How much force is the seat exerting on the person?

Question

A 59.8kg person is riding on a Ferris wheel that has a radius of 16.3m. When passing through the bottom of the Ferris wheel the person is moving at 8.2 m/s. How much force is the seat exerting on the person ?

Answer 1

Kayla,

You will be using using Newton's 2nd Law which is the sum of all forces acting on an object will be equal to the mass of the object times it's acceleration.

In this case we are going to you the variable y to represent motion in the vertical direction.

∑F_y = ma_y

Where ∑F_y is the sum of the forces in the y direction, m is the mass of the object and a_y is the y component of the acceleration.

Now, we will take upwards vertical as the positive y direction and downwards as the negative y direction.

What are the forces acting on the rider? 1) there is the force F_seat the seat exerts on the rider which will point upwards. 2) the weight of the rider, mg, which will point downwards.

so the sum of the forces acting in the y direction, ∑F_y will equal F_seat - mg

so

∑F_y = ma_y becomes

F_seat - mg = ma_y

but.. what is a_y? Well it turns out the y component of the acceleration will be the radial or centripetal acceleration of the revolving Ferris wheel and at the bottom of the point of the Ferris wheel, that will point upwards vertically towards the axis of the Ferris wheel. so a_y = a_centripetal

a_centripetal = v²/R so

F_seat - mg = ma_y = m(v²/R)

so

F_seat - mg = m(v²/R)

F_seat =mg + m(v²/R) = m(g+v²/R)

F_seat = (59.8 kg) (9.8 m/s² +((8.2 m/s)²/16.3 m) ≅ 833 N