Chapter 11, Problem 51A

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 100-g mass was placed at the 25-cm mark, and a 20-g mass at the 10-cm mark, where should a 500-g mass be placed to balance the system?

To determine

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

Answer to Problem 51A

The marking distance where the mass 500g is placed to balancing the system is 56.6cm.

Explanation of Solution

Given:

The given masses are 100g, 20g, and 500g and the marking distance are 25cm and 10cm.

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 basis 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,

  $\begin{array}{l}100\times \left(25-\text{distance}\text{from}\text{fulcrum}\right)+20\times \left(40-\text{distance from fulcrum}\right)\\ =500\left(x-\text{distance from fulcrum}\right)\\ 3300=500x\\ x=\frac{3300}{500}=6.6\end{array}$

Now, add this value to 50cm because it placed in the opposite side of the fulcrum. Therefore, the marking distance is $6.6\text{cm}+50\text{cm}=56.6\text{cm}$ . At the mark 56.6cm, the 500g mass is placed.

Conclusion:

Thus, the marking distance is 56.6cm.