Finding tension between two objects with different fs

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The discussion focuses on calculating the tension in a strap used by a boy to hold a pack while on an incline. Given the masses of the boy and pack, the coefficients of kinetic friction, and the angle of the incline, the equations of motion are set up using free body diagrams. The user expresses confusion over solving for both acceleration and tension, ultimately arriving at the correct tension value of 15N. The solution involves applying Newton's second law and combining equations for both the boy and the pack. The final calculations confirm that the tension in the strap is indeed 15N.
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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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