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Going up a frictionless ramp

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data


    A wedge of mass m= 1kg that is initially at rest is driven up a friction less ramp by a horizontal force
    F=50Ndue to a small attached rocket. The ramp is equilateral, with sides of length 50m

    What is the magnitude of the normal force exerted by the ramp on the wedge?
    What is the speed Vt (top of the ramp i assume) of the wedge as itl eaves the ramp?




    2. Relevant equations
    F=ma
    Fx=mgsin(60)
    Fnety=n-mgcos(60)
    (Final velocity)^2 – (Initial velocity)^2 = 2 * acceleration * distance



    3. The attempt at a solution
    So i know an equilateral triangle has angles of 60*

    The normal force of the block on the incline is:
    F(normal) = mg*cosθ
    = (1)(9.8)cos(60°)
    = 4.9.

    (b) The force of gravity acting on the block on the incline is:
    F(gravity) = mg*sinθ
    = (1)(9.8)sin(60°)
    = 8.487

    This implies that the net force is:
    F(net) = F(push) - P(graivty)
    = 4.9-8.487
    = -3.58 N. Is this the same as the magnitude?

    Using F = ma ==> a = F/m, the acceleration of the block is:
    a = F/m
    = -3.58/1
    = -3.58 m/s^2 (down the incline).

    Is this more along the right track?
    If i have the acceleration of -3.58 i can just find the final velocity with (Vf)^2 – (Vi)^2 = 2(a)(d)=18.9m/s

    I just want to make sure i have the right idea
     
  2. jcsd
  3. Oct 6, 2013 #2

    Dick

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    Quantities in physics often have units and directions. You should make both clear. For the normal force (only from gravity) it is 4.9N and it's pointing perpendicular into the ramp. The force of gravity along the ramp is 8.487N pointing down the ramp. You did this by splitting the 9.8N force of gravity into components parallel to the ramp and perpendicular to the ramp. Now you need to do the same with the applied 50N horizontal force. The total normal force is the sum of the normal force from gravity and the normal force coming from the 50N push. Same for the total force pushing the block along parallel to the ramp. Draw force diagrams.
     
  4. Oct 6, 2013 #3
    So for the applied force horizontally would i take it as 50(sin(60) as well as 50(sin(60)? Or am i thinking a little to hard
     
  5. Oct 6, 2013 #4

    Simon Bridge

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    That's a good assumption.

    You mean:
    ##\text{1. }F=ma\\ \text{2. }F_x=mg\sin\theta\\ \text{3. } F_{net,y}=N-mg\cos\theta\\ \text{4. } v^2=u^2+2ad##
    I'm guessing from this that you have adopted a coordinate system where the +x direction points directly up the ramp and the +y direction points upwards perpendicular to the ramp?
    ... I'd take another look at equations 2 and 3, they are incomplete.

    Well done. If you didn't you could figure it from the "sum of the angles" rule.

    Good. Don't forget units. What about the contribution from the rocket?
    Note: cos(60)=1/2

    ... this is the x component of the force of gravity?
    Check the direction.
    If +x points up the slope, which direction does the x-component of gravity point?
    Don't forget units.
    Note: sin(60)=(√3)/2

    This is why you have to understand what you have done.
    What was the whole point of working out those trig things?
    What does that minus sign mean?

    Isn't 4.9N the normal force? How does that push the cart?

    Take care when you mix negative numbers in your descriptions - does the minus sign mean the block is picking up speed moving down the incline?

    You have the wrong units for (d).
    If the initial velocity is zero, then the final velocity will be negative.
    What does that mean?

    I think you need to be more careful in your setup - paying attention to directions.
    Try sketching the situation, and drawing x and y axis on your sketch as a guide.
     
  6. Oct 6, 2013 #5

    Dick

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    One part of the force points perpendicular into the ramp (which adds to the normal force) and another part points up and parallel along the ramp. One is 50Ncos(60) and one is 50Nsin(60) in magnitude. Which is which? You really have to draw the force diagram.
     
  7. Oct 6, 2013 #6
    Drawing out the forces on the wedge i would say that it would be using cos perpendicular to the ramp to add on to the normal force. I do see what you are saying though

    As for acceleration F=ma The horizontal force + 4.9 and 8.487 and the other two components acting on the wedgedevided by the mass (1kg) = the acceleration
     
    Last edited: Oct 7, 2013
  8. Oct 7, 2013 #7

    Dick

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    Ooo. I'm wrong. Do as I say, not as I do. Draw the diagram. In your answers to the initial question, the force along the ramp is mgsin(60) and the force normal to the ramp is mgcos(60). I didn't draw the right diagram. Gravity points down. You have to do better. Think about what would happen if the angle were 0. Don't be as sloppy as I was. Now do the 50N horizonal force.
     
  9. Oct 7, 2013 #8
    I know if the angle would be flat (0) Sin would be 0, gravity wouldn't matter that much. I've drawn out the right triangle, but for some reason i just hit a brick wall when i try to think past that. I know why i get the normal force from gravity, I'm just struggling with how to get it from the force from the rocket.

    From what i would think is that the normal force from the rocket would be 50(cos(60)). Still looking for an example from my text book though.All i find from it are examples with only gravity acting on something. I apologize, i'm trying to think it out carefully.
     
    Last edited: Oct 7, 2013
  10. Oct 7, 2013 #9

    Dick

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    Don't apologize. Think it out all you want. I think if the angle were 90 then the normal force would be 50sin(90)=50. If the angle were 0 there would be no normal force 0=50sin(0). So I think the normal force is 50sin(60). Keep thinking.
     
  11. Oct 7, 2013 #10
    I probably drew the right triangle wrong, or i'm just looking at it incorrectly . I was thinking backwards between normal force and gravity. I do agree with you that there would be no normal force with no incline.

    This is just a shot in the dark, but would the be another right triangle with the adjacent side parallel to the ramp for the horizontal force?
     
    Last edited: Oct 7, 2013
  12. Oct 7, 2013 #11

    Simon Bridge

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    If you point the +x axis up the ramp...
    the angle to the horizontal is 60deg.

    Because gravity acts vertically downwards, it has a component in the -x direction with magnitude ##mg\sin\theta## and a component in the +y direction with magnitude ##mg\cos\theta##

    do the same for the force from the rocket - bearing in mind that the rocket acts horizontally instead of vertically.

    Then you can sum the forces in the +x and +y directions
     
  13. Oct 7, 2013 #12
    Drawn out a triangle

    Lookslike it is 50sin(60) for the normal force and 50cos(60) in the +x direction

    Magnitude of normal force is 43.4N+4.9N= 48.2N

    Netforce would be 4.9+(-8.487)+43.3+25=64.72

    And for the acceleration Fnet=1kg*a 64.72/1= 64.72m/s^2

    I think this is right
     
  14. Oct 7, 2013 #13

    Simon Bridge

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    The rocket supplies (50N)sin(60) perpendicular to the ramp - well done.

    You keep using the same words to refer to different things. You need to be careful about that.
    The normal force is supplied by the ramp, and it is equal and opposite to the net applied force perpendicular to the ramp. If you are careful with your descriptions you are less likely to get confused, and it is easier for people to give you more marks for your work.
     
  15. Oct 7, 2013 #14
    Thanks so much! Now i actually have a much better understanding of this.

    For the top i have the Final Velocity (i think) 80.44 m/s.
     
    Last edited: Oct 7, 2013
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