How Does a Ferris Wheel Affect the Forces on a Rider?

AI Thread Summary
A Ferris wheel with a diameter of 16.0 m rotates three times per minute, affecting the forces experienced by a 43.0 kg rider. The net force exerted by the seat on the rider, when halfway between the top and bottom and moving upward, is calculated to be approximately 427 N. The rider experiences two forces: the upward force from the seat and the downward gravitational force, which together create the required centripetal force. The angle of the net force direction, measured inward from the vertical, is determined using arctan(33.95/421.4), correcting earlier miscalculations. The discussion highlights the importance of ensuring calculations are in the correct mode, as errors in radians can lead to confusion.
chooch
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A Ferris wheel that rotates three times each minute and has a diameter of 16.0 m

What force [magnitude and direction (measured inward from the vertical)] does the seat exert on a 43.0 kg rider when the rider is halfway between top and bottom, going up?

Ok, the magnitude is easy...

Fnet = m(centripetal acceleration^2 + acceleration due to gravity^2)^1/2 = 422.77 N

What I can't figure out to save my life is the direction "measured inward from the vertical"..

My thinking is to basically make a triangle (similarly to the method used to find Fnet) with 33.95 N being the opposite side (pushing in), and the adjacent side being 421.4 N (pushing up). This gives 0.08 degrees in from vertical, which is incorrect.

What am I doing wrong here? :confused:

Thanks!
 
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opposite is 33.95. adjacent is 421.4. What is arctan(33.95/421.4) ?
 
The rider is experiencing two forces, the seat (diagonally up and inwards) and his weight (down) these two together produce the (horizontally inwards) centripetal force.

Fnet (the diagonal I get a bit different from you 427 N). The angle that is required is therefore between the weight and Fnet vector.

Also I do not get a vector of 33.95 N in my calculations.
 
andrevdh said:
The rider is experiencing two forces, the seat (diagonally up and inwards) and his weight (down) these two together produce the (horizontally inwards) centripetal force.

Fnet (the diagonal I get a bit different from you 427 N). The angle that is required is therefore between the weight and Fnet vector.

Also I do not get a vector of 33.95 N in my calculations.

I'm getting 33.95N. v = 3.14(16)*3/60 = 2.513m/s. mv^2/r = 43*(2.513)^2/8 = 33.95
 
OK... This is the second time this semester I've made myself look like a fool because I was in radian mode...

Sorry... the answer is indeed arctan(33.95/421.4).

:frown:
 
chooch said:
OK... This is the second time this semester I've made myself look like a fool because I was in radian mode...

Sorry... the answer is indeed arctan(33.95/421.4).

:frown:

Don't feel bad. I've made that same mistake a couple of times right here on the forum!
 
Ok, I took the diameter for the radius in my calcs.
 
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