# Force required to move a crate

#### bakin

1. The problem statement, all variables and given/known data
You drag a trunk of mass m across a level floor using a massless rope that makes an angle with the horizontal (figure below). Given a kinetic-friction coefficient (mu).

Find the minimum force needed to move the trunk with constant speed.

2. Relevant equations
F=ma
Fk=µk(Fn)

3. The attempt at a solution
What's really getting me is that the site says the answer doesn't depend on theta.

I found the normal force being Fn=mg-FsinΘ

the total friction force being µk(mg-FsinΘ)

For the force equation,
F=ma
F=0
FcosΘ=µk(mg-FsinΘ)

Solving for F, I came up with:

F= (-µkmg)/(µksinΘ-cosΘ)

Again, says no theta in the answer though. Any pointers?

#### tiny-tim

Homework Helper
Hi bakin!
F= (-µkmg)/(µksinΘ-cosΘ)

Again, says no theta in the answer though. Any pointers?
Yes that's the correct formula …

now find the value of Θ for which it's a minimum.

#### bakin

Well, to get the smallest force value, you want the largest denominator, correct? That would be when cosΘ=0, and sinΘ=1, being pi/2. That would make the answer just -mg. Doesn't make sense to me :(

edit: actually, no. Because if sinΘ=1, and cosΘ=0, the answer would be less than one on the denominator, because it would be µk times sinΘ, giving an answer of 0.XX . I guess then it would be when Θ=0. then, the denominator would be equal to -1, and the answer would be just µkmg?

edit2: I tried putting in µkmg in as an answer before, though, and it wasn't correct. Could I have entered it in wrong or something?

edit3omg: or maybe the angle is pi/4, giving you less pull, but also making the normal force less. not 100% sure.

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#### tiny-tim

Homework Helper
oops!

F= (-µkmg)/(µksinΘ-cosΘ)
Yes that's the correct formula …

should be a plus on the bottom …

can you see why?

#### bakin

Aw boo, looks like I need to review the distributive property

Ok, so redoing my formula I have the same thing, except all the negatives are now positive.

But, making the denominator equal to one, I'm still left with µkmg, which it said was wrong..

#### tiny-tim

Homework Helper
But, making the denominator equal to one …
uhhh? why?

Differentiate the bottom, and put it equal to zero.

#### bakin

Doing that gives me tan-1(µk)=Θ ????

But, how would you find this out without differentiating? I'm working on this with a friend of mine, and I don't think her class uses any calculus.

#### tiny-tim

Homework Helper
But, how would you find this out without differentiating?
without differentiating …

you can always rewrite asinΘ + cosΘ in the form b(sinφsinΘ + cosφcosΘ), = b cos(Θ-φ), which is easy to minimise.

#### bakin

Would plugging in tan-1(µk) in for theta be it, or is there another step to that method?

without differentiating …

you can always rewrite asinΘ + cosΘ in the form b(sinφsinΘ + cosφcosΘ), = b cos(Θ-φ), which is easy to minimise.
Where did you get φ? Not exactly sure how you went from asinΘ+cosΘ to the other form :uhh:

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