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Amusement ride physics problem

  1. Oct 6, 2004 #1
    Hi everyone :smile: ! I need some help :redface: .

    1.) An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. The combined weight of the car and riders is 5 kN. At the top of the circle the car has a speed of .5 m/s which is not changing at that instant. What is the force in kN of the boom on the car at the top of the circle? The unit vector j is directed vertically upwards.

    MY WORK:
    V=5 m/s
    mg=5
    r=10m

    My teacher said:
    ma= mg + Fb(force of boom)
    [m(v^2)]/r = mg + Fb
    So he said to solve for Fb. When I did this I got .5.

    ANSWER: -3.7j

    2.) A highway curve has a radius of .14 km and has no banks. A car weighing 12 kN goes around the curve at a speed of 24 m/s without slipping. What is the magnitude in kN of the horizontal force of the road on the car?

    MY WORK:
    Fs = (mv^2)/r
    Fs= [(122.45)(24^2)]/140
    Fs= 503.79
    What did I do wrong?

    ANSWER: 5

    3.) A 1500 kg is moving at 17 m/s. How fast is the car moving in m/s 2.2s after the brakes are applied if the coefficient of friction between the car and the road is .32?

    MY WORK:
    mass= 1500/9.8
    mass= 153.06

    F=ma
    .32=153.06a
    a=.0021

    Vf= Vi +at
    Vf = 17 + .0021(2.2)
    Vf= 17

    ANSWER: 10

    Any help would be greatly appreciated
    ~Thanks
     
  2. jcsd
  3. Oct 6, 2004 #2

    Gokul43201

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    mg = 5 ? What does the problem say ?
     
  4. Oct 6, 2004 #3

    Gokul43201

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    122.45 ? Check that. And check the units. Do you know what kN means ?
     
  5. Oct 6, 2004 #4
    It says that the weight is 5 so I need to find the mass. So would the weight be the force?
     
  6. Oct 6, 2004 #5

    Gokul43201

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    The mass is given to you, not the weight. When the unit is kg, this is the mass. Make sure you learn your units first. They are very important.

    0.32 is not the foirce of friction. It is the coefficient of friction. The force of friction is the product of the coefficient and the weight (not the mass).

    Also, will the accelration be positive or negative? Does the car speed up or slow down ?

    This doesn't make sense.

    Okay, you have Vf = Vi + at.

    First write down which of these quanties you know, and which one you don't know. Now sunstitute the known values and determine the unknown.
     
  7. Oct 6, 2004 #6
    Oh thanks. Yeah kN is kilo-newtons which is a measure of weight.
     
  8. Oct 6, 2004 #7

    Gokul43201

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    What are the units ? What does 5 kN mean ? You want to use consistent units throughout, so you should convert this to Newtons.

    But do you understand why you are doing the things that your teacher has suggested ? Do you understand the physics or are you blindly following instructions ?
     
  9. Oct 6, 2004 #8
    :redface: I'm blindly following instructions. Why would you add mg to the force of the boom?
     
  10. Oct 6, 2004 #9

    Gokul43201

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    You should read what your text has to say.

    Learn how to draw free body force diagrams, resolve forces and apply Newton's Second Law.
     
  11. Oct 6, 2004 #10
    For # 3 would I only need to find the the acceleration? I found the acceleration to be 7.73.

    So:
    Vf=Vi +at
    Vf= 17 + (7.73)(2.2)
    Vf=34

    I really don't understand this. Would I even use kinematics? Do I even need to find the weight?
     
  12. Oct 6, 2004 #11
    When I drew the FBD there were 2 forces. And since it is in a circle:
    F=ma(radial)
    F=(mv^2)/r
    mg + Fb = (mv^2)/r

    So I understand all of this but how do you solve for Fb because when I did it I got the wrong answer.
     
  13. Oct 6, 2004 #12

    Gokul43201

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    How did you get 7.73 ? It's not correct. And should the acceleration be positive or negative ? You are saying that Vf = 34 m/s which is faster than 17 m/s. Does it make sense for the car to travel faster, after hitting the brakes ?
     
    Last edited: Oct 6, 2004
  14. Oct 6, 2004 #13

    Gokul43201

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    Because you substituted mg = 5 when you should have substituted mg = 5000 N.
     
  15. Oct 6, 2004 #14
    negative. Yeah it does not make sense. What did I do wrong.

    a= delta v / delta t
    a= 17/2.2
    a= 7.73

    Would it be -7.73 because Vi is 17 and Vf is 0. So (Vf-Vi)/2.2 = -7.73?
     
    Last edited: Oct 6, 2004
  16. Oct 6, 2004 #15

    Gokul43201

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    No, this is not correct. You do not know what delta v is. Vi is 17 m/s but delta v is not known.

    You must find the acceleration using the frictional force, like you attempted to do the first time. Your only mistake then was that you should have multiplied the coefficient of friction with the weight to determine the force of friction (feels like I've said this before). Then find the acceleration from the force, using Newton's Second Law.
     
  17. Oct 6, 2004 #16
    Why would it be 5000? Why would you change 5 kN into 5000 N and then change it back into N when it asked for the answer in kN? Do you always change it into N for all problems?
     
  18. Oct 6, 2004 #17

    Gokul43201

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    Because when you calculate mv^2/r, the units are N. If you convert that to kN, that would also work. But you cant have N on one side of the equation and kN on the other.
     
  19. Oct 6, 2004 #18
    Thanks Gokul43201 :smile: ! I understand now!
     
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