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Using Advanced Dynamics to find the Force Applied of a Truck

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    A car is moving at an acceleration such that a 5kg block in the back of the remains still on a ramp at 35*. The ramp is frictionless, whereas the coefficient of friction between the truck and the road is μ=0.1. If the truck weighs 12000kg, what is the force the trucks engine is exerting?


    2. Relevant equations
    Dynamics formula
    Fnet Formulas


    3. The attempt at a solution
    We know that the Ramp only has Force Parrallel pushing it down.

    Parrallel= mgSinθ
    = (9.8)(5)(Sin35)
    =28.105

    We also know that the only thing that would push the block UP the ramp is the Fnet of the truck. For the Block to remain still, it has to have an Fnet of 0. Since its on a Ramp, you have to find the Fx.

    Fx= Cos35*28.105 = 23.02.

    From there, we know the Fnet of the truck is 23.02.

    Fnet = Fa - Ff
    23.02= Fa - (0.1)(9.8)(12000)
    Fa = 11783.02

    Does that seem right? I feel like it is a little bit off. Any help would be much appriciated.
     
  2. jcsd
  3. Mar 3, 2013 #2

    haruspex

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    Let the normal force on the block be N. Write out the ƩF=ma equations for the vertical and horizontal directions.
     
  4. Mar 3, 2013 #3
    Dont you already know the normal force on the block? Its the Perpendicular Force right? (9.8*5*Cos35)

    As for the Vertical and horizontal directions:

    Fx=(9.8)(5)Sin35 = 28.105
    Fy=(9.8)(5)Cos35 = 40.138
     
  5. Mar 3, 2013 #4

    haruspex

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    That's the wrong value. Please write out the equations as I suggested.
     
  6. Mar 3, 2013 #5
    My Calculations in the above post (#3) are off.

    Fx= (28.105)Sin35 = 23.02
    Fy= (28.105)Cos35 = 16.12

    That might be better, cause those are the forces acting in the Y and X component (Fparallel broken up into X and Y components)

    Am I on the right track?

    EDIT: Okay, doing as you asked with the Fn.

    Fx= FnSin35
    Fy= FnCos35

    Like that?
     
    Last edited: Mar 3, 2013
  7. Mar 3, 2013 #6

    haruspex

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    Yes, those are the horizontal and vertical components of the normal force. Now get each into a ƩF = ma equation by bringing in g and the known horizontal acceleration.
     
  8. Mar 3, 2013 #7
    Fy=(9.8)(5)Cos35 = 40.1385

    Is the known horizontal acceleration just (9.8)(Sin35)=5.6 m/s^2?

    Thus the Fx= (5.6)(5) = 28.105?

    I think im wrong, as thats just the answer I got before.

    EDIT: We dont really know the horizontal acceleration, as it is just stated that it is enough that the block appears motionless. From this, we know that the Fnet of the truck is the Fa on the block, Right?
     
    Last edited: Mar 3, 2013
  9. Mar 3, 2013 #8

    haruspex

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    In your previous post you wrote, correctly, that the vertical component of the normal force, Fn, is Fn cos(θ). (It's best to work entirely in symbols, only plugging in the numbers at the end.) The only other force on the block is gravity. What's the vertical component of that? What is the resulting vertical acceleration? Write an equation relating those three things.

    The horizontal component of the normal force is Fn sin(θ). What other horizontal forces act on the block? Let the resulting horizontal acceleration on the block be a. Write the horizontal ƩF=ma equation.
    Yes, sorry, I was wrong to say the horizontal acceleration is known... that's part of what we're trying to determine.
     
  10. Mar 3, 2013 #9
    It was my understanding that the only force that was acting down the ramp in this situation would be Force Parrallel (mgSinθ), is that not correct? (There is no friction on the ramp,only between the truck and the road.) After finding the force parralele, you could break it up into X and Y components and use that to find the Fa of the truck. I thought Fn would be irrelevant as there was no force of friction.


    Onto what you said, The verticle component of gravity is just mg, correct? The resulting verticle acceleration is just the addition of the Vectors of normal force and gravity. So that would mean the verticle acceleration would be a = ((FnCosθ) + mg) / 5)

    For the horizontal component, there is also the Fnet of the truck pushing it towards the ramp. So the X component of the Fn is = to the Fnet of the truck.

    Sorry if this is all wrong, im very confused ATM. If you could answer the top in detail and explain what forces are pushing where, I would much appriciate it.
     
    Last edited: Mar 3, 2013
  11. Mar 3, 2013 #10

    haruspex

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    Yes, that's the only force acting down the ramp, but we need the horizontal force, and the normal force is involved in that.
    Right, and since there is no net vertical acceleration we can write FnCosθ = -mg
    No, that's committing the sin of using a non-inertial frame. Don't worry about the truck at the moment. There is only one horizontal force on the block, Fn sin(θ), and that produces an acceleration a, so we have Fn sin(θ) = ma. Combine that with FnCosθ = -mg to eliminate Fn. What do you get?
     
