What is the Solution to a Classic Tension Problem Involving Ropes and Trees?

[tictactony]

Homework Statement

A camper hangs a 22 kg pack between two trees, using two separate pieces of rope of different lengths, as shown in the figure below. (I don't know how to post the picture but I'm sure you can visualize two ropes and two trees with angle measures 71 left and 28 right)

Homework Equations

Weight=mg

The Attempt at a Solution

I'm pretty sure I know how to do this problem because I've done it before with easier numbers. I'm actually getting quite frustrated and I'm on my last guess wondering what the heck I'm doing wrong.

Using T1 as the tension in the left rope I calculated: T1*sin(71)=T1y
T2 tension in the right: T2*sin(28)=T2y

then I know both these y components have to add up to 215.6 N because (22kg)(9.8m/s^2) yields the gravitational force on the pack.

Next I did the x components and got:
T1*cos(71)=T1x
T2*cos(28)=T2x

Since the pack is stationary I know the x components have to be equal.

I'm left with a system of equations and after eliminating T2 I'm left with

T1=(215.6)/(sin(28)) / ((sin(71))/(sin(28))+((cos(71))/(cos(28)))
T1=(215.6)/(sin(28)) / (sin(71)+tan(28)*cos(71))
T1=766.696 which I KNOW is incorrect because that exceeds the weight of the pack

 

[diazona]

A camper hangs a 22 pack

That's 22 kg you meant, right?

Using T1 as the tension in the left rope I calculated: T1*sin(71)=T1y
T2 tension in the right: T2*sin(28)=T2y

Where did 71 and 28 come from?

then I know both these y components have to add up to 215.6 N because (22kg)(9.8m/s^2) yields the gravitational force on the pack.

Next I did the x components and got:
T1*cos(71)=T1x
T2*cos(28)=T2x

Since the pack is stationary I know the x components have to be equal.

I'm left with a system of equations and after eliminating T2 I'm left with

T1=(215.6)/(sin(28)) / ((sin(71))/(sin(28))+((cos(71))/(cos(28)))

I don't follow how you got that by combining the previous equations.

T1=766.696 which I KNOW is incorrect because that exceeds the weight of the pack

Why can't the tension be greater than the weight of the pack?

 

[fzero]

I'm left with a system of equations and after eliminating T2 I'm left with

T1=(215.6)/(sin(28)) / ((sin(71))/(sin(28))+((cos(71))/(cos(28)))
T1=(215.6)/(sin(28)) / (sin(71)+tan(28)*cos(71))

The sin(28) in the second line shouldn't be there anymore.

 

[tictactony]

Answering your questions diazona,

yes 22 kg my mistake

the 28 and 71 are the angle measures

and lastly, they can it just looks like a ridiculously high amount for a 22kg.

And fzero, when I eliminated the T2 from the system of equations I had to divide all the parts of the 1st equation by sin(28) and all the parts of the second by cos(28)

 

[diazona]

Were 28 and 71 were given in the problem? If so, you should have included them in the problem statement.

Anyway, I misformatted my post so you might have missed the main question: how exactly did you get this equation?

T1=(215.6)/(sin(28)) / ((sin(71))/(sin(28))+((cos(71))/(cos(28)))

 

[tictactony]

T1sin(71) + T2sin(28) = 215.6N
T1cos(71) - T2cos(28) = 0N

Then divided the top equation by sin(28) and the bottom by cos(28) to get

T1[sin(71) / sin(28)] + T2 = 215.6 / sin(28)
T1[cos(71) / cos(28)] - T2 = 0

Then added them together to get

"T1=(215.6)/(sin(28)) / ((sin(71))/(sin(28))+((cos(71))/(cos(28)))"

 

[diazona]

T1[sin(71) / sin(28)] + T2 = 215.6 / sin(28)
T1[cos(71) / cos(28)] - T2 = 0

Then added them together to get

"T1=(215.6)/(sin(28)) / ((sin(71))/(sin(28))+((cos(71))/(cos(28)))"

That's not what you get when you add the two equations together. Go back and do that step again, more carefully this time.

 

[tictactony]

I just did it again and got the same thing, what am I doing wrong? The T2's cancel...

 

[diazona]

Looking at it again, I might have miscounted the parentheses. You have far more than necessary in there so it gets rather confusing. (Last time I double-checked my count but I guess I made the same mistake every time I read it)

Anyway, if the parentheses are in the right places, everything seems to be in order.