Friction and tension in a string

[Jennings]

Homework Statement

Mass m1 = 5.00kg is tied to the wall with a horizontal piece of string and is at rest onto of mass m2 = 10.00kg. A horizontal force F= 50.0N is applied to m2. The static coefficient of friction between all surfaces is μs = .350 and the kinetic coefficient of friction is μk = .250. Find the tension in the rope attached to m1 and the acceleration of m2

Homework Equations

Fk = μkN
Fsmax = μsN
ΣF = ma

The Attempt at a Solution

Trying to draw a free body diagram for the m1 and the tension, i know that the sum of forces must be 0 because m1 does not move I know that the weight and normal act on it in the y direction. But I am confused about the x direction. Is friction and tension the only forces in the x direction? Or is there another force?

I am just confused

 

[Chestermiller]

But I am confused about the x direction. Is friction and tension the only forces in the x direction?

Yes.

Or is there another force?

No.

Chet

 

[Jennings]

Does friction oppose the direction of tension? . The force of friction and tension must cancel to equal 0 right?

 

[Jennings]

Yes.

No.

Chet

How to determine how much is the friction, since there are two frictions and we do not even know if thr mass m2 will move

 

[Jennings]

Is it kinetic or static friction? How can we determine this?

 

[Chestermiller]

How to determine how much is the friction, since there are two frictions and we do not even know if thr mass m2 will move

In this problem, the problem statement implies that you are dealing with kinetic friction and that m2 is moving. However, this may not be the case. See my next post below.

Chet

 

[Chestermiller]

Is it kinetic or static friction? How can we determine this?

You solve the problem assuming that static friction prevails, and that you are just on the verge of exceeding the coefficient of static friction on both surfaces. Then you check to see whether the frictional forces are sufficient to balance the applied force F. Make sure you do a free body diagram on each of the two masses; this is very important.

Chet

 

[Chestermiller]

I have given this problem significantly more thought, and, in my judgement, it is extremely tricky. It really requires a great deal of thought and "what if" calculations (even though it looks deceptively simple). I'm not sure the person who posed this problem realized how tricky it is.

Chet

 

[Jennings]

I have given this problem significantly more thought, and, in my judgement, it is extremely tricky. It really requires a great deal of thought and "what if" calculations (even though it looks deceptively simple). I'm not sure the person who posed this problem realized how tricky it is.

Chet

Hmm.. well i hope ill be able to figure it out regardless.. with your help of course.

 

[Jennings]

You solve the problem assuming that static friction prevails, and that you are just on the verge of exceeding the coefficient of static friction on both surfaces. Then you check to see whether the frictional forces are sufficient to balance the applied force F. Make sure you do a free body diagram on each of the two masses; this is very important.

Chet

The normal force between the mass 1 and the mass 2 i will call N1 , and the maximum static friction ill call it fsmax1

The normal force between mass 2 and the surfacd i will call N2 and likewise the maximum static friction ill call it fsmax2

Therefore

N1 = mg = (5.0kg)(9.81m/s^2) =49.1N
N2 = mg = (15kg)(9.81m/s^2) = 147N

fsmax1 = (.305)(49.1N) = 17.2N
fsmax2 = (.305)(147N) = 51.5N

Im thinking this means that the force applied is not enough therefore the tension of the string and acceleration mass 2 will both be zero.

Please correct me if I am wrong but i believe that this is what it is ? Right ?

 

[Jennings]

The normal force between the mass 1 and the mass 2 i will call N1 , and the maximum static friction ill call it fsmax1

The normal force between mass 2 and the surfacd i will call N2 and likewise the maximum static friction ill call it fsmax2

Therefore

N1 = mg = (5.0kg)(9.81m/s^2) =49.1N
N2 = mg = (15kg)(9.81m/s^2) = 147N

fsmax1 = (.305)(49.1N) = 17.2N
fsmax2 = (.305)(147N) = 51.5N

Im thinking this means that the force applied is not enough therefore the tension of the string and acceleration mass 2 will both be zero.

