Finding Force for Backpacker in Bear Sling Equilibrium

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The discussion focuses on calculating the force a backpacker must exert to hold a 16-kg backpack suspended between two trees 7.6 meters apart, with the rope sagging at different midpoints of 1.5m and 0.15m. Initial calculations suggested a force of 57.6N, but further analysis revealed that the correct tension must account for the vertical distribution of weight, leading to a revised force of 427N. Participants emphasized the importance of using trigonometry and free body diagrams to accurately determine the angle and tension in the rope. There was clarification that the tension in the rope is consistent on both sides, negating the need to sum forces. The conversation highlights the necessity of precise calculations and understanding of physics principles in solving such problems.
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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


\SigmaFy = 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


\SigmaFy = 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.
 
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(a)
(3.8m)2 + (1.5m)2 = x2
x = 4.085m

Fg = Ftsin\theta
(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\theta
(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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