Newton's Third Law: A Car Pushes A Truck

In summary, a 1000 kg car pushing a 2000 kg truck with a dead battery has a net force of 4500 N pushing to the right. By using Newton's 2nd Law and considering the forces on the car and truck individually, it is determined that both objects will accelerate at the same rate and that the force on the truck is also 3000 N.
  • #1
vertabatt
30
0

Homework Statement



A 1000 kg car pushes a 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N. Rolling friction can be neglected.

Homework Equations



F=ma

acceleration of car = acceleration of truck

The Attempt at a Solution



I have determined that both the car and the truck must move at the same acceleration. The car pushes off the ground with 4500N, so that:

F = ma
4500 = (1000)a
a = 4.5 m/s^2I attempted to solve for F on the truck:

F = ma
F = (2000)(4.5)
F = 9000

But this is incorrect. Should I be combining the masses so that F = (3000)(4.5)? Or am I just way off here?

I am struggling to put together the proper free-body diagram and I think its messing up my understanding of what force is on what object. A push in the right direction is appreciated! Pun intended.
 
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  • #2
You might want to draw a FBD of the enture truck-car system. What is the net force acting on the system? What is the mass of the system? Your FBD's are not correct since you have not properly identified all forces acting individually on the isolated FBD of the car or truck.
 
  • #3
Hmmm, not exactly sure what you mean by a FBD of the entire system. I assume it would be a particle with a normal force, a weight force (consisting of the 1000 kg car + the 2000 kg truck), and the force of the car's tires on the ground (pointing to the right). That should be it since friction is neglected.

I haven't seen anything that combines the system into one FBD in my book... not sure how it helps me isolate the force on the truck?
 
  • #4
vertabatt said:
Hmmm, not exactly sure what you mean by a FBD of the entire system. I assume it would be a particle with a normal force, a weight force (consisting of the 1000 kg car + the 2000 kg truck), and the force of the car's tires on the ground (pointing to the right). That should be it since friction is neglected.

I haven't seen anything that combines the system into one FBD in my book... not sure how it helps me isolate the force on the truck?
It appears that you have correctly identified the forces acting on the car-truck system. The only unbalanced force acting on the system is the 4500N force acting to the right. So what's stopping you from using Newton 2 for the system? That will give you the acceleration, then you can look at the truck by itself to find the net force acting on it. NOTE: You draw an FBD of the system by drawing a circle around the car and truck and identifying the forces acting external to it.
 
  • #5
Ok, I think I figured it out:

Fnet Car:

4500 - Ftruck = (mass of car)(acceleration)

Fnet Truck:

Fcar = (mass of truck)(acceleration)

Solve both for acceleration and set them equal to each other, remembering that Fcar = Ftruck from Newton's 3rd Law.
(4500-Ftruck)/(mass of car) = (Fcar)/(mass of truck)

Ftruck = Fcar = 3000N
 
  • #6
vertabatt said:
Ok, I think I figured it out:

Fnet Car:

4500 - Ftruck = (mass of car)(acceleration)

Fnet Truck:

Fcar = (mass of truck)(acceleration)

Solve both for acceleration and set them equal to each other, remembering that Fcar = Ftruck from Newton's 3rd Law.



(4500-Ftruck)/(mass of car) = (Fcar)/(mass of truck)

Ftruck = Fcar = 3000N
Yes, that will do it. As a double check, look at the system, Fnet = (m + M)a ; 4500 = 3000a, solve a =1.5 and then Fcar on truck = 2000(1.5) = 3000.
 

1. How does Newton's Third Law apply to a car pushing a truck?

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that when a car pushes a truck, the truck also pushes back on the car with an equal force in the opposite direction.

2. How does the mass of the car and truck affect the application of Newton's Third Law?

The mass of the car and truck does not affect the application of Newton's Third Law. The law states that the forces are equal and opposite, regardless of the mass of the objects.

3. Can the truck push back on the car with a greater force?

No, according to Newton's Third Law, the forces are always equal and opposite. This means that the force exerted by the truck on the car is the same as the force exerted by the car on the truck.

4. Does Newton's Third Law only apply to objects in motion?

No, Newton's Third Law applies to all objects, whether they are in motion or at rest. The law describes the relationship between forces, not just between moving objects.

5. Can the car and truck have different accelerations even though the forces are equal and opposite?

Yes, the car and truck can have different accelerations even though the forces are equal and opposite. This is because the mass of the objects also affects their acceleration, according to Newton's Second Law of Motion.

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