Please correct me if I am wrong but i believe that this is what it is ? Right ?

meant for Ms to be .35 not .305. My fault

 

[Chestermiller]

The normal force between the mass 1 and the mass 2 i will call N1 , and the maximum static friction ill call it fsmax1

The normal force between mass 2 and the surfacd i will call N2 and likewise the maximum static friction ill call it fsmax2

Therefore

N1 = mg = (5.0kg)(9.81m/s^2) =49.1N
N2 = mg = (15kg)(9.81m/s^2) = 147N

fsmax1 = (.35)(49.1N) = 17.2N
fsmax2 = (.35)(147N) = 51.5N

Im thinking this means that the force applied is not enough therefore the tension of the string and acceleration mass 2 will both be zero.

Please correct me if I am wrong but i believe that this is what it is ? Right ?

As you said, the frictional forces you calculated are certainly the maximum values that can be tolerated at each surface without slippage. However, if we assume that m2 is totally rigid, we don't know what the actual frictional forces will be prior to slippage. This is because the system is statically indeterminate. Prior to slippage, we only know that their sum is 50 N. We would have to determine how the frictional forces are distributed between the upper and lower surfaces in order to ascertain whether slippage would occur. In order to do that, we would have to solve a problem in elasticity. For the present problem, with the 50 N force applied at the middle of m2, if we assumed no slippage at the interfaces, the forces at the two interfaces would be 25 N each. And 25 N at the upper surface would be enough for slippage to occur there, but 25 N at the bottom surface would not be sufficient for slippage to occur there. So, with m2 being elastically deformable (not totally rigid), the top surface of m2 would begin to slip in the x direction while the bottom surface of m2 would still be non-slipping. Continuing the solution to this problem would involve elastic deformation with friction and dynamics. This would be a very complicated problem. Do you see now what I meant by this all being very tricky?

Chet

 

[Jennings]

I do see now how this simple question has turned into a devastatingly complex problem when you look into it in depth. Wow. We haven't even gotten to elasticity yet. I can't imagine my professor would expect this from me at this point in time. I think you were correct when you said that the one who wrote the problem had no idea how tricky this was. I have only a basic knowledge of friction and forces at the current time. Is there any way we could dumb it down to a level where we can disregard all this complexity while still being theoretically correct with our answer?

 

[Jennings]

I do see now how this simple question has turned into a devastatingly complex problem when you look into it in depth. Wow. We haven't even gotten to elasticity yet. I can't imagine my professor would expect this from me at this point in time. I think you were correct when you said that the one who wrote the problem had no idea how tricky this was. I have only a basic knowledge of friction and forces at the current time. Is there any way we could dumb it down to a level where we can disregard all this complexity while still being theoretically correct with our answer?

Or is that not possible with this

 

[Chestermiller]

I do see now how this simple question has turned into a devastatingly complex problem when you look into it in depth. Wow. We haven't even gotten to elasticity yet. I can't imagine my professor would expect this from me at this point in time. I think you were correct when you said that the one who wrote the problem had no idea how tricky this was. I have only a basic knowledge of friction and forces at the current time. Is there any way we could dumb it down to a level where we can disregard all this complexity while still being theoretically correct with our answer?

I would assume that he didn't recognize the complexity and just continue solving the problem assuming kinetic friction at both interfaces, based on the rationale that you presented.

Chet

 

[Jennings]

I would assume that he didn't recognize the complexity and just continue solving the problem assuming kinetic friction at both interfaces, based on the rationale that you presented.

Chet

So we cannot solve the problem if it were to be static friction? I understand that if static friction were to occur then the acceleration of M2 would be 0, but your saying in order to find out what occurs with mass 1 would require more complex calculculations if i understood you correctly

 

[Jennings]

I would assume that he didn't recognize the complexity and just continue solving the problem assuming kinetic friction at both interfaces, based on the rationale that you presented.

