The problem is much simpler than you are doing. Due to symettry, each side is supporting half the weight (1/2 mg= 2.5g)
Now, using your angles
SinTheta=2.5g/Tension
Now substitute for sintheta, 2/sqrt(104) and solve for tension
A horizontal clothesline is tied between 2 poles, 20 meters apart.
When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 2 meters.
What is the magnitude of the tension on the ends of the clothesline?
Alright, so i have this a little bit started but not sure if I'm going in the right direction. Here's what i have:
sin 2/sqrt(104)
cos 10/sqrt(104)
T1 = -|T1| cos 10/sqrt(104)i + |T1|sin 2/sqrt(104)j
T2 = |T2| cos 10/sqrt(104)i + |T2|sin 2/sqrt(104)j
T1 + T2 + F = 0
Force: -5(9.8)
Do i have this set up right? And if i do, why are certain things positive and negative? What's the next step?
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