Problem in Newton's third law and rope tension

In summary: F is the force and B...F is the force for a system with mass M, the tension in the rope is F2MIn summary, Newton's third law states that if a force is applied to a body, then the body will apply a force to the same magnitude and in the opposite direction. This law is applicable to systems with mass, such as a system with a rope and two bodies. The system will move in the direction of the applied force, and the only force acting on the system is F.
  • #1
tmn50
30
0
Hello hope you're doing well :smile:

Homework Statement


I'm having a great problem understanding Newton's third law
let's say that we have 2 bodies with masses m1 & m2 where m1>m2
the 2 bodies are connected with a rope body1________body2----->F
a force F is applied on body 2 as shown below
body1________body2----->F
friction is ignored
how does the 2 bodies move when the rope is massless and when it mass can't be ignored

Homework Equations



there is none

The Attempt at a Solution


according to Newton third law
when b2 aplies a force on the rope, the rope aplies a force on b2 so that it is equal to F in the opposite direction F2
and the rope aplies a force on b1 in the same direction as F let's call it F3
and b1 aplies a force on the rope equal to F in the oppsite direction let's call it f4

F4<---body1--->F3________<---F2 body2--->F
so there are 2 forces on B2 "F" and "F2" which are identical how will it move ?
body 1 has only 1 force acting on it "F3" so I guess it will move toward Body 2
also what's the diffrence when the rope has a mass and when it is massless
in short I'm stuck
 
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  • #2
Hi, Tmn50, and welcome to Physics Forums!:smile:

Before you move on to the rope with mass, you first need to clarify your understanding of Newton 3
tmn50 said:
Hello hope you're doing well :smile:

Homework Statement


I'm having a great problem understanding Newton's third law
let's say that we have 2 bodies with masses m1 & m2 where m1>m2
the 2 bodies are connected with a rope body1________body2----->F
a force F is applied on body 2 as shown below
body1________body2----->F
friction is ignored
how does the 2 bodies move when the rope is massless and when it mass can't be ignored

Homework Equations



there is none

The Attempt at a Solution


according to Newton third law
when b2 aplies a force on the rope, the rope aplies a force on b2 so that it is equal to F
it is not equal to F
in the opposite direction F2
and the rope aplies a force on b1 in the same direction as F let's call it F3
and b1 aplies a force on the rope equal to F
it is not equal to F
in the oppsite direction let's call it f4

F4<---body1--->F3________<---F2 body2--->F
so there are 2 forces on B2 "F" and "F2" which are identical how will it move ?
body 1 has only 1 force acting on it "F3" so I guess it will move toward Body 2
also what's the diffrence when the rope has a mass and when it is massless
in short I'm stuck
You are confusing force pairs (per Newton3) which act on different objects, with net forces acting on the same object. You must draw a free body diagram of the system and of each block separately, identify the forces acting on them, and use Newton's laws. The system will be accelerating in the direction of the applied force F, where F is the only force acting on the 2-block system, and hence, is the net force acting on the system. Solve for the acceleration of the system using Newton 2.. Now move on to each block to identify the forces acting on them, and use Newton 2 to solve for the tension in the rope. The rope pulls on body 2 with a force F2, and that same rope pulls on body 1 with the same force F2 but in the opposite direction.
 
  • #3
PhanthomJay said:
Hi, Tmn50, and welcome to Physics Forums!:smile:

Before you move on to the rope with mass, you first need to clarify your understanding of Newton 3
it is not equal to F it is not equal to F You are confusing force pairs (per Newton3) which act on different objects, with net forces acting on the same object. You must draw a free body diagram of the system and of each block separately, identify the forces acting on them, and use Newton's laws. The system will be accelerating in the direction of the applied force F, where F is the only force acting on the 2-block system, and hence, is the net force acting on the system. Solve for the acceleration of the system using Newton 2.. Now move on to each block to identify the forces acting on them, and use Newton 2 to solve for the tension in the rope. The rope pulls on body 2 with a force F2, and that same rope pulls on body 1 with the same force F2 but in the opposite direction.

Newton's third law state that if A aplies a force F on B than B aplies a force equal to F in magnitude and have an oposite direction
here A is B2 and b is the rope so why it isn't equal to F ??
 
  • #4
You are mixing up your 'F's'. There is an applied force F acting to the right on block m2. If you draw a free body diagram of m2, you should see that in addition to the applied force F acting to the right on m2, there is also an internal tension force, T, in the rope, acting to the left on m2 (tension forces always pull away from the body on which they act). These are 2 separate forces, F and T, acting on the same object, m2. This has nothing at all to do with Newton 3. Applying Newton 2 to this block, then F_net = m2(a), or F - T = m2a. F is greater than T. Otherwise, the blocks could not accelerate.
Now draw a free body diagram of block m1. There is only 1 force acting on it, the rope tension acting to the right. For a massless rope, this tension force must equal T (since m1 pulls on m2 with a force T to the left, then m2 must pull on m1 with a force T to the right; this is Newton 3, force pairs acting on different objects.

