Dynamics Problem

  • Thread starter jam12345
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  • #1
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I have a homework problem that I'm having absolute difficulty with. The problem reads,
A train is running smoothly along a curved track at the rate of 50 m/s. A passenger standing on a set of scales observes that his weight is 10% greater than when the train is at rest. The track is banked so that the force acting on the passenger is normal to the floor of the train. What is hte radius of curvature of the track?

Can anyone help me please.
 

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  • #2
Gokul43201
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Yes, we can help you. But first you must show us what you've thought of/tried.

Any ideas about what are the forces are acting on the man ? Have you drawn a free-body force diagram ?
 
  • #3
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yes I drew a force diagram and also have tried using the equation dv/dt = v^2/r ...but I'm not getting the correct answer, which I found out was 556 m.
 
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  • #4
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I've tried using gravity plus 10% as the dv/dt and I've also tried using .98 as the dv/dt but nothing works out.
 
  • #5
Doc Al
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identify the forces

What forces act on the man? I see two: the man's weight (downward) and the normal force (perpendicular to the floor). Draw the free body diagram as Gokul43201 suggested.

Hint: What force does a scale measure?
 
  • #6
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Ok, you are right about a = v^2/r, but you have to do a lot more. I had to find the angle that the train is at compared to the track too to get the answer. Heres how I approaced the problem.

Draw free-body diagram. You should have a diagonal line, i drew from top left to bottom right. You dont know what angle its at right now but it doesnt matter. Then from the center of that line I draw a line downward that is mg, the gravitational force. Theres the force due to the track, the v^2/r that is : mv^2/r in the left direction. And the force normal to the track, which goes up and to the right, which i call for now Fn.

Then what you do is sum the forces in the X and Y directions, both will be == 0 because the personal isnt moving inside the train.

Sum Forces X : mv^2/r + FnCos(theta) = 0

the Cos(theta) is the X component of the normal force.

Forces Y dir : -mg + FnSin(theta) = 0

Now one thing before I let you work out the rest, actually two things :

Remember that the visible weight is in the opposite of the normal direction so
-Fw = Fn
and the Force do to the weight is 10% mroe than "g"
-m*g*(1.1) = Fn
dont forget the negative.
Then plug this all in, g=9.81, solve the Y one first for theta, then plug that into X and solve for r, and youre done.
 
  • #7
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Thanx guys for all your help, i wouldnt have gotten it without you :smile: :smile: :smile:
 

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