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Kinetic friction, dragging a box

  • Thread starter dudeman
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  • #1
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A box with mass M is dragged across a level floor having a coefficient of kinetic friction (greek mu)k by a rope that is pulled upward at an angle Theta above the horizontal with a force of magnitude F.

What is the expression of the mag. of force needed to move the box with a constant speed?

I dont' really know what to do on this. What I've tried so far is spliting the angle's force into acos(theta) and asin(theta), then somehow applying friction. Does anyone know where I can read up on friction, maybe I just dont' understand the concept fully.
 

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  • #3
Andrew Mason
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dudeman said:
A box with mass M is dragged across a level floor having a coefficient of kinetic friction (greek mu)k by a rope that is pulled upward at an angle Theta above the horizontal with a force of magnitude F.

What is the expression of the mag. of force needed to move the box with a constant speed?

I dont' really know what to do on this. What I've tried so far is spliting the angle's force into acos(theta) and asin(theta), then somehow applying friction. Does anyone know where I can read up on friction, maybe I just dont' understand the concept fully.
Is the box accelerating in the either the vertical or horizontal directions? So what are all the vertical forces and what do they add up to? What are the horizontal forces and what do they add up to? Set out those equations and you will easily see how to solve for the applied force. You have to know that the force of kinetic friction is: [itex]F_f = \mu_k N[/itex] and is in the horizontal direction (N is the normal force of the box on the surface).

AM
 
  • #4
I see you posted a formula Ff=UkN.. what do those variable stand for?
 
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  • #5
Andrew Mason
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rkslperez04 said:
I see you posted a formula Ff=UkN.. what do those variable stand for?
Ff is the force of friction. [itex]\mu_k[/itex] is the coefficient of kinetic friction (ie. [itex]F_f/N = \mu_k[/itex]).

AM
 

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