Massless ropes serve to transmit forces from one object to another and the tension in such ropes are uniform throughout. This concept is sometimes confusing. It might be better to pretend that the blocks are connected by a small coupling (like a train) to get a better feel for Newton 3 .

When the rope has mass, the tension in the rope is not uniform, and the problem complexity deepens.
 
  • #5
PhanthomJay said:
You are mixing up your 'F's'. There is an applied force F acting to the right on block m2. If you draw a free body diagram of m2, you should see that in addition to the applied force F acting to the right on m2, there is also an internal tension force, T, in the rope, acting to the left on m2 (tension forces always pull away from the body on which they act). These are 2 separate forces, F and T, acting on the same object, m2. This has nothing at all to do with Newton 3. Applying Newton 2 to this block, then F_net = m2(a), or F - T = m2a. F is greater than T. Otherwise, the blocks could not accelerate.
Now draw a free body diagram of block m1. There is only 1 force acting on it, the rope tension acting to the right. For a massless rope, this tension force must equal T (since m1 pulls on m2 with a force T to the left, then m2 must pull on m1 with a force T to the right; this is Newton 3, force pairs acting on different objects.

Massless ropes serve to transmit forces from one object to another and the tension in such ropes are uniform throughout. This concept is sometimes confusing. It might be better to pretend that the blocks are connected by a small coupling (like a train) to get a better feel for Newton 3 .

When the rope has mass, the tension in the rope is not uniform, and the problem complexity deepens.

I'm a bit confused
So when we say a rope is massless it means it will only transfer force right ?
so if it has a mass the tension in it's points will be diffrent right ?
so if we directly apply Newton third law on this one B2 will apply a force on B1 and B1 will apply a force equal to Fb/a in the opposite direction right ?
i have one more question
F is applied on B2 and B1 is attached to B2 wouldn't this force affect B1 in the same way as it affects B2
thanks for your help ^_^
 
  • #6
tmn50 said:
I'm a bit confused
So when we say a rope is massless it means it will only transfer force right ?
so if it has a mass the tension in it's points will be diffrent right ?
so if we directly apply Newton third law on this one B2 will apply a force on B1 and B1 will apply a force equal to Fb/a in the opposite direction right ?
i have one more question
F is applied on B2 and B1 is attached to B2 wouldn't this force affect B1 in the same way as it affects B2
thanks for your help ^_^
Well this lettering is getting messy, see attachment and ask again. Don't get too hung up on Newton 3...it just says that if A exerts a force on B, then B exerts an equal and opposite force on A...think Newton 2.

http://img854.imageshack.us/i/rope.pdf/
 
  • #7
PhanthomJay said:
Well this lettering is getting messy, see attachment and ask again. Don't get too hung up on Newton 3...it just says that if A exerts a force on B, then B exerts an equal and opposite force on A...think Newton 2.

http://img854.imageshack.us/i/rope.pdf/

exactly since the body exerts a force on the rope the rope should exert a force on the body right ?
and than the force exerted by the body on the rope should be equal to F cause there is no other source and there is no friction or anything to reduce it so now the body has two forces as you showed but why aren't they equal I mean what makes them diffrent please don't use Newton 2 law to explain it
 
  • #8
tmn50 said:
exactly since the body exerts a force on the rope the rope should exert a force on the body right ?
Yes, correct, the body exerts a force on the rope, and the rope exerts an equal and opposite force on the body. This is Newton's third law. And that is all it says, it says nothing about the magnitude of the force. If you knew the force that the body exerts on the rope, you would know the force that the rope exerts on the body, but you don't know either force...yet
and than the force exerted by the body on the rope should be equal to F cause there is no other source and there is no friction or anything to reduce it so now the body has two forces as you showed but why aren't they equal I mean what makes them diffrent please don't use Newton 2 law to explain it
Unfortunately, one must use Newton 2 to explain it. We've already established that the blocks are accelerating, which means there must be a net force acting on the block. If the applied force F was equal to the rope force T, then the body would not accelerate.
There are 2 types of forces in Intro Physics...action at a distance forces, like gravity and magnetic forces, and contact forces (electromagnetic in nature), like applied forces, normal forces, tension forces, and friction forces. In this problem, the 2 contact forces on the 2nd block in the horizontal direction are the applied contact force, F, and the rope contact tension force, T2. F must be greater than T2 inorder for the block to accelerate.

