Finding tension between two objects with different fs

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SUMMARY

The discussion focuses on calculating the tension in a strap used by a boy to hold a pack on an incline. Given the boy's mass of 45 kg, the pack's mass of 25 kg, an incline angle of 42°, and coefficients of kinetic friction of 0.35 for the boy and 0.48 for the pack, the tension was determined to be 15 N. The solution involved applying Newton's second law and constructing free body diagrams for both the boy and the pack to derive the equations necessary for solving the problem.

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  • Basic trigonometry for resolving forces on an incline
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Homework Statement


What is the tension in the strap which the boy uses to hold onto the pack?
Mass of Boy=45kg
Mass of Pack=25kg
θ=42°
Coefficient of kinetic friction between boy and incline=0.35
Coefficient of kinetic friction between pack and incline=0.48
Acceleration of boy=4m/s^2
Capture.JPG

Homework Equations


Fnet=F1+F2+F3...+Fn
F=ma

The Attempt at a Solution


Mass of boy=M
Mass of pack=m
Fpb=parallel force of boy
etc...

Fpb=Mgsinθ
Fpp=mgsinθ
fsb=0.35Mcosθ
fsp=0.48mcosθ

a(M+m)=Fpb+Fpp-fsb-fsp-T

I'm really confused as to how to solve this since both acceleration and tension are unknowns. I've tried doing a lot of things to try and get a value for acceleration but none of them have given me the correct answer which is T=15N.
 
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Draw free body diagrams for the boy and the pack separately.
Apply ##\sum \vec F = m\vec a##, and you'll end up with four equations with four unknowns which have to be true simultaneously.
 
Last edited:
Simon Bridge said:
Draw free body diagrams for the boy and the pack separately.
Thanks so much! Can't believe I messed that up haha. If anyone wants the solution

1. Fpb-T-Fsb=Ma
2. Fpp+T-Fsp=ma

Add them together

FpB+Fpp-Fsb-Fsp=Ma+ma
a=(FpB+Fpp-Fsb-Fsp)/(M+m)
After putting values in
a=3.67m/s^2

Rearranging first equation

T=Fpb-fsb-Ma
T=15N
 

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