I have a disagreement with my teacher, we recently reviewed a problem where a car was pushing a disabled truck up to speed on the highway, and then pushes the truck at a constant speed, and asked what the forces exerted on the truck by the car were and vice versa.

  • I argued that when the car was pushing the truck up to speed, the car must be exerting more force on the truck than the truck on the car, because in order to have an acceleration there must be net force.

  • My teacher argued that due to Newtons 3rd law that the forces must be the same.

  • We both agreed that during the second part when there was constant speed the forces were equal and opposite.

Who is right?

  • 1
    $\begingroup$ Action and reaction act on different bodies, you are misusing the second law $\endgroup$
    – user126422
    May 11 '17 at 23:42
  • 1
    $\begingroup$ Forget about the question of whether the force on the car by the truck is equal to the force on the truck by the car for a moment. There is a force on the truck by the car, right? What other forces are acting on the truck? In the context of this problem, there are no other forces acting on the truck. So the force on the truck due to the car is unbalanced. Therefore, the truck accelerates forward. The End. (Now back to the question of the force on the truck by the car versus the force on the car by the truck. By Newton's 3rd Law, those two forces are equal and opposite. ) $\endgroup$
    – user93237
    May 12 '17 at 1:51
  • $\begingroup$ Newtons laws are applicable in inertial frames ie when the observer is moving without acceleration.When there is acceleration a pseudoforce due to inertia has to be taken into consideration. $\endgroup$
    – Chappy
    May 12 '17 at 4:38

Your teacher is right that the force exerted on the truck and the car are equal, according to Newton's 3rd law.

However, you are not wrong in thinking that, in order for there to be an acceleration, there needs to be a net force. When the truck pushes back on the car, that does not detract from the force that the truck feels; in other words, when the truck exerts a force on the car, that force is NOT included in the truck's free-body diagram.

Following this, the truck is happily accelerating because it has a net force on it. The car is also accelerating though, right? It has a force being applied to it in the opposite direction, so there must be some other force to offset that. That force is from the wheels on the pavement, pushing the car forward like you are used to.



Free-body diagrams:



Note that the force exerted by the car on the truck (--1-->) and the force exerted by the truck on the car (<--1--) are equal and opposite, following Newton's 3rd law. Force 2 is the car pushing against the ground.

Note: In this case, we do not need the net forces on the car and truck to be equal to each other. In fact, since they are accelerating at the same rate, and the car has a smaller mass, the net force the car feels will in face be smaller than the net force the truck feels.

$$ a_t=\frac{F_t}{m_t} $$ $$ a_c=\frac{F_c}{m_c} $$


$$ a_c=a_t, $$

so we get $$ \frac{F_c}{m_c}=\frac{F_t}{m_t} $$


If $F$ is the force exerted by the engine of the car, then let the contact forces between the car and the truck be $N$. So if the car was pushing the truck up to speed and both were accelerating at an acceleration of $a$. Then according to the free body diagram of the car, $$F-N=M_ca$$ Then according to the free body of the truck $$N=M_Ta$$ Hence $$F=(M_c+M_T)a$$ Herein, as the car and the truck were speeding up the highway, the forces exerted were equal and opposite, $N$.

When they are up to speed and maintaining a constant speed, $a=0$, hence $N=0$ and also $F=0$. The forces acting on each other are always $N$ (equal and opposite) according to Newton's 3rd Law.

  • 1
    $\begingroup$ $N=0, F=0$ when $a=0$ assumes that there are no other forces acting on the disabled truck (and the car). If there is rolling friction and air resistance on the truck, then $N \ne 0$. $\endgroup$ May 11 '17 at 23:26
  • $\begingroup$ yeah, but I was just trying to ease the questioner into the topic. If there's friction, the FBD evolves and hence so does the equations of motion. $\endgroup$ May 12 '17 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.