Tension of a Rope Connected to Two Identical Blocks on a Rough Table

[Benzoate]

Homework Statement

two identical blaocks each of Mass M are conneted by a light inextensible string and can move o the surface of a rough horizontal table. The blocks are being towed at constant speed in a straight line by a rope attached to one of them. The tension in the tow rope is T(0) or T naught. What is the tension of the connecting string

Homework Equations

The Attempt at a Solution

Two equations:

-F+T(!)=m(1)*a, T(1) being the tension of the connecting string, F being the frictional force

-F-T(1)+T(0)=m(2)


-F+T(1)=m(1)*a==> -F=m(1)*a-T(1) therefore -F-T(1)+T(0)=m(2)*a==> (m(1)*a-T(1)-T(0)=m(2)*a==> 2*T(1)-T(0)=a*(m(2)-m(1))==>

==> T(1)= a*(m(2)-m(1))-T(0)/(2).

Are my calculations correct , because I wasn't sure whether I should include the gravitational force in my equation. The reason I left out the gravitational force is because the problems says the two blocks connected by a string are being towed in the horizontal direction and the gravitational force is in the vertical direction

 

[Redbelly98]

Two equations:

-F+T(!)=m(1)*a, T(1) being the tension of the connecting string, F being the frictional force

-F-T(1)+T(0)=m(2)

Hi,

Your calculations are correct, but I have a couple of observations:

It may be easier for you just to type m1, T1, etc. instead of typing in all those parentheses.
Eg:
-F + T1 = m1*a

And, more importantly,

From the problem statement information, what do we know about the acceleration "a"? This can greatly simplify the solution.

(And you're right, we do not need to use gravity here.)

 

[Benzoate]

I would like to add the last segment of the problem. Now the tension in the tow rope has increased by 4*T(0). What is the instantaneous acceleration of the block and what is the instantaneous tension of connected string/

My_solution
-F+T1=m*a.

-F-T1-4T0=ma

T1=2*T0

ma-4T0+4T0=ma. I can figure out what's wrong with this calculation.

 

[HallsofIvy]

If the two boxes are moving at constant speed, and the only force is the friction from the "rough" table. Are you given a coefficient of friction or other way of calculating that force? Without friction, the two boxes would continue at the constant speed with no need for a rope. Without friction, there is no tension in the rope.

 

[Benzoate]

If the two boxes are moving at constant speed, and the only force is the friction from the "rough" table. Are you given a coefficient of friction or other way of calculating that force? Without friction, the two boxes would continue at the constant speed with no need for a rope. Without friction, there is no tension in the rope.

I am not given a coefficient of friction. I didn't really need to know the coefficient of friction for the first part of the problem to calculated the Tension of the string that was connecting the two masses together . The equation for friction was written in terms of mass, acceleration and T1.

the answer is 3*T0/2*m. I can really closed to this answer but the acceleration disappeared from the equation since the masses were the same. I will show you what I mean. From the first problem , I know that :

T1=-(a*(M-M)-T0)/(2)= T0/2

Therefore, when I multiply T0 by 4 and do lot more algebra, I eventually get: 4T0=T0-M*a+M*a==> -M*a+M*a=3*T0. if the forces of both masses have the same sign, I have my acceleration.

 

[HallsofIvy]

Again, if there were no friction, there would be no tension in the line. The tension in the line depends upon the friction force. If you are not given that, I don't see how you can calculate the tension.

 

[Benzoate]

Again, if there were no friction, there would be no tension in the line. The tension in the line depends upon the friction force. If you are not given that, I don't see how you can calculate the tension.

I am given a frictional force. The problem stated at the beginning that the surface of the table is rough, therefore, there is a frictional component involved in my equation. As you can see from above, I've written my friction in terms of f= T1-M*a.

 

[Redbelly98]

Well, you are given that there is a frictional force, but you are not given it's value. Nor are you given the coefficient of friction.

It seems this information is necessary in order to calculate the acceleration.

 

[Benzoate]

Well, you are given that there is a frictional force, but you are not given it's value. Nor are you given the coefficient of friction.

It seems this information is necessary in order to calculate the acceleration.

I am given the friction. but I can write the frictional force in terms of T1 and the net force. For the first part of the problem, I did not need to know the coefficient of friction in order to calculate the tension of the connected rope. I show you what I mean:


Force equations:
(1) -F+T1=M*a

(2) -F-T1+T0=M*a.

F is the frictional force.
Now the only thing I have to do is move T1 to the other side of the equation and now I have my equation for the force of friction: -F=T1+M*a and plugging -F inton equation (2) I have:

(M*a-T1)-T1+T0=M*a

I did not need to know the coefficient of friction in order to find the T1, and I am sure I don't need to know the coefficient of friction to find the acceleration of the two blocks now that the Tension has increased to 4*T0.

I am not sure if I should used the same two equations or do I need to formed an entire new set of equations that would assist me in finding the acceleration, because I have dealt with this force equation equations repeatedly and the acceleration keeps disappearing.

 

[Redbelly98]

I am given the friction...

It would be beneficial to include all the given information in your posts when you ask for help.

 

[Benzoate]

It would be beneficial to include all the given information in your posts when you ask for help.

I did include all the information. I just posted the last part of the problem in another post. What information do you think is still missing?

