Newton's 3rd Law Dynamics Problem

In summary, the problem involves a 12.0-kg object being pushed by a 24-N force with a frictional force of 6.0 N opposing its motion. The acceleration of the object is found to be 0.587 m/s^2. The normal force acting on the object is calculated to be 9 N. When the object collides with a 4.0-kg object, the overall friction increases by 3.0 N, resulting in a new acceleration of 0.652 m/s^2. The force exerted by the 4.0-kg object on the 12-kg object is found to be either -0.652 N or 0.652 N, depending
  • #1
Abu

Homework Statement


A 12.0-kg box is pushed along a horizontal surface by a 24-N force as illustrated in the diagram. The frictional force (kinetic) acting on the object is 6.0 N
physproblem1.PNG

A) What is the acceleration of the object
B) Calculate the value of the normal force acting on the object
C) If the 12.0-kg object then runs into a 4.0-kg object that increases the overall friction by 3.0 N, what is the new acceleration?
D) What force does the 4.0-kg object exert on the 12-kg object when the two are moving together? (Issue I am having)

Homework Equations


a = fnet/m
fnet = Force applied - Force of Friction

The Attempt at a Solution


D)
Fnet = Force applied - Force of Friction
m*a = Force applied - Force of Friction
Force applied = m*a + Force of Friction
Force applied = 4*0.587 + (3 + -18.4)
Force applied = -13.052 N

The reasons why I chose those numbers:
  • 4 is the mass of the object that is exerting the force on the 12kg object
  • If they are moving together, then I need to use the acceleration found in part C, which is 0.587 (if you need me to show my work for that part I can)
  • Assuming the force applied is to the right, and that right will be positive, then the frictional force which is opposing the 4kg's movement to the left will be positive
  • the 18.4 N that is pushing the entire set of objects will be negative because it is pushing to the left
I know I am wrong, can anyone tell me how I can improve? Thank you.
 
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  • #2
Abu said:
If they are moving together, then I need to use the acceleration found in part C, which is 0.587 (if you need me to show my work for that part I can)
Think again. The pushing force in part C accelerated only the 12.0 kg mass. In part D it is accelerating a 16.0 combined mass. Do you really think that the acceleration can be the same? You need to find the common acceleration anew.
 
  • #3
kuruman said:
Think again. The pushing force in part C accelerated only the 12.0 kg mass. In part D it is accelerating a 16.0 combined mass. Do you really think that the acceleration can be the same? You need to find the common acceleration anew.

Sorry, should have clarified. This is how I got my acceleration:

a = fnet/m
a = Fappl - Ffr/m
a = 24cos40 - 9/16
a = 0.587 m/s^2

That should be the acceleration for the 16 kg combined mass, right? And that means that the 4kg mass is moving with an acceleration of 0.587m/s^2 because the two objects are moving together?
 
  • #4
OK, thanks for the clarification.
Abu said:
Force applied = 4*0.587 + (3 + -18.4)
This is incorrect. Please explain
1. Is ma in the same or in the opposite direction to friction?
2. What is the -18.4 doing in this equation? Where does it come from? (This is an important point for you to understand)
 
  • #5
kuruman said:
OK, thanks for the clarification.

This is incorrect. Please explain
1. Is ma in the same or in the opposite direction to friction?
2. What is the -18.4 doing in this equation? Where does it come from? (This is an important point for you to understand)
1. Actually now that I think about it, ma is in the opposite direction to friction, right? Since it is the net force and the net force is pointed to the left while the friction is pointed to the right. Does that mean that ma should be negative? Or maybe should ma be positive and the friction should become negative...
2. So the 18.4 should not be in the equation. This is because I am focusing on the forces that the 4kg object is exerting on the 12kg, not the other way around? So I should take out 18.4 entirely? Since it comes from the 12kg and not the 4kg
 
  • #6
Your reasoning is correct in both cases. So what's the answer?
Abu said:
... and the friction should become negative...
Friction is opposite to the direction of motion. You have already specified "to the right" as positive. So what sign does friction have?
 
  • #7
kuruman said:
Your reasoning is correct in both cases. So what's the answer?

Friction is opposite to the direction of motion. You have already specified "to the right" as positive. So what sign does friction have?
So the friction is positive because it is to the right, ma is negative because I said to the right is positive and the net force is directed to the left.

So Fapplied = -4*0.587 + 3
= 0.652 Newtons

Am I on the right track? I am pretty sure ma and the frictional force have opposite signs so its either 0.652 Newtons or -0.652 Newtons, right?
 
