A 52 Kg child is riding the Ferris wheel of radius 12m and period 18s. What force does the seat exert on the child at the top and bottom of the ride?
f=m w^2 *r
w=2pi/18 radians/sec
solve for force f
I am not sure what W stands for/is :/
To determine the force exerted by the seat on the child at the top and bottom of the ride, we need to consider the forces acting on the child in those positions. The two major forces are the gravitational force and the normal force exerted by the seat.
At the top of the ride:
1. Gravitational Force (Fg): The child's weight is given by the formula Fg = m * g, where m is the mass of the child (52 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, Fg = 52 kg * 9.8 m/s^2.
2. Normal Force (Fn): At the top of the ride, the normal force exerted by the seat on the child opposes the force of gravity. Therefore, the normal force is equal in magnitude and opposite in direction to the gravitational force. So, Fn = -Fg.
At the bottom of the ride:
1. Gravitational Force (Fg): The child's weight is the same as before: Fg = 52 kg * 9.8 m/s^2.
2. Normal Force (Fn): At the bottom of the ride, the normal force exerted by the seat on the child helps support the child against the force of gravity. Therefore, the normal force and the force of gravity act in the same direction. So, Fn = Fg.
Therefore, the force exerted by the seat on the child at the top of the ride is -Fg (opposite direction to the force of gravity), and the force exerted at the bottom of the ride is Fg (same direction as the force of gravity).
To calculate the numerical values of these forces, you can substitute the value of the child's mass (52 kg) into the formulas shown above.