How Is Force Minimized When Dragging an Object on a Plane?

[ace123]

[SOLVED] minimum and maximum values

An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle $$\vartheta$$ with the plane, then the magnitude of the force is

F= Z(W)/Z sin $$\vartheta$$ + cos $$\vartheta$$

Where Z is a positive constant called the coeffecient of friction where 0< Z<$$\pi$$/2.

Show that F is minimized when tan $$\vartheta$$ =Z

The theorem's I am allowed to use:

The extreme value theorem
Fermat's theorem


I don't really understand what the problem is asking me. What do they mean by F is minimized? Maybe if I knew that I could do it.

 

[chaoseverlasting]

For F to be minimized, z sinv +cos v must be maximized. You can do that using extreme value theorem...

 

[HallsofIvy]

"minimized" means it takes on it's minimum value. That is it takes the smallest value for any $\vartheta$.
The standard way of finding maximum and minimum values is to take the derivative of the function and set the derivative equal to 0.

What is Z(W)? Is it possible you mean F= Z(W)/(Z sin $\vartheta$+ cos$\vartheta$)?

 

[ace123]

Yes, that is what I meant. Should of made it clearer. I was thinking about doing it like that but I was worried that if i used 0 it would give me value undefined. But i guess i will try. Thank you.

Edit: I also wasn't sure if it would be absolute minimum..

 

[ace123]

What about my W won't my answer contain it?

 

[ace123]

Well after i did the derivative of F and substituted for $$\vartheta$$ = 0. I got just negative W. Would that be correct? Thanks in advance

 

[ace123]

So does that mean I'm right and can mark this as solved?

 

[HallsofIvy]

You still haven't said what you mean by Z(W). Is that just Z times W?

Well after i did the derivative of F and substituted for = 0. I got just negative W. Would that be correct?

??Why would you set $$\vartheta$$= 0? A function has a critical point (and so possibly a max or min) when its derivative is 0. Set your derivative equal to 0 and solve for $$\vartheta$$.

 

[ace123]

but isn't a critical point either a mximum or minimum? I thought I'm looking for just the minimum?

 

[HallsofIvy]

Okay, how do you find a "minimum" without first finding the critical points?

 

[ace123]

Yea, I just hoped their would be a quicker way that would give just the min or max. Also the derivative of F comes out really big do you have a suggestion on how to make it smaller?

edit: nvm combining my numerator cancels most terms

 

[ace123]

Nvm. I think i got it. My answer was that $$\vartheta$$ = 90. But that's only one critical point. So would that be the answer? Thanks for any input