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  • #8
Abu said:
Am I on the right track? I am pretty sure ma and the frictional force have opposite signs so its either 0.652 Newtons or -0.652 Newtons, right?
You are still missing the subtlety of Newton's Second law that says Fnet = ma. You are required to add all the forces acting on the object and get a number which you set equal to mass times acceleration. In this case you have
Fnet = Funknown + friction
Set that equal to ma to get Funknown + friction = ma*. Substitute remembering that anything to the right is positive and anything to the left is negative. Solve only after you substitute. This will help you keep the positive and negative signs sorted out.

__________________
* In this equation ma and friction must have opposite signs.
 
  • #9
kuruman said:
You are still missing the subtlety of Newton's Second law that says Fnet = ma. You are required to add all the forces acting on the object and get a number which you set equal to mass times acceleration. In this case you have
Fnet = Funknown + friction
Set that equal to ma to get Funknown + friction = ma*. Substitute remembering that anything to the right is positive and anything to the left is negative. Solve only after you substitute. This will help you keep the positive and negative signs sorted out.

__________________
* In this equation ma and friction must have opposite signs.
Okay, so:

Fnet = Funknown + friction
-ma = Funknown + friction
-4*0.587 = Funknown + 3
Funknown = -4*0.587 - 3
Funknown = -5.348 Newtons

Okay so that is the final answer! Thank you for your patience with me I really appreciate it.. I just have one more question:

I am just confused about how Fnet = Funknown + Ffriction. I was always taught that Fnet = Fapplied minus Ffriction, not plus

Can you explain to me how come in this case it is addition and not subtraction? I thought Funknown was Fapplied, and thus you use the equation with the minus but you say I must add all of the forces.

*edited* looking back on what I just asked you.. is that just a misconception, and that it is always Fnet = Fapplied + Ffr, but force of friction is typically negative and not positive?

I hope my question is clear
 
Last edited by a moderator:
  • #10
sorry.. accidentally made this post not sure how to delete it.. i replied to your response right above this post
 
  • #11
Abu said:
Can you explain to me how come in this case it is addition and not subtraction? I thought Funknown was Fapplied, and thus you use the equation with the minus but you say I must add all of the forces.
I'm glad you asked this question because it is at the core of Fnet, the subtlety that I mentioned earlier. Fnet is indeed the sum of all the forces, but not all of them are positive. In this specific case, friction is positive and Funknown is negative. Therefore,
Fnet = (- Funknown) + friction. This is a sum. Of course, ma is also negative, so Newton's 2nd law is
- Funknown + friction = -ma
It might be useful to think of this as vector addition. Any vector to the left is negative and any vector to the right is positive and you add positive and negative vectors in one dimension. So when we symbolically put in an equation -ma, we mean "a vector of magnitude (i.e. a positive number) ma pointing to the left."
 
  • #12
kuruman said:
I'm glad you asked this question because it is at the core of Fnet, the subtlety that I mentioned earlier. Fnet is indeed the sum of all the forces, but not all of them are positive. In this specific case, friction is positive and Funknown is negative. Therefore,
Fnet = - Funknown + friction. Of course ma is also negative, so Newton's 2nd law is
- Funknown + friction = -ma
It might be useful to think of this as vector addition. Any vector to the left is negative and any vector to the right is positive and you add positive and negative vectors in one dimension. So when we symbolically put in an equation -ma, we mean "a vector of magnitude (i.e. a positive number) ma pointing to the left."
Thank you so much for your help, I really learned some valuable lessons from this. You're a great tutor! Hopefully I won't bother too many people with my questions.
 
  • #13
Thank you for you kind words. We're here to help, so don't worry about bothering us.
 

Related to Newton's 3rd Law Dynamics Problem

1. What is Newton's 3rd Law of Motion?

Newton's 3rd Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object will exert an equal and opposite force back on the first object.

2. How does Newton's 3rd Law apply to dynamics problems?

In dynamics problems, Newton's 3rd Law helps us understand the relationship between the forces acting on an object and its resulting motion. By considering the equal and opposite forces acting on an object, we can determine the net force and acceleration of the object. This is crucial in solving dynamics problems involving motion and forces.

3. Can you provide an example of a dynamics problem that involves Newton's 3rd Law?

One example is a car accelerating on a flat road. The car exerts a force on the ground in the backwards direction, and the ground exerts an equal and opposite force on the car in the forward direction, propelling it forward. This demonstrates how Newton's 3rd Law helps us understand the forces involved in an object's motion.

4. Are there any limitations to Newton's 3rd Law?

One limitation of Newton's 3rd Law is that it only applies to interactions between two objects. It does not account for situations where there are more than two objects interacting with each other. Additionally, the forces must be acting on different objects in order for Newton's 3rd Law to apply.

5. How does Newton's 3rd Law relate to conservation of momentum?

Newton's 3rd Law is closely related to the concept of conservation of momentum. When two objects interact, their total momentum before the interaction is equal to their total momentum after the interaction. This is because the forces they exert on each other are equal and opposite, resulting in a change in momentum that balances out.

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