  12. Mar 3, 2013 #11
    Ahh, that makes more sense. So when we put the FnSinθ = ma and solve for Fn, we get Fn=ma/Sinθ

    Doing the same with FnCosθ = -mg, we get Fn = -mg/Cosθ.

    Equating the two, we get

    ma/Sinθ = -mg/Cosθ

    Now we have all the Values to solve for A.

    (5)(a)/Sin35 = -(5)(-9.8)/Cos35

    a = (49/Cos35) / (5/Sin35)

    a = 6.86 m/s^2

    Correct? That gives us our Acceleration in the X.

    Expanding on that, we now know the acceleration in the X required to keep the block upwards, so that must be the acceleration of the truck. From there, we can calculate the Fnet of the Truck (6.86 * 12000) = 82344.41

    Fnet = Fa - Ff
    82344.41 = Fa - (9.8)(12000)(0.1)
    Fa = 94104.41 N

    I rounded when writing the numbers down here, but did the math without rounding in my calculator, so if youre doing the calculations ive written down, they might come out a tad off.
     
  13. Mar 3, 2013 #12

    haruspex

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    Right. And you can immediately simplify that to a = -g tan(θ). (The mass of the block doesn't matter.)
    Yes.
    No. It's a common error to think that the friction between the tyres and the road acts to slow a vehicle. That's only true when the vehicle is braking. When a vehicle is accelerating (positively), the friction provides that acceleration. (If there were no friction, the vehicle could not accelerate.) For a vehicle (or whatever) on a horizontal surface, if the only horizontal force acting is friction then frictional force = mass * acceleration.
    The question asks for the force from the engine. That's rather an odd question. Strictly speaking, a car engine does not exert a linear force; it produces power and torque. The wheels convert the torque into force. How well that force converts to forward acceleration (as opposed to merely accelerating the spin of the wheels) depends on available friction.
    I would give 82344N as the answer (or, more accurately, 82430N or 82400N, depending on the precision appropriate.)
     
  14. Mar 3, 2013 #13
    This seems odd to me, as that means the μ would be irrelevant. Your engine would have to generate a lot more force to get out of quicksand than accelerating on standard pavement, correct?
     
  15. Mar 3, 2013 #14

    haruspex

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    The relevance of the friction is to whether it is actually possible to keep the block stable on the ramp. Seems it is not. The maximum acceleration of the truck is gμ =0.98m/s2. But as you calculated, the acceleration needs to be 6.87m/s2.
    Reading the question again, I see there are two things that are not clear. Which way is the ramp facing? If it faces backwards, the truck is decelerating, so it doesn't require the engine to be doing anything - it could be just on the brakes. But that doesn't get past the 0.1 friction problem. Secondly, it doesn't say the road is level. But if we don't know the angle of the road then we cannot answer the question.
    Yes, but largely because getting out of quicksand involves going up as well as along. Also, spinning the wheels at a great rate to get out of sand means the propulsion is more like a jet than a normal rolling tyre; the forward force comes from the momentum of the sand thrown backwards.
     
  16. Mar 3, 2013 #15
    The Ramp is indeed facing the same way the car is accelerating, and the road is level.

    Making a little bit more sense, Thanks for everything!

    Is there somewhere I can like give you a thumbs up or something?
     
  17. Mar 3, 2013 #16

    haruspex

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    That's good, but it bothers me that the question no longer makes sense to me. The short answer is that with friction so low it is not possible for the block to stay in equilibrium on the ramp, no matter what the engine does. I'm going to post this thread to the homework helpers forum.
    I believe there is, but I don't know how it's done. Thanks for the thought.
     
  18. Mar 4, 2013 #17

    tiny-tim

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    Hi QuickSkope! :smile:
    This has all been very complicated for a simple question. :redface:

    As with most dynamics problems, you simply apply good ol' Newton's second law (F = ma) in the appropriate direction.

    In this case (as you've correctly noticed) the appropriate direction is along the slope.

    The only force (in that direction) is gravity.

    Call the acceleration of the truck (or is it a car? :confused:) "a".

    Then what is the acceleration of the block? And what is the component of acceleration of the block along the slope?

    Now do F = ma for the block to find a. :wink:

    (Alternatively, have you done accelerating frames? If so, the effective gravity is the vector sum of the actual gravity and the fictitious force, and it has to be normal to the slope)
     
  19. Mar 4, 2013 #18

    haruspex

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    QuickSkope, I passed this problem around the team and the consensus is that although the question says the friction between the car and the road is 0.1, that's not what it means. Instead, they believe it is referring to rolling resistance. If so, your earlier treatment of it (adding to the force the engine needs to produce) is correct. Is it possible you copied this wording inaccurately? If not, I think it's a very badly worded and misleading question.
    tiny-tim's solution is a little more direct, but - how can I put this? - I feel that you would do better sticking to a standard approach of resolving vertically and horizontally every time. Yes, it can be a bit more work, but you're less likely to outsmart yourself. OTOH, if you feel comfortable with tiny-tim's method, go right ahead.
     
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