Chet

Assuming the friction is kinetic .. do you have any hints you can give to me to set me up on the right path. I am assuming that value for kinetic friction will be less with M1 since M1 has a smaller normal, but M1 itself does not move, so its kind of confusing on what exactly to do

 

[haruspex]

As you said, the frictional forces you calculated are certainly the maximum values that can be tolerated at each surface without slippage. However, if we assume that m2 is totally rigid, we don't know what the actual frictional forces will be prior to slippage. This is because the system is statically indeterminate. Prior to slippage, we only know that their sum is 50 N. We would have to determine how the frictional forces are distributed between the upper and lower surfaces in order to ascertain whether slippage would occur. In order to do that, we would have to solve a problem in elasticity. For the present problem, with the 50 N force applied at the middle of m2, if we assumed no slippage at the interfaces, the forces at the two interfaces would be 25 N each. And 25 N at the upper surface would be enough for slippage to occur there, but 25 N at the bottom surface would not be sufficient for slippage to occur there. So, with m2 being elastically deformable (not totally rigid), the top surface of m2 would begin to slip in the x direction while the bottom surface of m2 would still be non-slipping. Continuing the solution to this problem would involve elastic deformation with friction and dynamics. This would be a very complicated problem. Do you see now what I meant by this all being very tricky?

Chet

It is a bit tricky, but I think it is all quite reasonable in the end. Suppose one surface just starts to slip. Is kinetic friction there plus static friction at the other interface enough to hold it? Try both cases. I think you'll find it holds.

 

[Chestermiller]

It is a bit tricky, but I think it is all quite reasonable in the end. Suppose one surface just starts to slip. Is kinetic friction there plus static friction at the other interface enough to hold it? Try both cases. I think you'll find it holds.

If it's rigid, you can't have one surface slipping while the other is not.

Chet

 

[Chestermiller]

So we cannot solve the problem if it were to be static friction? I understand that if static friction were to occur then the acceleration of M2 would be 0, but your saying in order to find out what occurs with mass 1 would require more complex calculculations if i understood you correctly

Yes. If the total force were <35.4 N, each surface would have <17.2 N, and static friction would be guaranteed to be intact on both surfaces. Otherwise, it's more complicated.

Chet

 

[Chestermiller]

Assuming the friction is kinetic .. do you have any hints you can give to me to set me up on the right path. I am assuming that value for kinetic friction will be less with M1 since M1 has a smaller normal, but M1 itself does not move, so its kind of confusing on what exactly to do

M2 is moving relative to M1, so there is slip. Get the two kinetic friction forces on the two surfaces. What values to you get? What direction do these forces act on M2?

Chet

 

[Jennings]

M2 is moving relative to M1, so there is slip. Get the two kinetic friction forces on the two surfaces. What values to you get? What direction do these forces act on M2?

Chet

Kinetic friction between M2 and surface is (.25)(147N) = 36.8N

Kinetic friction between M1/M2 =

(.25)(49.1N) = 12.3N

The sum of forces acting on M2

F - Fk1 - Fk2 = 50N -36.8N - 12.3N
= 0.9N

0.9N = M2(a)
0.9N = (10kg)a
a = .09m/s^2

Is it correct?

Tension = Fk(mass1/2) because mass 1 does not accelerate therefore the must cancel so tension = 12.3N

Please let me know if the is the correct way to solve.. Thanks!

 

[Chestermiller]

Kinetic friction between M2 and surface is (.25)(147N) = 36.8N

Kinetic friction between M1/M2 =

(.25)(49.1N) = 12.3N

The sum of forces acting on M2

F - Fk1 - Fk2 = 50N -36.8N - 12.3N
= 0.9N

0.9N = M2(a)
0.9N = (10kg)a
a = .09m/s^2

Is it correct?

Tension = Fk(mass1/2) because mass 1 does not accelerate therefore the must cancel so tension = 12.3N

Please let me know if the is the correct way to solve.. Thanks!

Yes. This looks right to me.

Chet

 

[Jennings]

Yes. This looks right to me.

Chet

Yay! Thank you!
Your a great helper this is the 2nd time in less than a week. Do you get points for this?

 

[Chestermiller]

I have an alternate scenario to offer that looks at the problem from a different perspective. I am going to assume that m2 is not rigid, but deformable, and is characterized by an elastic shear modulus G. I know how to do problems like this because of my prior experience with solid mechanics. What I'm going to show in this scenario is that, for a deformable m2, it is possible for slip to occur at the upper interface with m1, while, at the same time, the lower surface of m2 does not move.

