Calculate Tension at Midpoint of Rope w/ Mass

[madah12]

Homework Statement

since no one is bothering to help me in my first thread I will make a new one so can you just give me in general the idea for calculating the tension in the midpoint of a rope with mass?

Homework Equations

The Attempt at a Solution

 

[Dadface]

Madah,you need to show an attempt at a solution.

 

[madah12]

Madah,you need to show an attempt at a solution.

https://www.physicsforums.com/showthread.php?t=451542
there was my attempt

 

[p21bass]

Your other thread was a little confusing as to what, exactly, was what, but it looked like everything up until part D was correct.

For D: simply make a free body diagram of the bottom part of the problem - midpoint of the rope and the block. Yes, assume that the mass of the rope, if cut at the midpoint, is 2kg. Since you already solved for acceleration of the system, the rope and block are accelerating at this same rate. So, you know the masses of the two components, the acceleration of them, and know g=9.81m/s². There are three forces to sum up, with T obviously one of them. Plug your knowns into Newton's second law.

 

[madah12]

I know that one of the three forces is the weight of the rope which is 2g down
but then I also know it has tension up by the higher point of the rope T1 and down by the lower part T2 so I have 2 unknowns and one equation
T1-2g-T2 = 2a
should I take T1 to be the same as the tension between the higher block and rope ?
or should I take T2 to be the tension between the lower block and the rope?

 

[p21bass]

I also know it has tension up by the higher point of the rope T1 and down by the lower part T2 so I have 2 unknowns and one equation
T1-2g-T2 = 2a

No. You apparently omitted the lower block from your calculations. You only have one unknown - the tension at the midpoint. As I said in my last post, draw a FBD from the midpoint of the rope down and use the information you've already calculated.

 

[madah12]

120-T-mg=ma
T=120-(3.53*2)-19.6=93.4N
is that right?
Edit
I did it by thinking that since the motion is up the tension will be down and took the force by the upper block on the rope to be up and the the weight of the rope to be down then the sum to be m*a but I am not sure if I should include the weight of the lower block or not

 

[p21bass]

That's the correct answer, but you did it in (seemingly) a much more difficult way. I'm guessing you cut the rope in the middle and used the entire upper portion? If how you did it works for you, great - it's not wrong. I just think the way I suggested is easier to visualize. And, as a general rule, try to use as few calculated values as possible for each problem. Had you done everything else correctly, but had the wrong answer for C, D would not have come out right for you. As it is, you did C correctly, so it worked out.

 

[madah12]

wait in your way you meant for me to say T-(m half rope + m lower block)g=ma?

T= 7g+7a = 7*9.8+7*3.53=93.31N I actually didnt get it till now lol

 

[p21bass]

That's it. Now that I look back over the problem, I'm betting that that's why, in Part A, they wanted you to draw a FBD of the rope. Using that FBD, but cut at the middle, is really what I was doing.