What is the tension in the string?

[middleearthss]

Homework Statement

A string is connected to a box A mass M at its left end and a box B mass 2M at its right end. Boxes A and B start pulling in opposite directions at the same time with forces which are time dependent and described by functions F(t)=Ct for box A and F(t)=Dt for box B. Values of A and B 1 N/s and 3N/s respectably. Given the maximum tension of the string which is 25 N find the value of t at which the string breaks.

Homework Equations

The Attempt at a Solution

Here the boxes are pulling in opposite directions so tension equals sum of forces emitted by two boxes and g in formula can refer to acceleration of chosen box rather than acceleration of gravity. Here gravity is not in use since this is set up on the ground and friction as well as air drag is negligible.

T=∑(i)Fi(t)
T=F1(t) + F2(t)
T=t(C+D)
t=T/(C+D)
t=6.25 seconds
This seems coherent when plugged into the equation and results in 25 N of force after 6.25 seconds and yet masses than make no use of themselves and my answer sure
doesn't equal the one in the book. Are my equations wrong or is my calculation wrong. This seems perfectly reasonable from my point of view, but yet it make be wrong. The problem seems easy so what's wrong? Can you share your thoughts on this?[/B]

 

[Kinta]

At first, I also thought what you did was reasonable. Whenever I have trouble with problems like you're experiencing here, I check to see if there are "extremes" to the situation. For this particular problem, imagine what would happen if the force from box A shrunk to zero ($C \rightarrow 0$). What would be the tension in the string as a function of time? Does your tension equation reflect what you'd expect in this scenario?

 

[middleearthss]

Good point, it seems as if when c equals 0 only box B pulls to the right. So the tension in that case would not be 0 which would be the case if no box is connected to the left but still some value as box A resists movement. What comes to my mind is that the acceleration of the two boxes in the case C equals 0 would be the same but force produces by A would be in opposite direction and half as strong due to M 2M case. So could in that case be T=F2(t) + F2(t)/2=3F2(t)/2 T=3/2(Bt) making t=50/9 seconds. The solution in the book gives higher value of t and in this C=0 case t is already greater that it would be if box A pulls. So what is this?

 

[Kinta]

Have you tried drawing a free-body diagram and writing out Newton's Second Law for the two blocks?

 

[middleearthss]

I don't know how to do that and it all seems confusing, i can't find a role for masses anywhere.

 

[Kinta]

I don't know how to do that and it all seems confusing, i can't find a role for masses anywhere.

I've attached the free-body diagrams for the two masses bellow.

I recommend you take some time to learn the principles how to draw free-body diagrams and apply Newton's 2nd Law using them. (Note that what I wrote for Newton's law here is a little bit more explicit than usual). They're of great help in introductory physics and even some upper level physics courses. Once you understand the process, this method can more than return the effort put into learning the general process.

Can you think of a way to use this equation to get the desired time?

 

[middleearthss]

So T1=m1a1 - F1(t) and T2=m2a2 - F2(t) so total tension is T1 + T2? What would in this case be a? Only thing i have are the forces and masses?

 

[Kinta]

So T1=m1a1 - F1(t) and T2=m2a2 - F2(t) so total tension is T1 + T2? What would in this case be a? Only thing i have are the forces and masses?

Because there are only masses at the ends of the string, the tension is the string is the same all throughout its length. So, the tension in the string is actually the magnitude of either tension force $T_\mbox{string} = |\vec{T}_1| = |\vec{T}_2|$. I'm purposefully trying to be consistent with referring to these tensions as vector quantities. Make sure you're taking direction into consideration when writing them in non-vector form. You'll need to choose a coordinate system (directions associated with positive and negative change in distance). Because all forces in this problem act along a single axis/line, this should be relatively simple. With these directions in mind, what you wrote in the quoted post is incorrect.

 

[middleearthss]

I do understand that directions should be taken into account but i cannot formulate the equation that allows me to solve for t. From the equations above it did occur to me that T1=T2 but how do you place everything into a form of an equation. What is a in this formula. I have F's and masses but i don't know a's and t's. I am stumped.

