What is the required tension in a high wire for a 50 kg person to safely cross?

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The required tension in a high wire for a 50 kg person, Arlene, to safely cross with a maximum sag of 10 degrees is calculated to be approximately 1412.34 N. This statics problem involves balancing the forces acting on Arlene, where her weight (mg) is countered by the tension (T) in the wire. The forces create a 'Y' shape, with the tension components acting at 10 degrees above the horizontal. Ignoring the elasticity of the wire simplifies the calculation, as it has minimal impact on the required tension.

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dynamics question? can you help?

Arlene is to walk across a high wire strung horizontally between two buildings 10 m apart. The sag in the rope when she is at the mid-point should not exceed 10 degrees. If her mass is 50 kg what must be the tension in the rope?

Thanks so much!
 
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This is actually a statics problem. You need to consider the forces on Alrene: her weight and the tension of the wire. The forces make a 'Y' type shape, so that mg is pointing down, and T is pointing 10 degreesabove the horizontal to the left and the other T is pointing 10 degrees above the horizontal to the right. Balance the forces in the vertical direction to solve for T; this will be the minimum tension for a 10 degree sag; higher tensions will have less sag.

If you want to make the problem more complicated, you can consider the elasticity of the wire; the tension on each side of Arlene will be slightly higher than T, the amount of which determined by the spring constant, k, of the wire. You can find the additional elongation of these "springs" using the geometry of the situation. My initial reaction is that it's safe to ignore the elasticity of the wire, though, since even if you knew k, it wouldn't effect the answer by much.
 


Originally posted by magystical
Arlene is to walk across a high wire strung horizontally between two buildings 10 m apart. The sag in the rope when she is at the mid-point should not exceed 10 degrees. If her mass is 50 kg what must be the tension in the rope?

Thanks so much!

Draw a FBD first.

The sum of forces in the vertical direction would look like this

[tex]0 = 2Tsin(10) - 490.5[/tex]

T = 1412.3384 N
 

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