Chapter 11, Problem 50A

The diagram below shows a ruler balanced with the fulcrum at the 50-cm mark. Copy the diagram onto a sheet of paper and answer Questions 50-52 below.

If a 200-g mass is placed at the 20-cm mark (30 cm from the fulcrum), at what mark should a 500-g mass be placed So that the system balances?

To determine

To calculate: The marking distance where the mass 500g is placed to balancing the system.

Answer to Problem 50A

The marking distance where the mass 500g is placed to balancing the system is 62cm.

Explanation of Solution

Given:

The mass is 200gm which is placed at the distance of 20cm and 30cm from the fulcrum and another mass is 500gm.

Formula used:

According to the principle of moments in equilibrium,

  $\text{Sum}\text{of}\text{anticlockwise}\text{moments}=\text{Sum}\text{of}\text{clockwise}\text{moments}$

Calculation:

Consider the figure shown below.

On the bases of given figure, the fulcrum is at 50cm and according to the principle of moments, the vector sum of moment of the forces acting on the rigid body when a rigid body is in rotational equilibrium. Therefore, the summing moments about this 50cm mark for balancing the system. Take the clockwise moments as positive and anticlockwise moments as negative. So, the distance is,

  $\text{Sum}\text{of}\text{anticlockwise}\text{moments}=\text{Sum}\text{of}\text{clockwise}\text{moments}$

  $\begin{array}{l}200\text{g}\times 30-500\text{g}\times L=0\\ L=\frac{200\text{g}\times 30}{500}=12\text{cm}\end{array}$

Therefore, the mark will be 50cm plus calculated distance. So, the marking distance is

  $50\text{cm}+L=50\text{cm}+12\text{cm}=62\text{cm}$

Conclusion:

Thus, the distance is 62cm.