Derivation of Applied Force in a Grade 12 Force Question

[Troutkibe]

Homework Statement

When an applied force elevated at angle θ is used to drag mass m along a surface with μ as the coefficient of friction, the mass accelerates horizontally with some value, a. Derive the cleanest and most simplified formula for the applied force, F_applied, that will give the required acceleration.

Since the F_applied is elevated at angle θ one could break it down into components F_aCosθ
[Horizontal]and F_aSinθ[Vertical]. Additionally, F_N will be mg-F_aSinθ because F_aSinθ is alleviating some of the force that F_N normally has to counteract F_g, meaning that F_f will be μ(mg-F_aSinθ).
I'm just so sure my derivation is correct but when I used sample numbers it didn't work at all, I'd really appreciate any advice or help, thanks!

Homework Equations

a=F_net/m
F_g=mg
F_f=μF_n

The Attempt at a Solution

a=F_net/m
a=F_aCosθ- μ(mg-F_aSinθ/m
a=F_aCosθ-μmg-μF_aSinθ/m
am+μmg=F_a(cosθ-μsinθ)
am+μmg/cosθ-μsinθ=F_a

 

[Big Red Car]

Hi troutkibe, you're reasoning seems sound. Your equations seem sound too (although the bracketing is a little funny in part "

The Attempt at a Solution

"[/B], but I'm guessing that's a typo?). Maybe give us the numbers you need to plug in and we can try plugging them in too (might just be the way you've inputted those values into the calculator you're using.)

 

[TomHart]

I withdraw my comment. See SammyS' post below.

 

[SammyS]

Homework Statement

When an applied force elevated at angle θ is used to drag mass m along a surface with μ as the coefficient of friction, the mass accelerates horizontally with some value, a. Derive the cleanest and most simplified formula for the applied force, F_applied, that will give the required acceleration.

Since the F_applied is elevated at angle θ one could break it down into components F_aCosθ
[Horizontal]and F_aSinθ[Vertical]. Additionally, F_N will be mg-F_aSinθ because F_aSinθ is alleviating some of the force that F_N normally has to counteract F_g, meaning that F_f will be μ(mg-F_aSinθ).
I'm just so sure my derivation is correct but when I used sample numbers it didn't work at all, I'd really appreciate any advice or help, thanks!

Homework Equations

a=F_net/m
F_g=mg
F_f=μF_n

The Attempt at a Solution

a=F_net/m
a=(F_aCosθ- μ(mg-F_aSinθ))/m
a=(F_aCosθ-μmg-μF_aSinθ)/m
am+μmg=F_a(cosθ-μsinθ)
am+μmg/cosθ-μsinθ=F_a

Hello Troutkibe. Welcome to PF !

You are somewhat cavalier regarding parentheses. They're important.

I've corrected a few in the above quote of you OP.

I'll leave fixing parentheses up to you in that last line. If that is what you are literally using for plugging in your numbers, then you will definitely get erroneous results.

 

[Troutkibe]

Yeah, that's my bad on not using brackets here, but I was using them in the actual calculation. I think I figured it out though, when I multiplied everything by -μ I neglected its negative value and that really threw everything off.

 

[SammyS]

Yeah, that's my bad on not using brackets here, but I was using them in the actual calculation. I think I figured it out though, when I multiplied everything by -μ I neglected its negative value and that really threw everything off.

(am+μmg)/(cosθ-μsinθ) = F_a

 

[Chestermiller]

(am+μmg)/(cosθ-μsinθ) = F_a

Shouldn't that be a + sign in the denominator?

 

[Troutkibe]

Shouldn't that be a + sign in the denominator?

Yeah that's what I realize I messed up on. Dumb mistake too :/

 

[Chestermiller]

Yeah that's what I realize I messed up on. Dumb mistake too :/

The hard part was expressing the physics of the problem in the form of equations. You did masterfully at that. Then you messed up on the simple part: solving the equations algebraically.

 

[gneill]

Bonus question: What would be the minimum force you can apply to achieve a given acceleration? In other words, if the force is applied at an optimum angle to minimize the force required, what is the magnitude of that force?

 

[UchihaClan13]

Set dF/dθ=0
get θ and substitute back to get F