Define the moment of a force - Leaving Cert Physics - Question 6 - 2011

When the toy is knocked over, its center of gravity acts below the level of the pivot point (point of support). This creates a moment that restores the toy to its upright position. The gravitational force acting on the center of gravity creates a torque that returns the toy to a state of equilibrium. The design ensures that any tipping will create a torque that increases stability.

For a body to be in equilibrium under the influence of co-planar forces, two conditions must be fulfilled:

1. The vector sum of the forces acting on the body must be zero. This means: $\sum \vec{F} = 0$ where all vertical and horizontal forces balance each other.

2. The sum of the moments about any point must be zero. This condition ensures that there is no net rotational force acting on the body: $\sum (\text{turning moments}) = 0$

To balance the see-saw, we can set up the equation for moments about the fulcrum:

Let the distance of the third child from the fulcrum be x m. The moments can be set up as:

$30 \cdot 1.8 = 45 \cdot 0.8 + 45 \cdot x$

Solving for x: $54 = 36 + 45x$ $45x = 18$ $x = \frac{18}{45} \approx 0.4 m$

Thus, the third child should sit approximately 0.4 m from the fulcrum.

A diagram should show the following forces acting on the child:

• Weight (W) acting downwards due to gravity.
• Normal reaction force (R) acting upwards from the merry-go-round.
• Frictional force (F_friction) acting towards the center, opposing the motion as the child is moving outward due to rotation.

To find the maximum angular velocity (\