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Newton's Third Law of a car and truck

  1. Sep 10, 2013 #1
    I have a question for which I am intensely confused.

    Lets say a car was pushing a big truck from behind. And the car (and truck) are accelerating, then there is a breakdown because the car applies all the force it has to the truck, which pushes back with equal force. So the car shouldn't be moving. Only the truck would experience a little acceleration.

    For example, using completely unrealistic numbers.

    Looking just at the truck, it has the force of gravity (unimportant), a force pushing forward (500N), and the force of friction (200N).

    Then the car must have a force of 500N forward, a force of 500N pushing backward due to the truck, and no force of friction. Think about a car hitting a wall.

    So in theory, if this situation occurs irl, then the car and truck would be playing ping pong. The truck moves forward a little, then the car moves forward (because no backward force anymore) hits the truck and stops (because it is applying force on the truck which is pushing back again), then the truck moves forward and so on.

    Of course, this could be happening in infinitely small time intervals so it LOOKS like the car never breaks contact.

    So can someone explain if my reasoning is correct or flawed?
  2. jcsd
  3. Sep 10, 2013 #2


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    You are right about the potential ping-pong nature of the interaction and also that if its happening on a fine enough time scale it looks continuous. We deal with this by using the average force across the entire ping-pong cycle.

    When you say that only the truck should be accelerating, you are mistaken because you're missing one of the forces (probably because you're thinking in terms of "all the force [the car] has" instead of all the forces acting on the car). By equal and opposite force, the truck is applying a 500N force (on average - because of the ping-ponging effect it may be varying between zero and quite a bit more than 500N, but it averages out to 500N) to the car. But that's not the only force; the tires the car are turning against the ground, causing the ground to exert a force on the tires which in exerts a force on the car. If that force is greater than 500N the car will accelerate forward as well.
  4. Sep 10, 2013 #3


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    The Newton third law pair of forces is different than the net force on each object. The Newton third law pair of forces includes reaction forces related to an object being accelerated by an external (to the object) force. The net force on an object determines it's rate of acceleration.

    Putting some numbers on this, it becomes a solvable problem. Say the car has mass 1600 kg and the truck has mass 4800 kg. If the car is pushing forward on the truck with 500N of force, which is opposed by 200N of friction, then the truck experience a net forward force of 300N, causing it to accelerate at (300 N) / (4800 kg) = 1/16 m / s^2.

    The car also accelerates forward at 1/16 m / s^2, so it experiences a net forward force of 100N. This would translate into 600N of forward force from the pavement against the tires, and 500N of opposing reaction force from the truck.

    At the pavement under the driven wheels of the car, there is another pair of Newton third law forces. 600 N of backwards force exerted by the tires onto the pavement, and 600 N of forward force exerted by the pavement onto the tires.
  5. Sep 10, 2013 #4


    Staff: Mentor

    Why? The truck has a net force of 300 N forward, so if it weighs 3 tonnes then it will have an acceleration of a=.1 m/s^2. If the car is pushing the truck then it also has an acceleration of a, and assuming a mass of 1 tonne a net force of 100 N. To achieve a net force of 100 N, the force pushing forward must be 600 N, not 500 N.
  6. Sep 10, 2013 #5
    The friction between the car and the road cannot be zero since it is the friction tha provides the 600 N forward force that makes the car go.
  7. Sep 10, 2013 #6
    In reality the force would be spread across the contact area and cause strains in the materials in that area of contact. Their probably wouldnt be much ping-ponging after the first initial contact had been made due to the deformations of the materials of the car & the truck.

    Unless of course they were both made from perfectly rigid materials but as this supposed to be real life then this wouldnt be the case.
  8. Sep 10, 2013 #7


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    Every tried using one vehicle to push another vehicle over a few miles of non-level roads? :smile:

    "Ping-pong" is a mild word for describing what happens, unless the deformation from the repeated blows gets bad enough to mechanically lock the vehicles together.
  9. Sep 10, 2013 #8


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    No, that is wrong. The car applies all of "its force" (i.e. the force created by the engine) to the ground, through the friction between its wheels and the road.

    It also applies some of its force to the truck.

    The forces on the car are the reaction of the ground on the wheels (in the forwards direction) and the reaction of the truck on the car (backwards) The difference between those two forces = the mass x acceleration of the car, from Newton's second law.

    The force on the truck = mass of the truck x its acceleration. The accelerations (and velocities, and distances moved) of the truck and the car are equal, so there is no "ping pong" effect.
  10. Sep 10, 2013 #9
    This is perfect for the question I am stuck on. I am just a first year so this may be stupid but...

    You are assuming that the car accelerates at the same rate as the truck. I think that is wrong in its own way but more to the conceptual point, let me write up a new example:

    Assume there is no friction, and the car moves forward with 500N of force via magic (no tires or anything). It hits the truck. The truck experiences acceleration (500N/4800kg). At this instance in time, the car experiences a backward force of 500N. It stops (as opposed to your assumption that the car accelerates at the same rate as the truck). The truck's velocity increases slightly. Distance is created. The car moves forward again and hits it. A cycle ensues until both velocities are equal.
    Conceptually, I was wondering if this is actually what happens (some people agree with me I see). I was thinking this may be happening at an infinitely small time intervals at a molecular level.

    But you showed me a second point. How do you solve for total acceleration of both the truck and the car? In real life, you would think they would move as if they were one body.

    In this respect, my answer would be:
    a = 500N/(1600kg+4800kg)
    = 5/64 m/s^2
    The truck experiences a net force of 4800*(5/64) = 375N
    the car 1600* 5/64 = 125N.

