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A small car (mass 380 \(\mathrm{kg}\) ) is pushing a large truck (mass900 \(\mathrm{kg}\) ) due east on a level road. The car exerts a horizontal force of 1200 \(\mathrm{N}\) on the truck. What is the magnitude of the force that the truck exerts on the car?

Short Answer

Expert verified
The force that the truck exerts on the car is 1200 Newtons.

Step by step solution

01

Understand Newton's Third Law

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means if the car is exerting a force on the truck, the truck will exert an equal and opposite force on the car.
02

Identify the Forces

In this situation, the car exerts a force of 1200 Newtons on the truck because it is pushing it. According to Newton's Third Law, the truck will exert an equal and opposite force back on the car.
03

Calculate the Reaction Force

Using Newton's Third Law, the magnitude of the force that the truck exerts on the car is equal to the magnitude of the force that the car exerts on the truck, which is 1200 Newtons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Action-Reaction Pairs
When discussing Newton's Third Law, the concept of action-reaction pairs often comes into play.
Imagine the car and the truck as two friends playing a game of push and pull. When the car pushes the truck, it applies a force in a specific direction.
This push is the 'action'. In response, the truck pushes back with an equally strong force, but in the opposite direction. This response is the 'reaction'. A critical takeaway is that these forces are always equal in magnitude but opposite in direction.
So, if the car exerts a force of 1200 N on the truck, the truck doesn't just sit there passively. Instead, it becomes an active participant by exerting a force of 1200 N back on the car.

It's essential to understand that action-reaction pairs are always present wherever forces occur. They are the silent dance between objects interacting through force. It reminds us that forces are always mutual.
Exploring Force Interaction
Force interaction is how objects communicate through forces. In the context of this problem, it's the car and the truck discussing who moves whom.
Forces are vectors, meaning they have both magnitude and direction. When the car exerts force on the truck, it does so directionally.
  • Horizontal Force: The car applies a force parallel to the ground.
  • Magnitude of Force: In this case, it is 1200 N.
  • Direction of Force: Due east.

But what's vital about force interaction isn't just who pushes who but how this push creates movement or resistance.
The truck could feel this interaction due to factors like friction or its mass, which may influence how much it moves in response. By grasping force interactions, we learn how objects affect each other in various physical situations, lending insight into the mechanics of everyday phenomena.
Mechanics Problem Solving with Newton's Third Law
Solving mechanics problems like the one about the car and truck gives us practical insights into applied physics. Here's a handy guide on tackling such problems:

1. **Identify the Forces:**
Understand which objects are interacting and what forces are involved. For example, the truck and car share a force interaction.
2. **Apply Newton's Third Law:**
Always remember, for every action, there is an equal and opposite reaction. This step often involves identifying action-reaction pairs such as the car exerting force on the truck and vice versa.
3. **Calculate the Magnitude:**
Once you recognize the pairs, calculating the force involved is straightforward. Since the forces are equal and opposite, you only need to know one to know the other.

  • In our exercise, if the car exerts a force of 1200 N, the truck exerts a force of 1200 N in the opposite direction on the car.
Following these steps helps in solving mechanics problems systematically and gains confidence in understanding forces and motions.

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Most popular questions from this chapter

CP A 4.80 -kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75.0 \(\mathrm{N}\) . If the bucket starts from rest, what is the minimum time required to raise the bucket a vertical distance of 12.0 \(\mathrm{m}\) without breaking the cord?

A 4.9 -N hammer head is stopped from an initial downward velocity of 3.2 \(\mathrm{m} / \mathrm{s}\) in a distance of 0.45 \(\mathrm{cm}\) by a nail in a pine board. In addition to its weight, there is a 15 downward force on the hammer head applied by the person using the hammer. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward. (a) Draw a free-body diagram for the hammer head. Identify the reaction force to each action force in the diagram. (b) Calculate the downward force \(\vec{\boldsymbol{F}}\) exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward. (c) Suppose the nail is in hardwood and the distance the hammer head travels in coming to rest is only 0.12 \(\mathrm{cm}\) . The downward forces on the hammer head are the same as in part (b). What then is the force \(\vec{\boldsymbol{F}}\) exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

A crate with mass 32.5 \(\mathrm{kg}\) initially at rest on a warehouse floor is acted on by a net horizontal force of 140 \(\mathrm{N}\) . (a) What acceleration is produced? (b) How far does the crate travel in 10.0 \(\mathrm{s} ?\) (c) What is its speed at the end of 10.0 \(\mathrm{s} ?\)

The upward normal force exerted by the floor is 620 \(\mathrm{N}\) on an elevator passenger who weighs 650 \(\mathrm{N}\) . What are the reaction forces to these two forces? Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration?

Two forces, \(\vec{\boldsymbol{F}}_{1}\) and \(\vec{\boldsymbol{F}}_{2},\) a point. The magnitude of \(\vec{\boldsymbol{F}}_{1}\) is \(9.00 \mathrm{N},\) and its direction is \(60.0^{\circ}\) above the \(x\) -axis in the second quadrant. The magnitude of \(\vec{\boldsymbol{F}}_{2}\) is \(6.00 \mathrm{N},\) and its direction is \(53.1^{\circ}\) below the \(x\) -axis in the third quadrant. (a) What are the \(x\) - and \(y\) -components of the resultant force? (b) What is the magnitude of the resultant force?

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