Let A be the area of m2 in contact with m1 and with the table, and let h be the thickness of m2. The displacement to the right at the bottom of m2 is going to be zero. The displacement to the right at the center plane of m2 is going to be δ₁ and the displacement to the right at the top surface of m2 is going to be δ₂. If we did not allow a displacement at the top surface (i.e., temporary slip with kinetic friction, followed by re-establishment of stick) the frictional force at the top surface would be 25N, which would exceed the critical force for release of static friction. Therefore, the top surface of m2 must slip at the kinetic friction force of 12.3 N (temporarily, until static friction is re-established). Once static friction is re-established, the force at the top of m2 will remain at 12.3 N. That is where we are going to examine what prevails.

Now for the analysis:
Shear strain in the lower half of m2 = δ₁/(h/2) = 2δ₁/h
Shear strain in upper half of m2 = 2(δ₁-δ₂)/h

Shear stress in lower half of m2 = 2Gδ₁/h
Shear stress in upper half of m2 = 2G(δ₁-δ₂)/h

Frictional force on top surface = 12.3 N = 2GA(δ₁-δ₂)/h
Frictional force on bottom surface of m2 = 50.0-12.3= 37.7 N = 2GAδ₁/h

If we solve for δ₁ and δ₂, we obtain:

##δ_1=\frac{37.7h}{2AG}##

##δ_2=\frac{25.4h}{2AG}##

So the bottom of m2 will not be displaced to the right at all, and the center plane of m2 will be displaced to the right about 50% more than the top surface.

In the end, m2 would not be accelerating.

Chet

 

[Jennings]

I have an alternate scenario to offer that looks at the problem from a different perspective. I am going to assume that m2 is not rigid, but deformable, and is characterized by an elastic shear modulus G. I know how to do problems like this because of my prior experience with solid mechanics. What I'm going to show in this scenario is that, for a deformable m2, it is possible for slip to occur at the upper interface with m1, while, at the same time, the lower surface of m2 does not move.

Let A be the area of m2 in contact with m1 and with the table, and let h be the thickness of m2. The displacement to the right at the bottom of m2 is going to be zero. The displacement to the right at the center plane of m2 is going to be δ₁ and the displacement to the right at the top surface of m2 is going to be δ₂. If we did not allow a displacement at the top surface (i.e., temporary slip with kinetic friction, followed by re-establishment of stick) the frictional force at the top surface would be 25N, which would exceed the critical force for release of static friction. Therefore, the top surface of m2 must slip at the kinetic friction force of 12.3 N (temporarily, until static friction is re-established). Once static friction is re-established, the force at the top of m2 will remain at 12.3 N. That is where we are going to examine what prevails.

Now for the analysis:
Shear strain in the lower half of m2 = δ₁/(h/2) = 2δ₁/h
Shear strain in upper half of m2 = 2(δ₁-δ₂)/h

Shear stress in lower half of m2 = 2Gδ₁/h
Shear stress in upper half of m2 = 2G(δ₁-δ₂)/h

Frictional force on top surface = 12.3 N = 2GA(δ₁-δ₂)/h
Frictional force on bottom surface of m2 = 50.0-12.3= 37.7 N = 2GAδ₁/h

If we solve for δ₁ and δ₂, we obtain:

##δ_1=\frac{37.7h}{2AG}##

##δ_2=\frac{25.4h}{2AG}##

So the bottom of m2 will not be displaced to the right at all, and the center plane of m2 will be displaced to the right about 50% more than the top surface.

In the end, m2 would not be accelerating.

Chet

Wow that's crazy. This problems not suppose ti be that hard

 

[haruspex]

If it's rigid, you can't have one surface slipping while the other is not.

Chet

You don't need to care whether it's rigid, or what the relative elasticities may be. There are three possibilities for slipping: top first, bottom first, simultaneous. All three can be ruled out at 50N.

 

[Chestermiller]

Wow that's crazy. This problems not suppose ti be that hard

Well, we don't know what your professor's analysis looks like. Maybe his analysis is oversimplified.

Chet

You don't need to care whether it's rigid, or what the relative elasticities may be. There are three possibilities for slipping: top first, bottom first, simultaneous. All three can be ruled out at 50N.

If the top slips, but not the bottom, this will affect the tension in the wire (which the problem statement asks us to determine). Also, for a non-rigid (deformable) block, all three can not be ruled out at 50N.

