Derivation of Applied Force in a Grade 12 Force Question

In summary, the conversation discussed deriving a formula for the applied force, Fapplied, needed to achieve a given acceleration when an object with mass m is dragged along a surface with coefficient of friction μ, at an angle θ. The formula was derived by breaking down Fapplied into horizontal and vertical components and using equations for net force and friction. A mistake in parentheses led to incorrect results, but correcting it allowed for the correct formula to be obtained. Further discussion was had about finding the minimum force needed for a given acceleration by setting the derivative of Fapplied with respect to θ equal to 0 and solving for θ, then substituting back into the formula to find the minimum force.
  • #1
Troutkibe
3
0

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, Fapplied, that will give the required acceleration.

Since the Fapplied is elevated at angle θ one could break it down into components FaCosθ
[Horizontal]and FaSinθ[Vertical]. Additionally, FN will be mg-FaSinθ because FaSinθ is alleviating some of the force that FN normally has to counteract Fg, meaning that Ff will be μ(mg-FaSinθ).
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=Fnet/m
Fg=mg
Ff=μFn



The Attempt at a Solution


a=Fnet/m
a=FaCosθ- μ(mg-FaSinθ/m
a=FaCosθ-μmg-μFaSinθ/m
am+μmg=Fa(cosθ-μsinθ)
am+μmg/cosθ-μsinθ=Fa
 
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  • #2
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.)
 
  • #3
I withdraw my comment. See SammyS' post below.
 
  • #4
Troutkibe said:

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, Fapplied, that will give the required acceleration.

Since the Fapplied is elevated at angle θ one could break it down into components FaCosθ
[Horizontal]and FaSinθ[Vertical]. Additionally, FN will be mg-FaSinθ because FaSinθ is alleviating some of the force that FN normally has to counteract Fg, meaning that Ff will be μ(mg-FaSinθ).
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=Fnet/m
Fg=mg
Ff=μFn

The Attempt at a Solution


a=Fnet/m
a=(FaCosθ- μ(mg-FaSinθ))/m
a=(FaCosθ-μmg-μFaSinθ)/m
am+μmg=Fa(cosθ-μsinθ)
am+μmg/cosθ-μsinθ=Fa
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.
 
  • #5
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.
 
  • #6
Troutkibe said:
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) = Fa
 
  • #7
SammyS said:
(am+μmg)/(cosθ-μsinθ) = Fa
Shouldn't that be a + sign in the denominator?
 
  • #8
Chestermiller said:
Shouldn't that be a + sign in the denominator?
Yeah that's what I realize I messed up on. Dumb mistake too :/
 
  • #9
Troutkibe said:
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.
 
  • #10
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? :smile:
 
  • #11
Set dF/dθ=0
get θ and substitute back to get F
 

1. What is the formula for calculating applied force in a grade 12 force question?

The formula for calculating applied force is F = m x a, where F represents applied force, m represents mass, and a represents acceleration.

2. How do you determine the direction of applied force in a grade 12 force question?

The direction of applied force is determined by the direction in which the force is acting. This can be determined by the angle at which the force is applied or by the direction of the object's movement.

3. Can you provide an example of a grade 12 force question that involves the derivation of applied force?

Sure, an example of a grade 12 force question involving the derivation of applied force could be: A 10 kg object is being pushed with a force of 20 N. What is the acceleration of the object?

4. How does the mass of an object affect the applied force in a grade 12 force question?

The mass of an object has a direct impact on the applied force required to move it. The heavier the object, the more force is needed to accelerate it at a certain rate.

5. Are there any other factors that can affect the calculation of applied force in a grade 12 force question?

Yes, other factors that can affect the calculation of applied force include friction, air resistance, and the angle at which the force is applied. These factors can alter the amount and direction of the applied force needed to move an object.

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