Asked by Michael
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?
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?
Answers
Answered by
Bobpursley
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
Now, using your angles
SinTheta=2.5g/Tension
Now substitute for sintheta, 2/sqrt(104) and solve for tension
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