Calculating the Angle of Static Friction for a Crane Lifting a Box

AI Thread Summary
The discussion revolves around calculating the angle at which a 10kg box begins to slide when lifted by a crane, given a coefficient of static friction of 0.4. Participants clarify that the net force is zero in static conditions, allowing for the rearrangement of equations to find the angle using the inverse tangent of the coefficient of static friction. It is noted that the mass of the box does not affect the sliding angle, as gravitational forces cancel out in the calculations. The conversation also highlights confusion regarding the problem's wording and the relevance of the box's contents. Ultimately, the conclusion is that both boxes, regardless of mass, will slide at the same angle due to the constant coefficient of static friction.
Nicolaus
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Homework Statement


A small crane is lifting a 10kg box of nails. If the coefficient of static friction is .4, calculate the angle at which the box begins to slide.


Homework Equations


Fg x component: mg sin
Ff = ukmgcos


The Attempt at a Solution


Fnet = Fgx - Ff
Would the net force be zero since there's no acceleration in its static state?
Do I then rearrange the equation to solve for the angle?
 
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Hi Nicolaus! :wink:
Nicolaus said:
Would the net force be zero since there's no acceleration in its static state?
Do I then rearrange the equation to solve for the angle?

(I don't understand the actual question, but …)

That's correct: for static friction, the acceleration is zero, and so the net force is zero.

And yes, rearrange and solve for the angle :smile:
 
Ok, would it then make sense that the result ends up being just the inverse of tan multiplied by 0.4 (which is the coefficient of the static friction): angle = tan(inverse) x .4
 
erm :redface:

i've already said i don't understand the question :confused:
 
Is it the way it was worded or..? Imagine the box was on a flat board and you slowly start to tilt the board.

EDIT: I think I did this wrong. I'm supposed to calculate the Force of static friction on the flat surface to find the minimum force required for it to move then place that value into the equation (ignoring the cos angle).
Can someone correct me on this.
 
Last edited:
Nicolaus said:
A small crane is lifting a 10kg box of nails. If the coefficient of static friction is .4, calculate the angle at which the box begins to slide.
Nicolaus said:
Is it the way it was worded or..? Imagine the box was on a flat board and you slowly start to tilt the board.

ah, in that case, the crane isn't lifting the box, it's lifting the board :wink:

and yes, tan-1µk would be correct :smile:
EDIT: I think I did this wrong. I'm supposed to calculate the Force of static friction on the flat surface to find the minimum force required for it to move then place that value into the equation (ignoring the cos angle).

oooh, lost me again :cry:
 
Yes, forgot to state that the box was on a plank. Ignore my EDIT. So, if there were 2 boxes (each of different mass) with that same coefficient of static friction, would the angles at which each box starts to slide be the same?
Thanks :)
 
Nicolaus said:
So, if there were 2 boxes (each of different mass) with that same coefficient of static friction, would the angles at which each box starts to slide be the same?

yes, mass and shape would make no difference :smile:

(btw, what have the nails to do with it? :confused:

is there a second part in which you're asked whether the box tips over?)
 
No, nothing. It's just a problem using a construction site and 2 boxes of nails as the setting and objects, respectively. It says that there are 2 boxes (each of different mass) on a plank being raised by a crane and the static friction is .4, and asks us to find the angle at which each box starts to slide. We concluded that both would, logically, start to slide at the same angle since the the mass and gravitational constant cancel out in both cases when solving for the angle.
:)
 
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