Calculating Force Applied to Box with Friction Coefficient \mu

AI Thread Summary
To calculate the force applied to a box with mass x being pulled at an angle θ on a surface with friction coefficient μ, the normal force (N) must counterbalance the weight (W) of the box. The discussion clarifies that N is indeed equal to W but acts in the opposite direction, ensuring no net vertical force. Participants confirm that understanding the balance of forces is crucial for solving the problem. The conversation emphasizes the importance of correctly identifying forces acting on the box to avoid confusion. Overall, the key takeaway is the relationship between applied force, friction, and normal force in this scenario.
Angello90
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Hi guys,

I got this thought in my head.

If I have a box of mass x, being pulled towards East and the angle \theta,
then force applied to the box in order to move it on the surface with friction coefficient \mu has to be in such a way:

attachment.php?attachmentid=25494&stc=1&d=1272632393.jpg


My question is... wouldn't N be:

attachment.php?attachmentid=25496&stc=1&d=1272632659.jpg


Or, N would be as usual W but in opposite direction? The drawing shows all the forces on the box:

Cheers and thanks for the thought

attachment.php?attachmentid=25495&stc=1&d=1272632393.jpg
 

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Angello90 said:
My question is... wouldn't N be:

attachment.php?attachmentid=25496&stc=1&d=1272632659.jpg
You got it.

Or, N would be as usual W but in opposite direction?
Nope. You were right the first time. The key is that there must be no net force on the box in the vertical direction.
 
Ok thanks a mil! I was bit confused
 

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