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Angles and forces/normal force

  1. Feb 7, 2008 #1
    [SOLVED] Angles and forces/normal force...

    1. A car coasts (engine off) up a 30 degree grade. If the speed of the car is 25 m/s at the bottom of the grade, what is the distance traveled by the car before it comes to rest?



    2. Umm...I have no idea what equations are used in this problem. I know we were using A=Fnet/M and I also know that Fnet=Tcos(angle). I can't seen to find any equations that were given that include this angle.


    3. The answer is 64 M. I just have no clue how to get this answer. Since it's asking for distance and it has a velocity I am completely thrown off.
     
  2. jcsd
  3. Feb 7, 2008 #2

    Dick

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    You can do it using acceleration if you want. The force tangential to the ramp is m*g*sin(30). So what's the acceleration on the ramp? If you've covered energy you can do it that way too. Equate the initial kinetic energy to the gravitational potential energy when the car comes to a stop on the ramp.
     
  4. Feb 7, 2008 #3
    m*g*sin(30) but it doesn't give a mass so I can't use that equation can I? Yeah, we just got done with kinetic energy. There are just way too many equations and my teacher doesn't explain it to us very well. She works the problems then goes to another problem. I've never felt so stupid before this class. I can do some of the work, but I'm not understanding how to weigh in the angles with the problem correctly.
     
  5. Feb 7, 2008 #4

    Dick

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    F=m*g*sin(30) and F=m*a. When you find the acceleration, the mass will cancel. To figure what to do with the angle draw a force triangle. mg is vertical. The tangential part is parallel to the ramp and the normal part is perpendicular and they sum to make the vertical mg.
     
  6. Feb 7, 2008 #5
    I feel so stupid...I've never heard of a force triangle before. I think the acceleration is 25m/sec. So how does the mass cancel because of that?
     
  7. Feb 7, 2008 #6
    No no..acceleration is in m/sec squared so 25 m/sec has to be the velocity. I think..
     
  8. Feb 7, 2008 #7

    Dick

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    How can you think that the acceleration is 25m/sec??? That doesn't even have the units of acceleration! It's a velocity. You might call the force triangle a 'free body diagram'. You should really review some inclined plane problems. Try using energy. What's the initial kinetic energy? (Just call the mass m). What's the potential energy when it is a distance L up the ramp?
     
  9. Feb 7, 2008 #8
    I have no clue man! Is there not just one equation to use? I see that I have an initial velocity of 25 m/sec. Okay, so I if I plug that in to F=m*g*sin(30) wait I can't plug that into that since I don't know m. Ugh...I'm so lost.
     
  10. Feb 7, 2008 #9

    Dick

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    F=m*a. F=m*g*sin(30). So m*a=m*g*sin(30). Cancel m. a=g*sin(30).
     
  11. Feb 7, 2008 #10
    Ok, then the acceleration is -9.68 m/sec squared. Now to find the distance..jesus I need a tutor or something..I've spent nearly 2 hours on this one problem..
     
  12. Feb 7, 2008 #11

    Dick

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    NOT 30 RADIANS. 30 degrees. Switch your calculator to degrees!
     
  13. Feb 7, 2008 #12
    Ah sorry. I forgot that I cleared the RAM earlier. So it's 4.9 m/sec squared. What do I do next?
     
  14. Feb 7, 2008 #13

    Dick

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    You tell me. It has an initial velocity of 25m/sec. It is decelerating at 4.9m/sec^2. Doesn't this sound at all like a familiar problem yet? How long till it decelerates to zero velocity?
     
  15. Feb 7, 2008 #14
    D=V/A? I've never seen that equation before. woah woah I'm getting closer...VF^2=VI^2+2A(D) so in other words D=VF^2-VI^2/2A.....VF=0 (since it's at rest)...VI would be 25 m/sec and A is 4.9 m/sec^2. So...I get arggggg!! I have a program that does it for me but it doesn't show what it's doing. I end up getting -63.7755102 on there. :(( Physics is killing me here!!! Ok ok..AHHHHH I have to take a break..from this
     
  16. Feb 7, 2008 #15

    Dick

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    You got it. -63.7755... rounds to -64m. It's negative because you got the velocities backwards or A with the wrong sign. I think you should go back and find some easier problems and get some tutoring as well. That was painful for both of us.
     
  17. Feb 7, 2008 #16
    Oh I forgot to square 25. So...would it be 625/9.8? Which gives me 63.77! My book usually rounds this has to be it!!! Please tell me it is!
     
  18. Feb 7, 2008 #17
    Man...sorry about all that. I can't seem to do any of these problems from this chapter without intense meltdowns or whatever you want to call it. I'm going to try from the beginning of this chapter and work my way through.
     
  19. Feb 7, 2008 #18

    Dick

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    That's it. Good luck!
     
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