Two ropes exert forces holding a suspended box having a mass of 100 N at rest. If one rope is at an angle of 30o to the horizontal and exerts a force of 40 N along the rope, what is the tension on the other rope and what is the angle between the rope and the horizontal?
I was wondering if you could check my work.
Solved for x-direction:
Sum F = ma = 0 = 40cos(30degrees) - T cos(theta)
Solved for y-direction:
Sum F = 40sin(30degrees) + T sin(theta) - 100N = 0
I then solved for T and set them equal to each other. For T I got...
T = 34.8107
Then I plugged that into original equation and got theta.
Theta = 5.6595
Are my T and theta answers correct?
2 answers
Incase you were wondering in the problem description, that "30o" was meant to be 30 degrees. I forgot to change that "o" to "degrees".
Your equations of equilibrium in the two components are correct. Working in degrees and Newtons:
40cos(30)-T cos(theta)=0 .....(1)
40sin(30)+T sin(theta)-100 = 0 ....(2)
Transpose the numerical terms to the right, we get
T cos(θ) = 40 cos(30) ...... (3)
T sin(θ) = (100-40 sin(30)) ....(4)
Divide (4) by (3) to eliminate T:
tan(θ) = (100-40 sin(30))/(40 cos(30))
=tan-1(4/√3)
from which we get θ= 66.587°
Substitute in (1) to get
T=40 cos(30)/cos(θ)
=20*sqrt(19)
=87.178 N
40cos(30)-T cos(theta)=0 .....(1)
40sin(30)+T sin(theta)-100 = 0 ....(2)
Transpose the numerical terms to the right, we get
T cos(θ) = 40 cos(30) ...... (3)
T sin(θ) = (100-40 sin(30)) ....(4)
Divide (4) by (3) to eliminate T:
tan(θ) = (100-40 sin(30))/(40 cos(30))
=tan-1(4/√3)
from which we get θ= 66.587°
Substitute in (1) to get
T=40 cos(30)/cos(θ)
=20*sqrt(19)
=87.178 N
1 answer