a man weighiing 7-kg lies in a hammock whose ropes make angles of 20 and 25 degrees with the horizontal. What is the tension in each rope?

Answer 1

the horizontal components are equal in magnitude or the guy would accelerate.
A cos 20 = B cos 25
.940 A = .906 B
A = .964 B

A sin 20 + B sin 25 = 7*9.8
.964 B (.342) + B (.423) = 68.6

I think you can solve for B and go back and get A

Answer 2

Rofl, I think you're in my calculus class. I have the same question for homework too.

I'm not sure if you're solving the problem using geometric or catersian vectors.

It's always best to draw out the problem. . .

Assuming you meant 70kg instead of 7kg. . .

I'll do this problem using geometric vectors.

You have a triangle with the following angles: 65, 45 and 70.

70 corresponds with |T1| (Tension 1)
65 corresponds with |T2| (Tension 2)
45 corresponds with 686 N (9.8 x 70)

Use sine law to solve for the tensions.

|T1|/sin70 = 686/sin45
|T1| = 911.6 N

|T2|/sin65 = 686/sin45
|T2| = 879.3 N

Therefore, the tensions are 911.6 N and 879.3 N

Answer 3

thx, i just have problems drawing the diagrams