 

[Redbelly98]

It sounded like you said you were given the value of the frictional force (Post #9). But never mind that, I have finally figured out what's going on.

The friction force F can be solved for from the first part of the problem. You have already shown that

-F + T1 = M*a
and
T1 = T0/2

Try to express F in terms of T0. Use this information:

The blocks are being towed at constant speed in a straight line

What does that statement tell you about "a"?

edit:
I'm logging off for now, by the way.

 

[Benzoate]

It sounded like you said you were given the value of the frictional force (Post #9). But never mind that, I have finally figured out what's going on.

The friction force F can be solved for from the first part of the problem. You have already shown that

-F + T1 = M*a
and
T1 = T0/2

Try to express F in terms of T0. Use this information:
What does that statement tell you about "a"?

edit:
I'm logging off for now, by the way.

That the magnitude of a is zero. that can't be right. only a is zero when the velocity vector is constant. Would I have to write the equation for friction only in terms of T0? Should I just ignore the net force of the rope? -F=M*a-T1= M*a -T0/2. In the first part of the problem, T1=2*T0. so wouldn't -F=M*a-2*T0?

 

[Redbelly98]

That the magnitude of a is zero. that can't be right. only a is zero when the velocity vector is constant.

... The blocks are being towed at constant speed in a straight line by a rope attached to one of them.

"Constant speed" → The magnitude of the velocity is constant.
"In a straight line" → The direction of the velocity is constant.

Since the magnitude and direction of the velocity are both constant, the velocity vector is constant. Acceleration is zero.

-F=M*a-T1= M*a -T0/2

Yes, and since a=0, this simplifies to F=___?

T1=2*T0

No, T1 = T0/2

 

[Benzoate]

"Constant speed" → The magnitude of the velocity is constant.
"In a straight line" → The direction of the velocity is constant.

Since the magnitude and direction of the velocity are both constant, the velocity vector is constant. Acceleration is zero.Yes, and since a=0, this simplifies to F=___?No, T1 = T0/2

my teacher says on his class webpage that a=3*T0/2*M. Why did you not acknowledge that the tension of the rope being towed is now 4 times what it was. wouldn't the tension of the rope have some influence on the acceleration of the rope?

 

[Redbelly98]

Sorry about the confusion. I was still discussing the 1st part of the problem, when a=0 (constant speed and direction). This was necessary in order to determine what F is. As you showed in Post #5, T1=T0/2 in the first part of the problem.

Yes, you're right about the 2nd part: the tension in the lead rope becomes 4*T0. Tension in the 2nd rope is half of that, or 2*T0. All that is left to do is account for F in terms of T0.

 

[Benzoate]

Sorry about the confusion. I was still discussing the 1st part of the problem, when a=0 (constant speed and direction). This was necessary in order to determine what F is. As you showed in Post #5, T1=T0/2 in the first part of the problem.

Yes, you're right about the 2nd part: the tension in the lead rope becomes 4*T0. Tension in the 2nd rope is half of that, or 2*T0. All that is left to do is account for F in terms of T0.

For the last of part a. T1=a*(-M+M)-T0/(2)

T0/2= a*(-M+M)-T0/(2)

even if I multilply T1 by 8 to obtained 4*T0 , my masses cancel and therefore have no idea what my acceleration could be. 4*T0=2*a*(-M+M)-2*T0 ==> a=6*T0=2*a*(-M+M)==> a=3*T0/(-M+M).

I told you . I am one sign error away from the correct answer. If my masses have been both positive , I would have the correct acceleration.

If both masses are moving in the same direction, wouldn't they have the same velocity vector and therefore have the same sign?

 

[Redbelly98]

For the last of part a. T1=a*(-M+M)-T0/(2)

T0/2= a*(-M+M)-T0/(2)

There is a sign error here, it should be +T0/2 on the right-hand-side.



I suggest we review a few facts we have established:

Part (a), a=0 case:

T1 = T0/2

Part (b), a≠0

Net force on 1st mass:

4T0 - T1' - F = M*a

Net force on 2nd mass:

T1' - F = M*a

Where T1' is the 2nd rope's tension in part (b) -- different than T1, the tension in part (a)


Okay, so now we have 2 equations with 3 unknowns (T1', F, and a). We need a 3rd equation ... perhaps the equation for F that I suggested in Post #14.

 

[Benzoate]

There is a sign error here, it should be +T0/2 on the right-hand-side.



I suggest we review a few facts we have established:

Part (a), a=0 case:

T1 = T0/2

Part (b), a≠0

Net force on 1st mass:

4T0 - T1' - F = M*a

Net force on 2nd mass:

T1' - F = M*a

Where T1' is the 2nd rope's tension in part (b) -- different than T1, the tension in part (a)


Okay, so now we have 2 equations with 3 unknowns (T1', F, and a). We need a 3rd equation ... perhaps the equation for F that I suggested in Post #14.

Yes! Thank you for all of your help. I found that if I would have realized all along that my acceleration was zero and that the tension of the connected rope changed when the rope of Tow rope change, I would have had my acceleration a long time ago. I just want to know that when there is a sudden increase in the tension force, their is a sudden increase in acceleration since acceleration is directly proportional to Net force of a particle , according to Newton's second law.

 

[Redbelly98]

Your welcome! Good luck.