How Does Newton's Third Law Apply When a Truck Pulls a Car?

In summary, the truck is driving up a mountain with a constant velocity, neglecting friction. However, the driver of the car left the emergency brake on, causing the truck to move at a faster speed to compensate.
  • #1
sciencecats
3
0

Homework Statement


A truck is pulling a car.
  • mimetex.gif
    is the magnitude of the force that the truck exerts on the car
  • mimetex.gif
    is the magnitude of the force that the car exerts on the truck
Consider the following scenarios independently.

1. The truck is driving up a mountain with a constant velocity, neglecting friction.
2. The truck is speeding up while driving up a mountain, neglecting friction.
3.The truck is driving with a constant velocity, but as it turns out, the driver of the car left the emergency brake on.

Homework Equations


F=ma

The Attempt at a Solution


I know that the answer to all of the scenarios are FT = FC > 0 (the key for this past homework problem is available to me), but I trying to explain the concept to myself. The 3rd one is the hardest one for me. I suppose since the mass of the car and truck are not changing, that the truck must be moving at a faster speed to compensate for the emergency brake? Am I thinking about this correctly?
 
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  • #2
FT is the magnitude of the force that the truck exerts on the car
FC is the magnitude of the force that the car exerts on the truck

Sorry that it did not post correctly in the first post!
 
  • #3
Have you drawn separate free body diagrams of the truck and of the car, showing the forces acting on each, or do you feel that you have advanced beyond the point where you need to draw free body diagrams? If you have drawn free body diagrams, please show them for case c.

Chet
 
  • #4
index.jpg


I am not sure if this is correct though!
 
  • #5
sciencecats said:
View attachment 97582

I am not sure if this is correct though!
The free body diagram for the car is correct. Nice job. The free body diagram for the truck is not separate (and should be). Please draw it separately. Then write down a horizontal force balance on the truck and a separate horizontal force balance on the car. (For case 3)
 

Related to How Does Newton's Third Law Apply When a Truck Pulls a Car?

1. What is F=ma and how is it related to Newton's Second Law?

F=ma is the mathematical representation of Newton's Second Law, which states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. This means that the greater the force applied to an object, the greater its acceleration will be, and the more massive an object is, the less it will accelerate for a given force.

2. What are the units of measurement for F=ma?

The unit for force is typically measured in Newtons (N), while mass is measured in kilograms (kg) and acceleration in meters per second squared (m/s²). Therefore, the units for F=ma are typically expressed as N = kg * m/s².

3. How can F=ma be applied in real-life scenarios?

F=ma can be applied in many real-life scenarios, such as calculating the force needed to accelerate a car or determining the impact force of a falling object. It is also used in the design and testing of structures and machines, as well as in understanding the motion of celestial bodies.

4. Can F=ma be used for objects in motion with changing acceleration?

Yes, F=ma can still be used for objects with changing acceleration, as long as the net force acting on the object is known at each point in time. This can be calculated by taking into account all the individual forces acting on the object, including friction and air resistance.

5. How does F=ma relate to the conservation of momentum?

F=ma is related to the conservation of momentum in that both principles involve the relationship between force, mass, and acceleration. While F=ma is used to calculate the force required to accelerate an object, the conservation of momentum states that the total momentum of a closed system remains constant, meaning that any change in momentum of one object must be offset by an equal and opposite change in momentum of another object.

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