right)\\ {\tau }_{\text{D}}=8.1\times {10}^{-3}\text{ N}\cdot \text{m}\end{array}$

By comparing the values, the torque of the blocks is ${\tau }_{\text{C}}>{\tau }_{\text{D}}>{\tau }_{\text{A}}>{\tau }_{\text{B}}$ .

Conclusion:

Thus, the rank of the torque of the blocks produce about the wheel axes from greatest to least is ${\tau }_{\text{C}}>{\tau }_{\text{D}}>{\tau }_{\text{A}}>{\tau }_{\text{B}}$ .