Webpage title: Solving for Tension in a Rope Problem

[jarmen]

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 565 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tensions in the rope (a) to the left and (b) to the right of the mountain climber.

Im sorry but i really don't know where to start.
(m)(g)=565

angle on 1st is 65
angle on 2nd is 80

 

[Sdawg1969]

If she weighs 565N that is how much gravity is pulling down on her, what is pulling her back up to keep her from falling?

 

[jarmen]

Normal Force, but isn't it the same as the 565 N.
How is the brought to tension

 

[Sdawg1969]

the normal force is right, which would act straight up, but is there anything holding her straight up?

 

[jarmen]

the tension on the rope is the only thing i can think of.

 

[Sdawg1969]

right, that is the only thing holding here up. now how are you going to turn a vertical force into one that is almost perpendicular?

 

[jarmen]

I am really not sure, this is where i am lost

 

[Sdawg1969]

hint: the answer is in the angles

 

[Sdawg1969]

also, I wish your picture would work so I could see the problem, that would help

 

[jarmen]

https://www.physicsforums.com/attachment.php?attachmentid=12516&d=1202263891

does this work?

 

[jarmen]

The only thing I can think of is taking the sin of the angles to get the vertical force, but how do i relate this.

 

[Sdawg1969]

no, it does not. but depending on where the angles are, you are going to use sin or cos to figure out your 2 tensions

 

[jarmen]

\ (65)... - (80)
.\..... -
...\ ... -
...\ -
...O

best picture i can make sorry.

person is hanging in between with a 65 degree angle on left and 80 degrees on right.

 

[jarmen]

hold on that didnt post correctly

ok the ...'s are empty space, needed it to make space.

the \\\\\'s are the rope being pulled 65 degrees
the ---'s are rope being pulled 80 degrees

sorry, best i can come up with

 

[Sdawg1969]

does the problem look like this?

A --------- c ----------------- B

---------- D

AB would be the rope, D would be the climber, 80* would be the angle between Dc and DB, and 65* would be the angle between Dc and DA?

 

[jarmen]

close, the 80* is the angle on the outside of DB, 65* is outside DA

A--------------c--------------------B
65*\
...\
--------D

 

[Sdawg1969]

so are the angles

-----------------
\ 65*...|...80*/

 

[Sdawg1969]

got it, ok
---------------
|65\.../80|

 

[jarmen]

Yes.., the outside angles

 

[Sdawg1969]

you don't need to figure out the vertical component of force, you should know that. What is the vertical component of force in Newtons?

 

[jarmen]

565 Newtons

 

[Sdawg1969]

right, so you have an angle, and side (Newtons). the rest should be simple

 

[jarmen]

so for the left it would be...

sin(65) (565)

right would be
sin(80) (565)

correct?

 

[Sdawg1969]

not quite- cos$$\Theta$$=adjacent/hyp, 565N is the adjacent, so...

 

[jarmen]

i thought 565 was the opposite?

 

[Sdawg1969]

SohCahToa

Sin=Opp/Hyp
Cos=Adj/Hyp
Tan=Opp/Adj

sohcahtoa

 

[jarmen]

so, cos(65)=adj/hyp

then hyp=565/cos(65) ?

 

[jarmen]

no i know that but since the 565 is force downward and the 65* is at one end, isn't the 565 the opposite side?

 

[Sdawg1969]

check out my pic

 

[jarmen]

i cant... But i just realized what u are saying.. Ok i get it

 

[jarmen]

so the triangles are on the outside of the ropes not inside... ok i understand

 

[Sdawg1969]

565 is the force upward! it is the Normal force (Fn). the sumation of all forces on her is zero (she is not falling ie. accelerating down, or up) so Fn=Fg and as you said Fg=565N=Fn

 

[jarmen]

right. so, cos=adj/hyp

so hyp=565/cos(65) right?

 

[Sdawg1969]

so let T1 = hyp1

then cos(80)=565N/T1

let T2 = hyp2

then cos(65)=565N/T2

 

[Sdawg1969]

right
I wish the pictures would work, as that would have made this so much easier, but hopefully that helped you