Finding Force for Backpacker in Bear Sling Equilibrium

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Homework Help Overview

The problem involves calculating the force exerted by a backpacker to hold a 16-kg backpack suspended between two trees that are 7.6 meters apart, with the rope sagging at its midpoint by specified amounts. The subject area includes concepts from mechanics, specifically tension in ropes and equilibrium conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating the angle of the rope and the components of force involved. There are attempts to apply trigonometric relationships to determine the tension in the rope. Some participants question the validity of the calculations and the interpretation of the problem statement regarding the force exerted by the backpacker.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and interpretations. There is a mix of agreement and disagreement regarding the approach to calculating the forces involved, particularly concerning the distribution of weight and the tension in the rope.

Contextual Notes

Participants note that the weight of the backpack is distributed equally to the rope sections on either side, and there is emphasis on drawing free body diagrams to clarify the forces at play. The problem constraints and the specific sag measurements are also under consideration.

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Homework Statement


Two trees are 7.6m apart. Calculate the magnitude of force a backpacker must exert to hold a 16-kg backpack so that the rope sags at its midpoint by (a) 1.5m, (b) 0.15m.

http://2.bp.blogspot.com/_qDvB7jRAneY/SrLIlGBVq1I/AAAAAAAAAVA/J8UdxspkfSk/s1600-h/bear+sling.bmp"


Homework Equations


[tex]\Sigma[/tex]Fy = 0


The Attempt at a Solution


I tried to figure the angle measure and get a component of the force but I think that's totally wrong. For part a I ended up getting 57.6N
 
Last edited by a moderator:
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Lightsout565 said:

Homework Statement


Two trees are 7.6m apart. Calculate the magnitude of force a backpacker must exert to hold a 16-kg backpack so that the rope sags at its midpoint by (a) 1.5m, (b) 0.15m.

http://2.bp.blogspot.com/_qDvB7jRAneY/SrLIlGBVq1I/AAAAAAAAAVA/J8UdxspkfSk/s1600-h/bear+sling.bmp"


Homework Equations


[tex]\Sigma[/tex]Fy = 0


The Attempt at a Solution


I tried to figure the angle measure and get a component of the force but I think that's totally wrong. For part a I ended up getting 57.6N
That's not right, please show your work in determining the angle and tension in the wire. Note that the weight of the 16 kg backpack is vertically distributed equally to the rope section on each side of it. The rope tension can then be calculated using simple trig.
 
Last edited by a moderator:
(a)
(3.8m)2 + (1.5m)2 = x2
x = 4.085m

Fg = Ftsin[tex]\theta[/tex]
(16kg)(9.8 m/s2) = Ft(1.5/4.085)
F = 427N

Correct?
 
Lightsout565 said:
(a)
(3.8m)2 + (1.5m)2 = x2
x = 4.085ml
yes, good
Fg = Ftsin[tex]\theta[/tex]
(16kg)(9.8 m/s2) = Ft(1.5/4.085)
F = 427N

Correct?
No, the backpack weight splits vertically 1/2Fg to each side. Draw a free body diagram of the rope joint to prove this using Newton 1. Your result is off by a factor of 2.
 
But the questions asks "Calculate the magnitude of force a backpacker must exert", which is the sum of the two tensions forces. So wouldn't my answer be correct?
 
Lightsout565 said:
But the questions asks "Calculate the magnitude of force a backpacker must exert", which is the sum of the two tensions forces. So wouldn't my answer be correct?
Why would you sum them? The rope tension around an ideal pulley is the same on both sides of the pulley. Would you say that the rope tension at the right hand tree is 427 N or 213.5 N? Draw a free body diagram of the rope at the tree connection.
 

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