Finding Force of an Object on Incline w/ Constant Speed & Coefficient

AI Thread Summary
The discussion revolves around calculating the force required to push a 100-lb box up a 20° incline at a constant speed of 30 ft/s, considering a coefficient of kinetic friction of 1/3. The user expresses confusion about the problem and indicates that the net force is zero, as the box is moving at constant speed. They have attempted to apply the equation F=ma but are unsure about the numerical solution and seek clarification. The user plans to revisit the problem later after focusing on other subjects. The thread highlights the importance of understanding the balance of forces in physics problems involving friction and inclined planes.
wmcbain
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Homework Statement



Hi guys and gals,

Having an issue with this one and was hoping to get some help. I'm going to write out the equation and include an attachment with my free body diagram.

Q:

How long a force parallel to a 20° hill does it take to push a 100-lb box up the hill at a constant speed of 30 ft/s if the coefficient of the kinetic friction between the box and the hill is 1/3.


Homework Equations



F=ma fk=uk(Fn) w=mg m=w/g v=30ft/s

The Attempt at a Solution



F=ma
F=(100lb/32.2fts^2)(0)
=0

I just can't seem to wrap my head around this one. There really isn't a good explanation in the book. I've included my free body diagram. Thanks for the help.
 

Attachments

  • incline_force.jpg
    incline_force.jpg
    28.4 KB · Views: 447
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Okay just did it on a paper and snapped a pic of it.


001.jpg


It appears not clear enough with flash >.<
 
So is there no numerical answer? Or just variables?
 
The idea is there are still forces acting on the objects, just that the net force (sum of all) is zero.
So your upward force is equal to your downward ones.
 
Okay, yeah I figured that. As F=ma equals zero. I appreciate your help. Just wanted to make sure there wasn't something else that I was missing
 
there is numerical answer...
I'm lazy to calculate it. Just told you the concept,ask if I'm not stating clearly
Try and post your calculations :)
 
okay good luck haha
 
I will try later. I got stumped, it's my last question for this week. I moved onto Chemistry. I will go back to it though and post.
 
cool :)
 
  • #10
Sorry about the wait on this one. Been super busy learning new stuff. Below is the attachment to the solution.
 

Attachments

  • Solution.jpg
    Solution.jpg
    52.6 KB · Views: 460
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