Chet

 

[haruspex]

If the top slips, but not the bottom, this will affect the tension in the wire (which the problem statement asks us to determine). Also, for a non-rigid (deformable) block, all three can not be ruled out at 50N.
Chet

Yes, sorry, I was making the classic blunder of equating maximum static friction with actual static friction. And I agree the tension cannot be determined, I was just arguing that slipping will not occur.
To correct my argument, microscopic slippage may occur at either interface, but not both, so we can say the acceleration is zero. The weakest resistance to movement is when the lower interface slips: 5g*0.35+15g*0.25 > 50.

 

[Chestermiller]

To correct my argument, microscopic slippage may occur at either interface, but not both, so we can say the acceleration is zero. The weakest resistance to movement is when the lower interface slips: 5g*0.35+15g*0.25 > 50.

I respectfully disagree. I maintain that the upper surface would slip first. Suppose we applied a force of only 30 N, rather than 50 N. Because of the symmetry of the system (i.e., the top surface is in contact with m1, which is held in place by an inextensible wire; this makes the contact of m2 with m1 equivalent to the contact of m2 with the table top), the friction force with m1 will be the same as the friction force with the table top, namely 15 N. So, the total force will be supported by equal frictional forces at the two surfaces. Now, it will take only about 17.3 N for slip to start occurring at the upper surface. So, if the force were increased to 35 N and m2 were deformable, that would be enough for slip to start occurring at the interface with m1. (Of course, at the interface with the table, m2 would not be slipping because the critical force of about 51.9 N would not be exceeded there). The slippage at the top will occur with a frictional force determined by kinetic friction (12.3 N), until the deformational stiffness of m2 brings the deformation to a halt. This will result in a tension of about 12.3 N in the wire.

Chet

 

[haruspex]

I respectfully disagree. I maintain that the upper surface would slip first. Suppose we applied a force of only 30 N, rather than 50 N. Because of the symmetry of the system (i.e., the top surface is in contact with m1, which is held in place by an inextensible wire; this makes the contact of m2 with m1 equivalent to the contact of m2 with the table top), the friction force with m1 will be the same as the friction force with the table top, namely 15 N. So, the total force will be supported by equal frictional forces at the two surfaces. Now, it will take only about 17.3 N for slip to start occurring at the upper surface. So, if the force were increased to 35 N and m2 were deformable, that would be enough for slip to start occurring at the interface with m1. (Of course, at the interface with the table, m2 would not be slipping because the critical force of about 51.9 N would not be exceeded there). The slippage at the top will occur with a frictional force determined by kinetic friction (12.3 N), until the deformational stiffness of m2 brings the deformation to a halt. This will result in a tension of about 12.3 N in the wire.

Chet

I think you misunderstood my point. I was looking at whether the mass can accelerate. The lowest total friction before that can happen is when the lower interface slips first (for whatever reason). 5*0.25+15*0.35 > 5*0.35+15*0.25. Yet even then it is greater than 50N. But now I realize there's an even simpler reason... the static friction between the lower mass and the ground is itself adequate. 15g*0.35 > 50. So I'm left puzzled as to what the author intended.

 

[Chestermiller]

We have no dispute as to whether the mass m2 can sustain significant movement; it can't. But what I'm saying is that, if we allow the mass m2 to be slightly deformable, the upper surface of m2 can move forward slightly (during which kinetic friction prevails), and we can thereby establish a value for the tension in the wire.

Chet

 

[haruspex]

We have no dispute as to whether the mass m2 can sustain significant movement; it can't. But what I'm saying is that, if we allow the mass m2 to be slightly deformable, the upper surface of m2 can move forward slightly (during which kinetic friction prevails), and we can thereby establish a value for the tension in the wire.

Chet

Certainly if the upper surface slips a little then the tension during the slip can be determined. Not entirely certain what happens when it stops again - maybe there's no reason for the tension to increase again at that point.
The conditions for the upper surface to slip are somewhat complex, involving not only the deformability of m2 but also of m1 and of the wire, and the relative heights of the blocks and points of attachment.

 

[Chestermiller]

The conditions for the upper surface to slip are somewhat complex, involving not only the deformability of m2 but also of m1 and of the wire, and the relative heights of the blocks and points of attachment.

I agree. I assumed for simplicity that the wire is inextensible and m1 is rigid. If either of these is relaxed, the problem becomes still more difficult.

Chet