# Need help on difficult Newton's First Law Question

1. Oct 17, 2004

### BinaryFinary

*Just noticed that I posted in the wrong thread. Sorry about that is there any way I can move it to the right thread?*

I have a difficult question and any help would be greatly appreciated. Here it is:
To keep Robin from being captured, Batman tosses him out a third storey window, knowing that 17.0 m rope hangs slack between hooks of equal height on adjacent buildings 13.0 m apart. Robin grabs the rope and hangs on a point 5.0 m from one end. Assuming that Robin's mass is 45.0 kg and the rope withstands the initial impulse, what is the tension in each part of the rope when equilibrium is established.

Ok we know fnet is 0. Therefore the sums of the forces vertically and the horizontally must equal 0. In this problem we have mass but we dont have acceleration of the force applied by Robin (variable is a) or the acceleration of the tension (variable is b). We don't have the angle of the tension (variable @) and we dont have the Force applied horizontally by Robin (Fa) or the Tension in the rope. We do have the force of gravity however (mg, 45*9.8 = 441N). I assume that :
Horizontally
fnet = Fa - Tcos@ (Horizontal component of tension)
Tcos@ = Fa
Tcos@ = ma
T = 45a / cos@

Vertically
fnet = Fg - Tsin@ (Vertical component of tension)
Tsin@ = Fg
Tsin@ = mg
T = 441N/sin@

Substitution:
45a/cos@ = 441/sin@
45asin@ = 441cos@
45asin@/cos@ = 441
45atan@ = 441
tan@ = 441/45a
@ = tan-1(441/45a)

That will give you angle if you can get a. I believe there is a way to get a from using the values given in the question (17m, 13m, 5m) or if not then there should be a way to get @ from the values given and then solve for a instead of @. Substitute @ back into one of the T equations to get T. As I said before any help would be greatly appreciated.

Last edited: Oct 17, 2004
2. Oct 17, 2004

### BinaryFinary

Nevermind, I figured out the process to find the answer.