Highschool help

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A guy wishes to swing across a hole, using a vine rope. In order to reach the other side he must swing so that the rope makes a maximum angle of 40 degres to the vertical. Regarding the guy as an 85 kg point mass, what is the minimum breaking tension of the rope if the rope is not to break?
 

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  • #2
Andrew Mason
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b_phys28 said:
A guy wishes to swing across a hole, using a vine rope. In order to reach the other side he must swing so that the rope makes a maximum angle of 40 degres to the vertical. Regarding the guy as an 85 kg point mass, what is the minimum breaking tension of the rope if the rope is not to break?
When the man is swinging, what are the forces on the rope? Where are they maximum? What are the forces on the rope at that point?

AM
 
  • #3
Crazy, is this a torque or like.. centripital force question?.. or.. haha sorry i'm trying to learn too.
 
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This is a centripetal force question. You can use the following equations:

F = ma
a = v^2/r

KE is at a max at the bottom of the swing, PE is at a max at the starting point. so mgh = (1/2)mv^2.

You can calculate the height, but you need the length of the rope to figure that out.

I can't really think of a way to do this without the length of rope. If you have it, just use trig, get the h, plug it in.
 
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Andrew Mason
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dboy said:
This is a centripetal force question. You can use the following equations:

F = ma
a = v^2/r

KE is at a max at the bottom of the swing, PE is at a max at the starting point. so mgh = (1/2)mv^2.

You can calculate the height, but you need the length of the rope to figure that out.

I can't really think of a way to do this without the length of rope. If you have it, just use trig, get the h, plug it in.
I don't think you need the length of the rope. The PE is converted to KE at the bottom:

[tex]mgRsin(40) = PE = \frac{1}{2}mv^2[/tex]

[tex]mv^2/R = F_c = 2mgsin(40)[/tex]

So the maximum force on the rope (at the bottom) is:

[tex]F_{max} = mg + F_c = mg + 2mgsin(40) = mg(1 + 2sin(40))[/tex]

[tex]F_{max} = 85*9.8(1 + 2(.6428)) = 1903 N[/tex]

AM
 

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