Tension of ropes when laying on a hammock

[Nathan777]

Homework Statement

A 50.0 kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 15° above the horizontal. Find the tension in the ropes.

Homework Equations

T=mg, F=ma, T= (Tcos15°)+(Tsin15°)

The Attempt at a Solution

What I did was I started with X and Y coordinates and I figured the coordinates in the X direction would cancel out to be 0 because in one X direction it is positive and the other is negative so the total tension would be the tension in the y direction multiplied by two. What I did was I got the total tension as ƩF= 2Tsin15° which is nowhere near the correct answer in the answer key. The correct answer is 948N which would be the cos of 15 degrees and not the sin. I don't understand how you know which one to us because if the ropes are 15 degrees above the horizontal I figured the x coordinates would cancel out.

 

[PeterO]

Homework Statement

A 50.0 kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 15° above the horizontal. Find the tension in the ropes.

Homework Equations

T=mg, F=ma, T= (Tcos15°)+(Tsin15°)

The Attempt at a Solution

What I did was I started with X and Y coordinates and I figured the coordinates in the X direction would cancel out to be 0 because in one X direction it is positive and the other is negative so the total tension would be the tension in the y direction multiplied by two. What I did was I got the total tension as ƩF= 2Tsin15° which is nowhere near the correct answer in the answer key. The correct answer is 948N which would be the cos of 15 degrees and not the sin. I don't understand how you know which one to us because if the ropes are 15 degrees above the horizontal I figured the x coordinates would cancel out.

What I did was I got the total tension as ƩF= 2Tsin15°

seems OK.

You do need to transpose to find T of course.

T = ƩF/2sin15° where ƩF is of course the weight of the person: 50*9.8

 

[Nathan777]

What I did was I got the total tension as ƩF= 2Tsin15°

seems OK.

You do need to transpose to find T of course.

T = ƩF/2sin15° where ƩF is of course the weight of the person: 50*9.8

Thank you so much. I don't know where I went wrong with that one.

 

[PeterO]

Thank you so much. I don't know where I went wrong with that one.

Note that the first line of my previous post was a "cut and paste" from your original post, so you started out OK.

 

[Nathan777]

Note that the first line of my previous post was a "cut and paste" from your original post, so you started out OK.

Yes but I figured out the answer with that specific equation. I guess I just didn't think of dividing it out. Error on basic Algebra.