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Crash Test Craig drives his 800kg car up a 100m ramp. Help?

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Crash Test Craig drives his 800kg car up a 100m ramp that is at an angle of 35 degrees above the horizontal. The car exerts a forward force of 12000N and the force of friction between the ramp and the car is 20N. He is trying to jump over a series of spinning helicopter blades that are each 8m across to reach another ramp down on the other side How many helicopter blades can he safely jump over?

    Variables:
    mass of car= 800 kg
    distance or length of ramp= 100m
    angle= 35 degrees
    Applied(?) force= 12000N
    Force of friction= 20N
    Normal force= 9.8 m/s^2 * 800 kg = 7,840N
    Force of gravity= 9.8 m/s^2 * 800kg = 7,840N


    2. Relevant equations
    Vf^2=Vi^2 + 2ad

    F=ma

    There could be a variety of kinematics equations to use.

    3. The attempt at a solution
    So, I thought since the car was exerting 12000N, that we could find the acceleration of the car using F=ma. I thought 12000N= 800kg*a and got a to be 15m/s^2. Then, assuming that initial velocity is 0 m/s, I solved for the final velocity using Vf^2= Vi^2 + 2ad. Vf^2= (0m/s)^2 + 2(15m/s^2)(100m). Vf= 54.77 m/s which is the launch velocity of the car as if leaves the ramp. I know I need to find the distance the car travels in the air and divide that by 8 to get the number of helicopter blades, but I am confused by what to do next.

    Would final velocity be 0 m/s? Is acceleration 9.8 m/s^2? Is it now a vector problem? Do I use the force in any way?
     
  2. jcsd
  3. Nov 8, 2014 #2

    Simon Bridge

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    ... using F=12000N?
    While accelerating up the ramp, how many forces are on the car?
    You are correct that the initial runup distance is not given.
     
  4. Nov 9, 2014 #3
    I'm not really sure what I am doing.
    Would I break down 12000N into vectors? Then sin(35)12000 would be Fy which would be added to Normal force to get Force of gravity. Then again, I probably got the normal force wrong. And cos(35)12000N would be Fx and I would subtract 20N from that. That would be my net force and I would solve for acceleration by dividing the mass, 800kg. Then I would find the final velocity which would in turn be the initial velocity. Using vi^2/g * sin 2 theta, I could find the range and divide it by 8 to get the number of helicopter blades.

    Am I right in my thought process or completely wrong?
     
  5. Nov 9, 2014 #4
    Draw a FBD. What is the net force acting on the car?
     
  6. Nov 9, 2014 #5

    vela

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    Newton's second law is ##\vec{F}_\text{net} = m\vec{a}##. You can't just plug in any old force into the lefthand side. You have to add up all of the forces on a body to get ##\vec{F}_\text{net}##, and only that sum can be set to ##m\vec{a}##.
     
  7. Nov 9, 2014 #6

    Simon Bridge

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    You have noticed that the driving force, while the car is on the ramp, acts at an angle to the horizontal ... because the ramp has an angle.
    Well done - but it is not always easiest to align the vertical axis with the direction of gravity.
    Always choose your x and y axes to make the maths simplest.

    Your original approach was almost there.

    The starting point is to draw a free-body diagram for the car as it goes up the ramp.
    In your first post, you used only one force. There is more than one force on the car.
    How many forces are acting on the car?

    Please answer the questions put to you - it is difficult for us to hep you if you don't.
     
  8. Nov 9, 2014 #7
    There are multiple forces (4), but the normal force and the force of gravity cancel out. There is the force exerted by the car which is 12000N and the force of friction which is 20N.
     
  9. Nov 9, 2014 #8
    Yes, I realize that now which is why I broke the 12000N into vectors. Sorry, we didn't really go over the laws in class. We only got the formulas from it.
     
  10. Nov 9, 2014 #9

    vela

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    The normal force and weight can't cancel because they don't point along the same line. Normal means perpendicular, as in perpendicular to the surface of the ramp. Did you draw a free-body diagram?
     
  11. Nov 9, 2014 #10
    Would the net force come from breaking 12000N into vectors and subtracting 20N from Fx (from the vector)? If so, the net force would be 9809.8N.
     
  12. Nov 9, 2014 #11
    Yes, but it is probably incorrect. I haven't gotten much of a handle on the angular free body diagrams. Freebodydiagram.png
     
  13. Nov 9, 2014 #12
    Sorry, gravity should not be at an angle.
     
  14. Nov 9, 2014 #13
    Would I be closer with this one? freebodydiagram.png
     
  15. Nov 9, 2014 #14

    vela

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    Pretty close. The forces appear to be pointing in the right directions now.

    The force of friction is kind of strange in this problem. It's the friction between the road and the car which propels the car forward, just like the friction between the soles of your shoes and the ground allow you to walk forward. In one part of the problem, you're told this force is 12000 N, and then you're told there's another force of friction which is 20 N. (Maybe it's air resistance or something like that.) So let's just assume it's slowing the car down. If we do that, the way you drew it on the free-body diagram is correct.

    Typically, what you do with inclined-plane problems is you orient the axes so that one is parallel to the plane and the other is perpendicular. This makes the math simpler. If you do that, only the weight doesn't lie in a direction parallel to the axes. So that's the only one you need to resolve into components. That's what you want to do next.
     
  16. Nov 9, 2014 #15
    I understand it now! So the net force would be 12000N- m*g*sin(35) - 20N (frictional force). From there, we divide by mass to get the acceleration and the find the final velocity which in turn would be the launch velocity. Then I can use the Range= vi^2 /g * sin 2 theta to find the distance and divide by 8 to get the number of helicopter blades.
    freebodydiagram.png
     
  17. Nov 9, 2014 #16
    ^^^
     
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