Moving a box with a force that is less than gravity

AI Thread Summary
The discussion centers on the feasibility of moving a box with a force less than its weight, focusing on the effects of friction and the angle of applied force. The equation for the force required to move the box, F_o(θ), incorporates the coefficient of kinetic friction and the angle of application. It is suggested that if the angle θ is chosen appropriately, specifically less than arctan(μ_k), the applied force can indeed be less than the gravitational force acting on the box. However, this condition may result in the box not moving if the frictional force exceeds the applied force. Ultimately, the relationship between friction and the weight of the box is crucial in determining whether it can be moved under these constraints.
kotchenski
Messages
16
Reaction score
0

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
 
Physics news on Phys.org
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) ...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top