Force required to move a crate

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In summary, the conversation discusses finding the minimum force needed to move a trunk with constant speed across a level floor using a massless rope and given a kinetic-friction coefficient. Different methods are suggested, including using calculus and rewriting the formula in a different form. Ultimately, it is determined that the minimum force can be found by setting the denominator of the formula equal to one.
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Homework Statement


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).
RW-05-64.jpg

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


Homework Equations


F=ma
Fk=µk(Fn)


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?
 
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  • #2
Hi bakin! :smile:
bakin said:
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. :wink:
 
  • #3
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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  • #4
oops!

tiny-tim said:
bakin said:
F= (-µkmg)/(µksinΘ-cosΘ)

Yes that's the correct formula …

Sorry, I misread it :redface:

should be a plus on the bottom …

can you see why?
 
  • #5
Aw boo, looks like I need to review the distributive property :blushing:

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..
 
  • #6
bakin said:
But, making the denominator equal to one …

uhhh? why? :confused:

Differentiate the bottom, and put it equal to zero. :smile:
 
  • #7
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.
 
  • #8
bakin said:
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. :wink:
 
  • #9
Would plugging in tan-1(µk) in for theta be it, or is there another step to that method?

tiny-tim said:
without differentiating …

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

Where did you get φ? Not exactly sure how you went from asinΘ+cosΘ to the other form :uhh:
 
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1. What is "force required to move a crate"?

"Force required to move a crate" refers to the amount of force or energy needed to move a crate from one location to another. This force is necessary to overcome the inertia and friction of the crate, as well as the force of gravity acting on the crate.

2. How is the force required to move a crate calculated?

The force required to move a crate can be calculated using the formula: Force = mass x acceleration. This means that the force required to move a crate is directly proportional to the mass of the crate and the acceleration required to move it.

3. What factors affect the force required to move a crate?

The force required to move a crate can be affected by various factors such as the mass of the crate, the surface it is being moved on, the force of gravity, and the presence of any obstacles or friction. Other factors that can affect the force required include the shape and size of the crate, as well as the speed and direction in which it is being moved.

4. How can the force required to move a crate be reduced?

The force required to move a crate can be reduced by decreasing the mass of the crate, using lubrication to reduce friction, and choosing a smooth and flat surface to move the crate on. Additionally, using pulleys or other mechanical aids can also reduce the force required to move a crate.

5. What are some real-world applications of understanding the force required to move a crate?

Understanding the force required to move a crate is important in various industries and fields, such as transportation, construction, and logistics. It can help in designing efficient and safe methods for moving heavy objects, as well as in calculating the power and energy needed for different tasks. Additionally, understanding this concept can also aid in preventing injuries and accidents related to moving heavy objects.

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