Newton's Third Law Problem: Two masses, a rope and a pulley

In summary, two blocks of masses 1.5 kg and x kg are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One block accelerates downward at (3/4)g. To find the mass of the other block, we can use the equation a=(m2gsin0-m1g)/(m2+m1) and treat the two masses separately, considering all the forces and defining variables as necessary. However, since the rope is massless and the pulley is frictionless, the tension in the rope must be equal throughout its length, making the net force in the upward direction equal to tension minus weight. Therefore, the mass of the other block
  • #1
Zeke Bevan
11
1

Homework Statement


Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks w/ a mass of 1.5 kg accelerates downward at (3/4)g.

A. What is the mass of the other block?

Homework Equations


There were no given equations.

The Attempt at a Solution


I tried using an equation that we solved for from the previous problem that was a=(m2gsin0-m1g)/(m2+m1).
I'm confused because I'm assuming there isn't an angle and I don't know wether to include tension in this problem or not.
This chapter has been really hard for me setting up problems. If there's any advice for setting/approaching problems for Newtons laws of motion that would be amazing!
 
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  • #2
Zeke Bevan said:

Homework Equations


There were no given equations.
This refers to standard equations, such as Newton's laws.
Zeke Bevan said:
I'm assuming there isn't an angle
Right, take the rope as hanging vertically both sides.
Zeke Bevan said:
I don't know wether to include tension in this problem
Treat the two masses separately. Consider all the forces on each and their accelerations. Define variables as necessary.
 
  • #3
So, the forces acting on each block is gravity pulling the blocks down and tension in the rope holding the blocks together? If there is tension in the rope, can I assume there are equal to each other as well?
I'm really sorry if am oblivious on how to do this problem, I'm finding myself to be not good at physics but I really wish to be.
 
  • #4
You are given that the rope is massless and that the pulley is massless and frictionless. This has implications for the tension in the rope.

For instance, if the tension at two ends of a length of massless rope were different, what acceleration would result?
 
  • #5
I can only think that the accelerations would be different? One object would be pull down while the other got pulled up.
So for this problem would I use f=ma to find the tension of the rope getting pulled down?
 
  • #6
Zeke Bevan said:
I can only think that the accelerations would be different? One object would be pull down while the other got pulled up.
So for this problem would I use f=ma to find the tension of the rope getting pulled down?
I am trying to get you to focus on a simpler exercise. No pulleys. No masses. Just a section of massless rope. What is the acceleration of the section of rope if the tensions at the two ends are different?
 
  • #7
jbriggs444 said:
What is the acceleration of the rope if the tension at the two ends are different?
I'm really sorry, I'm going through my textbook right now and I can't find anything on this, I can only find that in massless strings the tension will be equal for each object. I'm really bad at assumptions because I don't think I'll be right but I'm thinking it will be equal to the object pulling it down times gravity or it will just be gravity. I'm sorry again for making this more complicated than it needs to be!
 
  • #8
For a massless string or rope, the tension will be the same throughout its length, yes. This includes the case where it runs over a massless and frictionless pully.

One can see that this must be true from first principles. The only external forces on a segment of rope is from the tension at its ends. From Newton's second law, ΣF=ma. If the tensions are unequal then ∑F will be non-zero. If the mass is zero then ma will be zero. That means that unequal tensions are impossible in a massless rope(*) by Newton's second law.

A similar line of reasoning works for massless ropes running over massless and frictionless pulleys.

(*) Subject to no forces other than from the tensions at its ends.
 
  • #9
Okay I believe I understand what you are saying. So, for this problem, the tension would equal the mass(1.5kg) times the rate at which it's accelerating downward(3/4g)?
 
  • #10
The tension in the rope is whatever it needs to be so that the block accelerates at the stipulated rate. Make yourself a free body diagram. What forces act on the 1.5 kg mass?
 
  • #11
jbriggs444 said:
The tension in the rope is whatever it needs to be so that the block accelerates at the stipulated rate. Make yourself a free body diagram. What forces act on the 1.5 kg mass?

The tension in the rope that is going upwards and the weight of the block(1.5 times 9.8). So to find the net force, would it be Tension minus the weight?
 