    But your methodology (and this is a flaw I think) is wrong.
    Here is your answer:
    Truck accelerates at 500N/4800kg = 5/48 m/s^2.
    The car therefore accelerates too at 5/48 m/s^2.
    Therefore, the car is experiencing a net force of (1600)(5/48) = 166.67N
    The truck (4800)(5/48) = 500N.

    I'm not sure. I have never heard that forces are conserved. In this respect, can force be split up? the 500N force is hitting the truck and making it accelerate. At the same time 166.67N is acting on the car.
    Your answer also makes sense if the 'ping pong' thing is happening. If the car breaks apart only for an infinitely small interval, the truck does indeed experience 500N of force for the remainder of the time (although the opposite could be argued I guess -- and the car would have to experience 500N as well. Impossible. The car would pass the truck).

    I'm so confused :s
  11. Sep 10, 2013 #10


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    Things don't "move with force". Forces are applied to things by other things, so it makes no sense to talk about the car "moving forward with 500N of force". You have to look at all the forces that are being applied to car, add them up - if the sum is zero the car will move at a constant speed and if the sum is non-zero the car will accelerate or decelerate.

    If there is no friction between tires and ground, the wheels will spin and nothing will accelerate.

    If the car is moving freely forwards (sliding on ice, maybe) then the total force on the car is zero so it will move at a constant speed unless and until it hits the truck. When it does, the truck will have to exert a force on the car to slow the car down, and by Newton's Third Law (which is how you started the thread) the car will exert an equal and opposite force on the truck. The truck will accelerate a bit and the car will slow a bit, just like what happens when a stationary billiard ball is hit by a moving one.
  12. Sep 10, 2013 #11


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    You don't seem to be stupid to me, but you are not approaching this methodically and as a result you are making incorrect assumptions that cause problems. While you can have scenarios where the vehicles repeatedly "bounce" that is certainly not required by the laws of nature and introductory physics classes will not have a problem like that.


    Not in general, no. Here is where you are not being methodical, you are simply assuming the force without actually calculating it. The force during a brief collision is unknown and can be very high. Unless you have detailed information about the material and structure of the vehicles there is actually no way to calculate it. You can make some simplifying assumptions to approximate it, e.g. by treating the vehicles like springs.

    However, usually there is no need to calculate the force during a collision explicitly. You will simply use conservation of momentum to describe the situation before and after the collision, without worrying too much about the details. Furthermore, typically you will assume either a perfectly elastic (vehicles bounce off each other with no damage) or a perfectly plastic (vehicles stick together) collision.

    Assuming that the collision is perfectly plastic then you can analyze the continued force between the vehicles if the 500 N magic force continues pushing the car. Since the collision was plastic they are stuck together and their acceleration after the collision will be the same. That is an additional constraint which can be used to determine the force on the truck.

    If your previous statement were correct, then this would be correct by Newton's 3rd law.

    No, this is incorrect (even if your assumption were correct). A car which experiences 500 N of magic force forwards and 500 N of "reaction" force backwards is experiencing no net force, and by Newton's second law it will continue at the same velocity as before. It will not stop.

    No distance would be created since the truck's velocity is less than the car's velocity which is unchanged.

    Yes, this is correct given those masses. The truck experiences 375 N, not 500 N. As a result the reaction force on the car is 375 N. So the net force on the car is 500 N - 375 N = 125 N, as you showed.

    They are not conserved.

    No, this is simply your assumption. The 500 N force is being applied to the car, not the truck. You need to approach this more methodically. I would recommend the following approach.

    1) draw free-body diagrams for every object in the problem (I always write the mass inside the free body diagram and the acceleration as a vector near but not touching it)
    2) draw all forces
    3) write down the magnitude of any known forces
    4) label any unknown forces with some variable
    5) use Newton's laws to write as many equations as you have variables
    6) solve
  13. Sep 10, 2013 #12
    If you want to balance all the forces, you must include the ground. The car moves forward by putting a force backward on the ground. If you make the car fly forward by magic, all physics breaks down.

    The car puts a force on the ground, and the car puts a force on the truck. If the car is accelerating forward, the force on the ground exceeds the force on the truck.
  14. Sep 10, 2013 #13


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    I'll be concise:
    There are two+ very common misunderstandings of Newton's laws here:

    1. Net force of 0 means no acceleration, not no motion (speed). That's Newton's 1st law.
    2. In the context of Newton's laws, "net force" is on an object and does not include the inertial reaction (from acceleration). If you include the force due to acceleration (N2), all forces always sum to zero. And at a point (the contact point), they always sum to zero whether or not the objects are accelerating. That is what N3 tells us.
  15. Sep 11, 2013 #14
    Ah yes. That was silly of me. Fell in a common trap (and thing is, I've heard this before).

    Are you implying that given perfect collision, the force on the truck would NOT be 500N and instead must be calculated based on conservation of momentum?

    mv = mv
    You know the car's velocity and mass.
    And you can solve for the truck's velocity.
    But that doesn't tell you how much force is experienced, which I would assume is vital for car companies or something.

    Of course, yes, you are right that building materials and the k constant for metal (if that is a thing) would be needed to find the amount of 'give' (again, I have no idea. I assume these things are quantifiable) to find the actual amount of force transferred and absorbed. I was just simplifying.

    Either way, I appreciate all the insight!
  16. Sep 11, 2013 #15


    Staff: Mentor

    Yes, the force on the truck would not generally be 500 N, even given an imperfect collision.

    No, if you want to actually calculate the force during the collision you need a lot of very detailed information about the materials and shape of the vehicle. Simple conservation of momentum principles can get you the final state after the collision, but cannot give you details about the forces during the collision.

    You are quite welcome, and it sounds like you are on the right track now!
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