# Tension in a rope with mass

1. Jul 30, 2014

### Govind_Balaji

1. The problem statement, all variables and given/known data
Two blocks shown in figure are connected by heavy uniform rope of mass 4kg. An upward force of 200N is applied as shown.
(a)What is the acceleration of the system?
(b)What is the tension at the top of heavy rope?
(c)What is the tension at the mid-point of the rope?

2. Relevant equations
I Took g=10m/s^2.

3. The attempt at a solution
See the attachment. I drew FBD.
The first question is easy.
Net gravity force=(7+4+5)g=160N downwards
Net Force in the system=200-160=40N upwards.
Acceleration of the system=40N/16kg=2.5m/s/s upwards.

For third question, I should break the rope into 2 identical rope and find the tension. Right?

First I tried second question. I get stuck in the second question due to the nonm-zero net force(40N).

In the diagram, F1-70=Net force=40N. Then F1=110N upwards. Right?
F2-F1+40=40. Right?
Then F2 should be 110N downwards right?

PS: I drew the FBD of three things. Did I miss any force in the FBD?

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2. Jul 30, 2014

### Nathanael

40N is the net force on the entire system. That is not the same as the net force on parts of the system.

3. Jul 30, 2014

### Govind_Balaji

Thank you.

That equation was the summation of froces on the 4kg rope.

\begin{align*} \text{Net force on 7kg block }&=7kg*2.5 m/s^2=17.5N=F_1-70N\\\Rightarrow F_1&=87.5 N \text{ upwards}\\\text{Net force on 4kg rope}&=4kg*2.5m/s^2\\10N \text{ upwards}&=F_1-F_2-40N \text{ upwards}\\10N \text{ upwards}&=-F_2+47.5N \text{ upwards}\\F_2&=37.5N \text{ downwards}\end{align*}

???????

4. Jul 30, 2014

### Nathanael

Ok a few things here, first of all (I should've said this in my first reply) your free body diagrams are wrong.

"F1" should be on the "FBD of 7kg block" and "F2" should be on the "FDB of 5kg block"

Secondly, you've ignored newton's 3rd law (for every force there is an equal and opposite force)
Because of newton's 3rd law, each force you've drawn should have another force to complete the pair.
However, you don't need to draw the "equal and opposite force" of gravity, because it acts on the Earth (which you don't care about).
You also don't need to draw the "equal and opposite force" of the 200N-force because it acts on whatever is pulling the system up (which you don't care about).
So that leaves you with just "F1" and "F2" that need an "equal and opposite force" drawn for them.

The rope pulls down on the 5kg block with a force "F2" ... therefore the block pulls UP on the rope with a force "F2"

Similarly, the rope pulls up on the 7kg block with a force "F1" ... therefore the block pulls DOWN on the rope with a force "F1"

So the FBD of the rope needs to be changed a bit.

5. Jul 30, 2014

### Govind_Balaji

I corrected it. I have a small confusion over drawing Force vectors on a FBD. What does the arrowhead and the base of arrow indicate. Do they indicate the object being pushed/pulled and the object which pushes/pulls respectively?

Now I get the equation right and got the answer. Now I am going to workout the third question after having lunch.

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6. Jul 30, 2014

### Nathanael

Good, good.

I will be asleep by then (it's 1 a.m. where I am)

There is no indication of whether the object is pushed or pulled (as there is not really any real physical difference)

The arrowhead indicates the direction of the force. The base of the vector typically indicates where the force is applied, although it doesn't have to be drawn where the force is applied. Often you will see the base of the vector drawn on the center of mass, even when it is applied elsewhere.

As far as I know, there are not any strict rules for where to draw the base of your vector. You just draw it in whatever way that shows the information that needs to be shown.

(If there are strict rules for it, I certainly don't follow them :tongue:)

7. Jul 30, 2014

### Govind_Balaji

I worked out my third question too. I got 112.5 N

8. Jul 30, 2014