Solving Friction Problem: Accelerating Wounded Soldier Out of Line of Fire

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Homework Help Overview

The problem involves a wounded soldier being pulled away from a line of fire by a comrade, with specific forces and weights provided. The subject area includes concepts of friction, force decomposition, and acceleration in a physics context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the decomposition of the applied force into vertical and horizontal components, and the impact on frictional force. Questions arise regarding the correctness of calculations and the implications of the results obtained.

Discussion Status

Some participants have offered insights on converting units and the implications of the calculated friction force. There is an ongoing exploration of whether the horizontal component of the applied force is sufficient to overcome friction, with no explicit consensus reached.

Contextual Notes

Participants note that the teacher prefers values to remain in the original units provided, which may influence the approach to solving the problem. There is also a mention of a negative acceleration result, raising questions about the feasibility of the scenario.

bob24
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Homework Statement


A wounded soldier caught in Line of fire is pulled away by his comrade with a force of 70lbs at 65 degrees. The wounded solider(with his gear) weighs 180 lbs and the coefficient of friction between him and ground is .55
How quickly will he be accelerated out of harms way?


Homework Equations


u=coefficient of friction
f=uF, w=mg, F-f=ma

The Attempt at a Solution


f=uFn
f=(.55)(180-64.44lbs)
f=64.108lbs

w=mg
180=m(32.2)
5.59kg=mass

I don't know what to do(correctly) after this. Have I done anything wrong in the steps shown above?
 
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There are two components to the force applied.

The vertical component lessens the frictional load.
The horizontal component is what you will have offset by the frictional drag.

Personally I would convert straight away to SI units.

1 lb force = 4.448 N
 
Our teacher wants us to keep the values like I have put there though. What should I do after finding the friction?
Do I use F-f=ma? If I do this, the answer I get is -6.176 ft/s(squared). So that means that he won't be accelerated away right?
 
If the horizontal component of the applied force is less than the maximum needed to overcome friction ... then whoever is doing the pulling should seek shelter.
 
Ok thank you!
 

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