How Does Angled Force Affect Maximum Box Weight on a Frictional Surface?

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Homework Help Overview

The discussion revolves around a physics problem involving the effect of an angled force on the maximum weight of a box on a frictional surface. The original poster describes a scenario where a box is pushed down at a 15° angle with a specified force and coefficient of static friction, seeking to determine the heaviest box that can be moved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to analyze the forces acting on the box, considering both horizontal and vertical components of the applied force. They question how the vertical component affects the box's weight and friction. Other participants suggest modifications to the equations to incorporate the vertical force component into the frictional force calculations.

Discussion Status

Participants are actively engaging with the problem, proposing equations and checking each other's reasoning. There is a collaborative effort to refine the approach to include the effects of the angled force, with some guidance provided on how to adjust the equations accordingly. However, there is no explicit consensus on the final formulation yet.

Contextual Notes

Participants are navigating the complexities of the problem, particularly the interaction between the applied force components and the frictional force. The original poster expresses uncertainty about how to factor in the vertical force, indicating a need for clarity on the assumptions made regarding the forces at play.

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Homework Statement



You push down on a box at an angle 15° below the horizontal with a force of 650 N. If the box is on a horizontal surface and the coefficient of static friction is 0.7, what is the heaviest box you will be able to move?

Homework Equations



F=ma

The Attempt at a Solution



So I've solved this problem using Fnet which was 650cos15 for the horizontal force and then subtracted [tex](\mu_s)(9.80)(m)[/tex] The ma part goes to zero since it's not accelerating.

Therefore I get [tex]650cos15 = (\mu)(9.80)(m)[/tex] and then solve for m.

I thought that was it. But now that I thinking about it more, there is a vertical component for the force. I'd imagine that extra force would make the box seem to weigh more, thereby increasing friction, making the box weigh less.

Do I factor that vertical force in? If so, how?
 
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Ah! Maybe like this?

[tex]650cos15 = [((\mu)(9.80)(m))+(650sin15)[/tex]

All I did here was add in the vertical force component to the friction.

Yeah...that isn't right since my box ends up weighing 2kg.
 
Last edited:
Your above equation is correct.Though you will multiply the coefficient of friction to both the weight as well as 650sin15
 
Nice.

So something like this?


[tex]650cos15 = [((\mu)(9.80)(m))+((\mu)(650sin15))][/tex]
 
Yeah,that's correct.
 
Thanks for the help!:approve:
 

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