Perhaps rather than looking at tension forces, consider instead 2 blocks, m1 on the left and m2 on the right, that are touching each other, resting on a frictionless table. A pushing force F is then applied to m2, acting to the left. Let's put some numbers on them, let m1 = 1 kg, m2 = 2 kg, and F = 9 N. Looking at the system of blocks, apply Newton 2,
F_net = Ma
9 = (1 + 2)a = 3a
a = 3 m/sec^2 to the left.

Now look at block 2 in a free body diagram. We have an applied force of 9 N acting left, and an unknown normal contact force, the force of block 1 on 2 (call it N), acting right. Apply Newton 2 again, and solve for the unknown normal force:
F_net = (m2)a
9 - N = 2(3) = 6
N = 3 N

Now look at block 1 in a free body diagram. From Newton's third law, the force of block 2 on block 1 must be equal and opposite to the force of block 1 on block 2, that is, 3 N acting left on block 1. That is the only force acting on block 1 in the horizontal direction. Thus, per Newton 2,
F_net = m1(a)
3 = (1)a
a = 3 N/m^2, which agrees with our original result (this serves as a check on your work).

The big thing to note is that the applied force on block 2, 9 Newtons, is NOT the same as the contact force of block 1 on block 2, which is only 3 Newtons.

Clear??
 
  • #9
PhanthomJay said:
Yes, correct, the body exerts a force on the rope, and the rope exerts an equal and opposite force on the body. This is Newton's third law. And that is all it says, it says nothing about the magnitude of the force. If you knew the force that the body exerts on the rope, you would know the force that the rope exerts on the body, but you don't know either force...yet Unfortunately, one must use Newton 2 to explain it. We've already established that the blocks are accelerating, which means there must be a net force acting on the block. If the applied force F was equal to the rope force T, then the body would not accelerate.
There are 2 types of forces in Intro Physics...action at a distance forces, like gravity and magnetic forces, and contact forces (electromagnetic in nature), like applied forces, normal forces, tension forces, and friction forces. In this problem, the 2 contact forces on the 2nd block in the horizontal direction are the applied contact force, F, and the rope contact tension force, T2. F must be greater than T2 inorder for the block to accelerate.

Perhaps rather than looking at tension forces, consider instead 2 blocks, m1 on the left and m2 on the right, that are touching each other, resting on a frictionless table. A pushing force F is then applied to m2, acting to the left. Let's put some numbers on them, let m1 = 1 kg, m2 = 2 kg, and F = 9 N. Looking at the system of blocks, apply Newton 2,
F_net = Ma
9 = (1 + 2)a = 3a
a = 3 m/sec^2 to the left.

Now look at block 2 in a free body diagram. We have an applied force of 9 N acting left, and an unknown normal contact force, the force of block 1 on 2 (call it N), acting right. Apply Newton 2 again, and solve for the unknown normal force:
F_net = (m2)a
9 - N = 2(3) = 6
N = 3 N

Now look at block 1 in a free body diagram. From Newton's third law, the force of block 2 on block 1 must be equal and opposite to the force of block 1 on block 2, that is, 3 N acting left on block 1. That is the only force acting on block 1 in the horizontal direction. Thus, per Newton 2,
F_net = m1(a)
3 = (1)a
a = 3 N/m^2, which agrees with our original result (this serves as a check on your work).

The big thing to note is that the applied force on block 2, 9 Newtons, is NOT the same as the contact force of block 1 on block 2, which is only 3 Newtons.

Clear??

yep very clear one more question why there is a diffrence in the force ?
is it because of inertia ?
 
  • #10
tmn50 said:
yep very clear one more question why there is a diffrence in the force ?
is it because of inertia ?
Yes, inertial mass, its resistance to change in motion.
 
  • #11
PhanthomJay said:
Yes, inertial mass, its resistance to change in motion.

thanks man :biggrin:
 

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 back on the first object.

2. How does Newton's third law apply to rope tension?

When a rope is pulled on by two objects, each object exerts a force on the rope in opposite directions. According to Newton's third law, these forces are equal in magnitude and opposite in direction, resulting in tension in the rope.

3. What is the problem with Newton's third law and rope tension?

The problem arises when one of the objects pulling on the rope is much heavier or stronger than the other. In this case, the force exerted by the heavier object will be greater, resulting in a net force in the direction of the heavier object. This can cause the rope to break or the lighter object to move much faster than expected.

4. How can the problem of Newton's third law and rope tension be solved?

This problem can be solved by using a pulley system, which changes the direction of the force and allows for equal tension in the rope. Additionally, using ropes with greater tensile strength or adding additional ropes can help distribute the force more evenly and prevent breakage.

5. Are there any real-life examples of this problem?

Yes, this problem can occur in activities such as tug-of-war or when towing a car with a rope. It can also be seen in more complex systems, such as bridges or cranes, where unequal forces can cause unexpected movements or failure of the structure.

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