  • #12
Zeke Bevan said:
The tension in the rope that is going upwards and the weight of the block(1.5 times 9.8). So to find the net force, would it be Tension minus the weight?
The net force in the upward direction will be tension minus weight, yes.
 
  • #13
Okay. Would the tension be equal to the rate a which it’s accelerating times the weight?
 
  • #14
Zeke Bevan said:
Okay. Would the tension be equal to the rate a which it’s accelerating times the weight?
No.

You do realize that "weight" denotes a force, not a mass, right? You can't multiply a force times an acceleration to get a force. The units are not sensible.
 
  • #15
Yes I realized what I said that was my bad! Weight is a force in Newton’s, I meant to say multiple the rate at which it’s accelerating times the mass, I assume that’s a force but I’m guessing that’s not the correct way to get the tension, I’m really confused as of how.
 
  • #16
Zeke Bevan said:
Yes I realized what I said that was my bad! Weight is a force in Newton’s, I meant to say multiple the rate at which it’s accelerating times the mass, I assume that’s a force but I’m guessing that’s not the correct way to get the tension, I’m really confused as of how.
In post #11 you decided on the net force. Have you changed your mind? Newton's second law says that mass times acceleration is equal to the net force. Not just the one force you are focusing on at the moment.
 
  • #17
Okay, what I am deciding on my net force to be is, tension minus weight. The tension which I am thinking is the acceleration (3/4g) times mass (1.5kg) minus the weight force (1.5kg times gravity.) This would give you 11.025-14.7= -3.675 N, was that the correct way of doing it or should it be weight minus tension?
 
  • #18
Zeke Bevan said:
Okay, what I am deciding on my net force to be is, tension minus weight. The tension which I am thinking is the acceleration (3/4g) times mass (1.5kg) minus the weight force (1.5kg times gravity.) This would give you 11.025-14.7= -3.675 N, was that the correct way of doing it or should it be weight minus tension?
You are getting a correct figure for tension (although with the wrong sign. Tension in a rope is always positive).

Now keep going. What forces act on the other block? What is the other block's acceleration? Can you write down an equation from Newton's second law applied to the other block?
 
  • #19
The forces acting on the other block are tension and weight as well. The acceleration is equal to the rate the other block is accelerating (3/4g). I think you can make the F=ma equation to be: Fnet=tension-weight. So, the tension be equal to 3.675N and weight is equal to mass times gravity with the mass is what were looking for. F net is equal to 11.025. So would it be 11.025=3.675-m(9.8)? I don't know if I set that up correctly.
 
  • #20
Since you do not know the mass of the other block, you do not know its weight. Since you do not know its weight, its acceleration does not determine F net. So no, Fnet is not equal to 11.025. You need to write down an equation.
 
  • #21
Okay, So fnet=3.675-mg?
 
  • #22
Zeke Bevan said:
Okay, So fnet=3.675-mg?
That's correct.

But you don't have a numerical value for ##\ F_\text{net} \,,\ ## so you need to express it in an additional way. (Hint: Newton's 2nd Law)
 

Related to Newton's Third Law Problem: Two masses, a rope and a pulley

1. What is Newton's Third Law?

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

2. How does Newton's Third Law apply to a system with two masses, a rope, and a pulley?

In this system, the two masses are connected by a rope which is attached to a pulley. When one mass exerts a force on the rope, the rope will exert an equal and opposite force on the other mass. This is an example of Newton's Third Law in action.

3. What is the role of the pulley in this system?

The pulley serves as a point of support and redirection for the rope. It helps to distribute the force exerted by one mass onto the other mass, allowing for smoother and more efficient movement.

4. How does the mass of the objects affect the forces involved in this system?

According to Newton's Third Law, the forces involved in this system will be equal and opposite regardless of the masses involved. However, the acceleration of each mass may vary depending on their individual masses and the overall force being applied.

5. Can this system violate Newton's Third Law?

No, this system cannot violate Newton's Third Law. The law is a fundamental principle of physics and applies to all interactions between objects. If it appears that one object is exerting a greater force than the other, it is likely due to differences in mass or other external factors, but the fundamental law still holds true.

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