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

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A wounded soldier weighing 180 lbs is being pulled by a comrade with a force of 70 lbs at an angle of 65 degrees. The coefficient of friction between the soldier and the ground is 0.55, which results in a frictional force of approximately 64.108 lbs. The vertical component of the pulling force reduces the effective weight, impacting the frictional load. The horizontal component of the applied force is insufficient to overcome friction, leading to a calculated negative acceleration of -6.176 ft/s², indicating the soldier will not be accelerated out of harm's way. The discussion emphasizes the need to analyze both force components and friction to determine the outcome.
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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=coefficent 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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