What is the Correct Formula for Static Friction on an Inclined Box?

AI Thread Summary
The discussion centers on determining the correct expression for static friction on an inclined box subjected to an applied force at an angle. The participants explore the relationship between the applied force, normal force, and frictional force, emphasizing that static friction must balance the applied force when the box remains stationary. They clarify that the normal force is affected by the vertical component of the applied force, which complicates the calculation. The confusion arises from differentiating between maximum static friction and the actual frictional force when the box is not moving. Ultimately, understanding how to correctly account for the normal force in relation to the applied force is crucial for solving the problem accurately.
physgirl
Messages
99
Reaction score
0

Homework Statement



On a box (of mass "m"), force "F" is applied at "theta" degree angle above the horizontal. The box has static friction coefficient of "u." When this force is applied, and the box remains stationary, what expression describes the frictional force?

Homework Equations



F=uN??
(N is normal force)

The Attempt at a Solution



I came down to 2 choices:
F=umg
or
F=uFcos(theta)

I initially thought the former, because I keep remember being drilled the equation F=uN, and about how it's all about the normal force when it comes to friction.

But then when I thought about it, forces must balance out... so if there's no net acceleration, applied force (i.e. Fcos(theta)) MUST balance out frictional force, no?

And now I'm just confusing myself.
 
Physics news on Phys.org
Right, yes the forces must balance out.

When you split the force into vertical and horizontal components what do you get?
 
right, so if the box isn't moving (yet), then for vertical, we just have normal force = gravity force (i think?), which is just mg=mg... then for horizontal, we have applied force (Fcos(theta))=frictional opposing force (u*F*cos(theta))...

i think i get the confusion. so the normal force equation is only relevant for *maximum* static frictional force equation, where that's the force at which the box moves b/c it overcomes static friction. yes?
 
yes but you see, your normal reaction is not just N=mg since the vertical component of the force Fy is present and depending how the force is applied, the normal reaction will change.
 
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