# What is the tension in each part of the rope

1. Jul 19, 2009

### donotremember

To keep Robin from being captured, Batman tosses him out of a third-story window, knowing that a 17.0m rope hangs slack between hooks of equal height on adjacent buildings 13.0m apart. Robin grabs the rope and hangs on at a point 5.0m from one end. Assuming that Robin's mass is 45.0kg and the rope withstands the initial impulse, what is the tension in each part of the rope when equilibrium is established?

I believe that this question can be simplified to the following:

(Think of this as a triangle)

Find T2 and T1:

.............................13m
\---------------------------------------------/
....\................................................./
..5m..\................................./
.............\.............../.........12m
..................\-/
....................|
....................|
....................|
....................V
....................45kg or (45kg)(9.8m/s^2) = 441N

Where the 5m rope is T2 and the 12m rope is T1

From the law of cosines we have the following angles

67.4 (or 112.6)..........................................22.6
\---------------------------------------------/
....\.............................................../
..........\.........................../
...............\............/
.....112.6........\-/...22.6
---------------------------------------------------------

Step 1 find a relation between T2 and T1:

The system is in equilibrium so the x components cancel

T2cos112.6 + T1cos22.6 = 0

T2(-0.384) = -T1(0.923)

T2 = 2.4T1

Step 2 use the relation found in step one to solve for T1

The system is in equilibrium so the y components added equal the weight:

T2sin112.6 + T1sin22.6 = 441N

2.4T1sin112.6 + T1sin22.6 = 441N

2.215T1 + 0.384T1 = 441N

T1 = 169.6N

and given step 1 T2 = 407N

The answers given in my book are

T1 = 1.5 X 10^2 N

T2 = 3.6 X 10^2 N

Note that they satisfy the relation T2 = 2.4T1

Also note that gravity is calculated as being 9.8m/s^2 in my book (which I have used)

Last edited: Jul 19, 2009
2. Jul 19, 2009

### flatmaster

you lost percision in your answer when you pluged in numbers too early. You want to have a final algebric expression, then plug. It's easier, more precise, and saves paper. Develop this skill now and it will make physics much easier.

3. Jul 19, 2009

### donotremember

show me how you would have done it

4. Jul 20, 2009

### songoku

flatmaster gave you advice not to plug the number in the beginning of the calculation.
you change the value of cos 112.6 and cos 22.6 at the second line of your calculation.
It's much better if you just leave them and plug the number at the end of calculation.

5. Jul 20, 2009

### maverick_starstrider

You are right. The answers in the back of the book are wrong. Although working exclusively algebraically is incredibly important and it is what you will be doing for the rest of your career. For example, in this case we find that the final expression for T2 is:

$T_2=\frac{mg}{\frac{sin(\theta_2)}{tan(\theta_1)}+cos(\theta_2)}$ where theta_1 and theta_2 are the inside of the triangle:

\theta_2|theta_1/
...\.........|........./
.....\.......|......./
..T2.\....|.../.....T1
.........\.|./
..........\|/

...And that's enough ASCII art for one lifetime.

6. Jul 20, 2009

### donotremember

Thanks for your help. I usually put numbers in early to eliminate variables and reduce the size of the expression. When I try to express the entire problem in terms of variables, sometimes but not always, I end up with a huge expression that takes several lines and I end up copying wrong. I guess its an art to finding the most elegant solution.

7. Jul 20, 2009

### queenofbabes

It's usually recommended to use your variables throughout, only putting numbers in at the final step. It's neater (really!), and it's much easier to see what's going on in your work.