 

[SammyS]

The magnitudes of the two tensions are equal, but their directions are opposite.

 

[middleearthss]

Ok, what would be you answers, what is the t at which string breaks and what was your calculation.

 

[Kinta]

Ok, what would be you answers, what is the t at which string breaks and what was your calculation.

I get $t = 15$ seconds. Does your book agree with this answer? If so, I have every intention of helping you arrive at this solution, but I'm not going to just give you my work.

 

[SammyS]

I get $t = 15$ seconds. Does your book agree with this answer? If so, I have every intention of helping you arrive at this solution, but I'm not going to just give you my work.

I would have simply said my answer is greater than say, 13 seconds.

 

[middleearthss]

Thats the solution! 15 sec. I respect that you can't give me the work and I am hoping you will be patient as i work my way through. Now, my main problem is setting everything up. In the free body interpretation i can formulate that if i choose right side as increasing of x and call m1 pulling to the left that m1a1=T1-F1(t) and 2m1a2=F2(t)-T2. As T2=T1=T=25 N i get m1a1=25-t and 2m1a2=3t-25. Am i mistaken here somewhere? I have taken directions into account and i have three variables. What now?

 

[SammyS]

Thats the solution! 15 sec. I respect that you can't give me the work and I am hoping you will be patient as i work my way through. Now, my main problem is setting everything up. In the free body interpretation i can formulate that if i choose right side as increasing of x and call m1 pulling to the left that m1a1=T1-F1(t) and 2m1a2=F2(t)-T2. As T2=T1=T=25 N i get m1a1=25-t and 2m1a2=3t-25. Am i mistaken here somewhere?

I have taken directions into account and i have three variables. What now?

How are a₁ and a₂ related?

 

[middleearthss]

From Newtons second law ma=F the a1= (T-F1)/m1 and a2= (F2-T)/2m1. Thats the only connection i see, is there more?

 

[SammyS]

From Newtons second law ma=F the a1= (T-F1)/m1 and a2= (F2-T)/2m1. Thats the only connection i see, is there more?

Is the motion of the two masses independent of each other?

 

[middleearthss]

No, the acceleration of either box depends on the mass of the other. The heavier its partner the lower its acceleration.

 

[SammyS]

No, the acceleration of either box depends on the mass of the other. The heavier its partner the lower its acceleration.

Can either move some additional distance away from the other ?

(Added in Edit:) the word additional.

 

[Kinta]

No, the acceleration of either box depends on the mass of the other. The heavier its partner the lower its acceleration.

The masses in this problem are fixed. Don't worry about considering variable masses (unless you have a solution and want to check to see that it applies to extremes like before).

Can either move some distance away from the other ?

I would say that the answer to Sammy's question above is key in this problem.

 

[middleearthss]

It cannot. Ohh now that means if i take A to be accelerating in the - direction and B to be accelerating in the + direction the accelerations depend on each other and are lessened. acc of the box 1 equals its original acceleration without the influence of box 2 - the acc of the box 2 and so for the box 2 also right? how do i use that in the equations i stated with three variables.

 

[Kinta]

We've determined that the two boxes cannot move with respect to each other, so they cannot be accelerating is different directions. We also know that the forces applied to either end of the string are unequal in magnitude. Can you make anything of this?

 

[SammyS]

Now, my main problem is setting everything up. In the free body interpretation i can formulate that if i choose right side as increasing of x and call m1 pulling to the left that m1a1=T1-F1(t) and 2m1a2=F2(t)-T2. As T2=T1=T=25 N

i get m1a1=25-t and 2m1a2=3t-25.

It's basic algebra from here on.

So, now you know that m₁a₁ can be used in place of m₁a₂ .

 

[middleearthss]

I see it 50 - 2t = 3t - 25
t = 15 and we used 2m1a1 = m2a2 because their accelerations are in basically the same as they cannot move away from each other. So m1a1=m1a2 from above. That was it right?

 

[Kinta]

Yes, that's it. The net acceleration on either block is equivalent to that on the other until the string snaps.

 

[middleearthss]

Thank you both for clear guidance, i am very much grateful.