Chapter 8.2, Problem 18PP

To determine

The mass of basket A.

Answer to Problem 18PP

  $0.056\text{kg}$

Explanation of Solution

Given:

The mass of another basket is $m_B=0.23\text{kg}$ .

Radius of the bigger pulley, $r_A=4.5\text{cm}$

Radius of the smaller pulley, $r_B=1.1\text{cm}$

Formula used:

The torque can be obtained by:

  $\tau =Fr\sin \theta$

Where, F is the applied force, r is the distance of the point of application of force from the axis of rotation, and $\theta$ is the angle between F and r .

Calculation:

The weight of the fruit baskets will generate the torque around the axis of rotation that passes through the center of both the pulleys and is parallel to the surface of the pulleys.

But, the torque generated by both the baskets will be balanced as the baskets are stationary. Thus,

  $\begin{array}{l}{\tau }_A={\tau }_B\\ W_A{r}_A\sin 90°=W_B{r}_B\sin 90°\\ m_{A}g{r}_A=m_{B}g{r}_B\\ m_A=\frac{m_B{r}_B}{r_A}\end{array}$

Using given values in above,

  $\begin{array}{c}{m}_A=\frac{m_B{r}_B}{r_A}\\ =\frac{0.23\times 1.1}{4.5}\\ =0.056\text{kg}\end{array}$

Conclusion:

Thus, the weight of the basket A is $0.056\text{kg}$ .