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problem solved

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Senior member
This seems like it should be very easy, but i can't figure out how it is done

A motorcycle and 58.1 kg rider accelerate at 2.3 m/s2 up a ramp inclined 6.8° above the horizontal. (a) What is the magnitude of the net force acting on the rider? (b) What is the magnitude of the force on the rider from the motorcycle?

The answers are
(a) 133.63 N
(b) (600 ± 10.0) N

I don't understand where part b comes from? any help?
 
Part one is easy. Rider accelerating at 2.3 m/s2, F=M*A, F=2.3 m/s2 * 58.1kg = 133.63N

Second part is a bit tougher. Not only is the seat making the rider go a bit up at 2.3 m/s2, its also preventing him from going down at 9.8m/s2.

So you get to chart all the forces the motorcycle is putting on this guy. You have 9.8*58.1 N directed straight, up, but then you have 133.63 N directed at 6.8 deg above the horizontal. Trig time. I don't have my grapher here so I'm gonna play this like hes on the horizontal (my number will be a bit off therefore)

So, do the pythagorean thing on them numbers. That comes out to 584 N. So, if you break the 6.8 degree force into its verticle and horizontal components, and then do the pythagoras thing, you will get the answer.

you have this sqrt((58.1*9.8+133.63sin(6.8deg))2+(133.63cos(6.8deg))2)

That will give you the right answer.
 
Originally posted by: LordMorpheus
Originally posted by: Heisenberg
Originally posted by: bleeb
The upper division physics is INSANE!!!
That's not upper division physics.

This is early newtonian. Upper division gets into relativity. Thats where all the fun stuff happens.
Relativistic mechanics really isn't that bad. You just have Lorentz transforms instead of Galilean ones. Relativistic quantum mechanics is where it really gets tough.
 
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