What is the tension on a high wire supporting an acrobat at a 5 degree angle?

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Homework Help Overview

The problem involves a 55kg acrobat standing on a high wire inclined at a 5-degree angle from the horizontal. The focus is on determining the tension in the wire while considering the forces acting on the acrobat.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equilibrium conditions of the acrobat, considering forces such as gravitational force and tension. There are attempts to draw free-body diagrams and identify the components of tension. Questions arise about how to label tension and whether to resolve it into perpendicular and parallel components.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the forces involved. Some have provided guidance on resolving tension into components and the relationship between the forces. There is a recognition that the normal force may not be necessary in this context, and discussions about the application of trigonometric functions are ongoing.

Contextual Notes

Participants note that the teacher has primarily taught forces using free-body diagrams, which influences their approach to the problem. There is also a mention of homework constraints regarding the use of specific methods and representations.

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Homework Statement


A 55kg acrobat is standing in the middle of a high wire that makes 5 degrees from the horizontal. What is the tension on the wire?


Homework Equations


Newton's 2nd law?


The Attempt at a Solution


------------ < horizontal
\ < 5 degree
 
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Remember that the acrobat in question is merely standing in the middle. There is no acceleration in the vertical or horizontal directions...ie equilibrium. Have you tried drawing a free body diagram yet? Identify all the forces and consider how those relate to each other for a condition of equilibrium.
 
@Galileo's Ghost:

I've tried the FBD but I'm not sure where to label the tension of the wire.
Fg = arrow down the acrobat
Fn = normal force arrow up the acrobat
Ftension = ?
 
charmedb4by said:
@Galileo's Ghost:

I've tried the FBD but I'm not sure where to label the tension of the wire.
Fg = arrow down the acrobat
Fn = normal force arrow up the acrobat
Ftension = ?

ignore force normal

draw a picture

bZ6wJ.png
 
OH, our teacher has only taught forces to us using free-body digrams.
If I represent the acrobat as a box then..
F=mg would be pointing down
& Do i divide up the tensions into perpendicular and parallel or no? S: I'm confused about that.
 
charmedb4by said:
OH, our teacher has only taught forces to us using free-body digrams.
If I represent the acrobat as a box then..
F=mg would be pointing down
& Do i divide up the tensions into perpendicular and parallel or no? S: I'm confused about that.

Tension in a wire is the same throughout the wire. The tension is acting along the length of the rope which is described as 5 degrees from the horizontal. Thus, a component of this tension is horizontal and a component is vertical...so yes, resolve the tension vectors into components.
 
Ohh, it's same throughout! I was thinking of your two tension arrows and how they would cancel each other out!

So if I divide up the tension into perpendicular and parallel, does it matter which way the perpendicular one goes? (Left/Right) The parallel arrow will face up, right? S:
How would I write out the formula to figure out the tensions? :/
 
charmedb4by said:
Ohh, it's same throughout! I was thinking of your two tension arrows and how they would cancel each other out!

So if I divide up the tension into perpendicular and parallel, does it matter which way the perpendicular one goes? (Left/Right) The parallel arrow will face up, right? S:
How would I write out the formula to figure out the tensions? :/

The horizontal components will "cancel each other out". The 2 vertical components of tension, however, do not cancel each other as they are both upwards. So, write an expression that sums up all the forces vertically and remember that all together they must result in ZERO acceleration.
 
Would the main equation be:

0 = Fg - Fn - Ften|| - Ften||?
0 = 55kg * 9.8 - Fn - 2(Ften||)?
 
  • #10
There really is no Normal force here. The upwards support is coming from the vertical components of the tension in the wire. If you eliminate the Fn in your equation, you should be good.
 
  • #11
oh so:

0 = Fg - 2(Ften||)
0 = 55kg * 9.8m/s^2 - 2(55kg * sin5)?

I'm not quite sure I understand the sin/cos either. :/ Are all vertical components sin and all horizontal components cos? D: sorry!
 
  • #12
charmedb4by said:
oh so:

0 = Fg - 2(Ften||)
0 = 55kg * 9.8m/s^2 - 2(55kg * sin5)?

I'm not quite sure I understand the sin/cos either. :/ Are all vertical components sin and all horizontal components cos? D: sorry!


55kg = mass NOT tension. Your question asks you to solve for tension...that should remain as your unknown variable in the equation.Regarding the sin/cos issue. The trig function comes about from an understanding of right triangle trigonometry. Whether to use sin or cos really depends upon which angle you know in a situation. The side OPPOSITE a known angle is related to that angle through the sine function. The side ADJACENT to a know angle is related to that angle through the cosine function. In this case, the angle you knew was with respect tot he horizontal. Thus the adjacent side is the horizontal component and the opposite side is the vertical component.
 
  • #13
don't know why you're subtracting anything.
 
  • #14
@Galileo's ghost: Alright, I think I get sin/cos now, thank you so much!
As for the question, do you think you can tell me the answer so I can double-check to make sure I got it right? :/

@Liquidxlax:

I just realized myself that it should be the upward force minus the downward force right?

0 = 2(Ften||) - Fg
0 = 2(Ften||) - 55kg * 9.8m/s^2
0 = 2(Ften||) - 539
2(Ften||) = 539
Ften|| = 269.5
Ften|| ~ 270 <- 2 sig. digits?
 

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