Moving a box with a force that is less than gravity

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SUMMARY

The discussion centers on the mechanics of moving a box with a force less than the gravitational force acting on it. The key equation derived is F_o(θ)=\frac{\mu_k mg}{cos(θ)-μ_k sin(θ)}, which relates the applied force to the angle of application and friction. The critical angle θ is determined as θ=arctan(μ_k), indicating the minimum force required to initiate movement. The relationship between the applied force, gravitational force, and friction is essential to understanding the conditions under which the box can be moved.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with friction coefficients, specifically kinetic friction (μ_k)
  • Basic knowledge of trigonometric functions and their applications in physics
  • Ability to interpret force diagrams and free-body diagrams
NEXT STEPS
  • Explore the concept of static vs. kinetic friction and their implications on motion
  • Learn about free-body diagrams and how to apply them to analyze forces acting on objects
  • Investigate the role of angles in force application and their effect on motion
  • Study the relationship between friction, normal force, and gravitational force in various contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of force and motion concepts.

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Homework Statement


The object is a box with a given mass m. Our person has the choice between pushing the box with a horizontal force, or pulling the box with a wire with an angle of θ=30o. The magnitude of Fo is the force vector he affects the box with in both cases.

Is it possible (If you can freely choose a value for θ) to keep the box moving without using a force fo that is greater than the gravity on the box


Homework Equations


A drawing of the situation:
http://myupload.dk/handleupload/64d68oBb8KHy8

I believe I need an equation that describes the force Fo of θ, which I've found is given as:
F_o(θ)=\frac{\mu_k mg}{cos(θ)-μ_k sin(θ)}

The Attempt at a Solution



I could choose to solve F'o(θ)=0 for theta which gives θ=arctan(μk)

This would describe the minimum force required but I don't know how to relate that to Fg
 
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What is the relationship between friction and the weight of the box?

But what would the turning point tell you?
Don't you want to know where F0 < Fg
 
Simon Bridge said:
What is the relationship between friction and the weight of the box?

But what would the turning point tell you?
Don't you want to know where F0 < Fg

So if I use that Fnet(x)=Fo*cos(θ)-μn

And that Fnet(x)<0, then the value of θ<arctan(μ), and therefor I'm applying a force Fo that is less than gravity? Wouldn't that technically mean the box is not moving and is being held back by the friction?
 
Wouldn't that technically mean the box is not moving and is being held back by the friction?
Why would that be? Gravity is mg... and force less than mg would be less than gravity. To overcome friction it just has to overcome μmg.cos(θ) ...
 

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