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

For a body to be in equilibrium under a set of co-planar forces, the following conditions must be satisfied:

1. Vector Sum of Forces: The sum of all vertical forces acting on the body must be zero.

   $ext{Sum of Forces} = 0 ext{ (i.e., } F_{ ext{up}} = F_{ ext{down}} ext{)}$

2. Sum of Moments: The sum of the moments about any point must also be zero.

   $ext{Sum of Moments} = 0 ext{ (i.e., } ext{CTM} = ext{ACTM} ext{)}$

To find the position where the third child should sit in order to balance the see-saw, we can set up the equation for moments about the fulcrum:

Given:

• Child of mass 30 kg is seated at 1.8 m (left side),
• Child of mass 40 kg is stationed at 0.8 m (right side),
• Third child of mass 45 kg is at position $x$ (to be determined).

Thus, we have:

$30g imes 1.8 = 40g imes 0.8 + 45g imes x$

Cancelling $g$:

$30 imes 1.8 = 40 imes 0.8 + 45x$

Calculating each side:

$54 = 32 + 45x$ $54 - 32 = 45x$ $22 = 45x$

Solving for $x$:

x = rac{22}{45} ext{ meters} \\ x hickapprox 0.488 ext{ m}

Thus, the third child of 45 kg should sit approximately 0.49 meters to the right of the fulcrum.

In the provided diagram of the child on the merry-go-round, the forces acting are:

• W (Weight): acting downwards, equivalent to the gravitational force on the child (W = mg).
• R (Reaction): acting upwards, equal and opposite to the weight.
• F (Friction): providing the necessary centripetal force to keep the child on the merry-go-round.

Label these forces in the diagram appropriately.

To find the maximum angular velocity, we can use the equation that relates centripetal force (F), mass (m), and angular velocity ($heta$):

For a child standing at the edge ($r = 2.2$ m), the centripetal force is given by:

F = m imes rac{v^2}{r}

In terms of angular velocity ($heta$), where $v = r heta$:

$F = m imes r heta^2$

Rearranging for $heta$ gives:

heta = rac{F}{mr}

Substituting values:

• $F = 50$ N,
• $m = 32$ kg,
• $r = 2.2$ m,

we find:

heta_{ ext{max}} = rac{50}{32 imes 2.2}

Calculating: heta_{ ext{max}} = rac{50}{70.4} hickapprox 0.710 ext{ rad/s}.

If there were no friction, the child would not be able to exert the necessary force to stay on the merry-go-round, and would therefore move outward due to inertia. Essentially, the child would continue in a straight line tangentially to the arc of the path taken by the merry-go-round at point of release, reflecting Newton's First Law of Motion. The movement would depend on the frame of reference observed.