1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Time to slide down a ramp w/ no friction

  1. Jun 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A wet bar of soap (m=150g) slides down a 2m long ramp at 7.3° relative to the horizontal. How long does it take to reach the bottom?

    2. Relevant equations
    d = 1/2at2

    3. The attempt at a solution
    The answer is apparently 1.81 seconds, although I can't figure out how.

    I can figure out the height of the ramp to be 0.25m, but using the equation above tells me that it should only take 0.22 seconds if you use gravity as the acceleration. What am I doing wrong?
  2. jcsd
  3. Jun 27, 2012 #2
    You have the equation just substitute.
  4. Jun 27, 2012 #3
    How do you figure? If I substitute 9.8 for acceleration and 2m for distance I get 0.63 seconds, and if I substitute 0.25m for distance I get 0.22 seconds for time....and neither of which are right.
  5. Jun 27, 2012 #4
    You should be able to differentiate between g and a.
  6. Jun 27, 2012 #5
    isn't gravity the only acceleration here?
  7. Jun 27, 2012 #6
    Hello TrpnBils,
    Resolve the acceleration due to gravity parallel to the incline because the actual path the soap travels is along the inclined plane.This should give you the the "a" in your equation and the "d" is already given.
  8. Jun 27, 2012 #7
    As you see the body is accelerating not towards the earth but slanting.
    If you drop a mass it will accelerate at the rate of g.
    In your case it is moving to the left or right as well as downward.
  9. Jun 27, 2012 #8
    okay, so if I assume 9.8 is pointing down (the Y axis) and I use sin(7.3) = 9.8/L then that gives me a diagonal component of 77m/s^2? That doesn't make sense....
  10. Jun 27, 2012 #9
    Analogy 6=4+2

    Since acceleration is vector, we can also make 2 components.
    BUT the total of these components should be equal to original vector.
    Evaluate you calculation.
  11. Jun 27, 2012 #10
    Like the other one I threw on here tonight, I see now that the acceleration is less than that of gravity....I just can't figure out why you do the calculation the way that you do. In this problem, if I take sin(7.3) * 9.8, I get an acceleration of 1.24m/s^2 which works out to be right. I just don't know WHY I multiply the sin(7.3) by gravity when the laws we all learned in trig were Sin(theta) = Opposite/Hypotenuse.....and in this case, the 9.8 is the "opposite" while the acceleration along the length of the ramp is the unknown (hypotenuse). In my mind, if you plug them in that way, you should be DIVIDING 9.8 by Sin(7.3).

    Why is it the other way?
  12. Jun 27, 2012 #11
    As you see from my analogy, 6 is bigger than any of its components.
    So in vector addition with right angle triangle, you should know where is the highest value and its 2 components.
  13. Jun 27, 2012 #12
    Yeah, the hypotenuse is always going to be the largest....but that still doesn't answer my question about why I'm multiplying by sine of the angle rather than dividing by it.
  14. Jun 27, 2012 #13
    Ok you get that right.
    Then how do you calculate the value of opposite and adjacent with the given hypotenuse?
  15. Jun 27, 2012 #14
    Isn't the value of gravity (9.8) the value for opposite since it's the vertical acceleration?
  16. Jun 27, 2012 #15
    No, the diagonal is acceleration due to gravity. A vector can be resolved into two components. When you find the resultant of the resolved vectors, you get the same vector back. See the figure below:
    (Sorry for an inaccurate figure, i know its dirty. :P)
    We have to find the component of g along the ramp.
    Do you now see why you use 9.8sin(7.3)?
  17. Jun 27, 2012 #16
    The diagram makes sense, yes, but why would the diagonal vector be 9.8 considering gravity isn't accelerating in the diagonal direction? This seems like it should be a pretty straightforward problem and I don't know why I'm having problems with this or the other one I asked about tonight....
  18. Jun 27, 2012 #17
    You should rotate the diagram anticlockwise 45°.
    Then you see the g is pointing down.
    And the opposite side becomes parallel to the inclined plane.
  19. Jun 27, 2012 #18
    I sketched that out on my paper, but logically I have absolutely no idea why you would do that. If I am standing in at the base of a ramp, gravity is not accelerating at the angle of that ramp down towards my feet.
  20. Jun 27, 2012 #19
    Now try standing on a slippery ramp .
    Do you feel something pulling you forward parallel to the inclined.
  21. Jun 27, 2012 #20
    Maybe it would help me if you all saw what I'm picturing in my head.....

    In my mind, I'm picturing this ramp at 7.3 degrees from the horizontal. Gravity acts vertically downward, not along the length of the ramp....

    I'm wrong, but I don't see the logic behind doing it so that gravity acts down the length of the ramp. To me, doing it that way just seems to arbitrarily rotate the ramp to point gravity in a new direction....why not just turn the whole thing upside down? It seems just as logical that way to me.....

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook