Max Incline for Equipment: What is the Optimal Slope for Preventing Sliding?

AI Thread Summary
A 10,000lb piece of equipment requires 2,000lbs of horizontal force to move it on a flat surface. The calculated incline at which the equipment would start sliding is 11.54 degrees, derived from the formula α = arcsin(2,000 / 10,000). This slope corresponds to a 20% incline, based on a coefficient of friction of 0.2. However, the force required to move the equipment can vary with the angle of incline, as the weight's parallel component changes. Understanding the distinction between static and kinetic friction is crucial for accurate calculations in real-world applications.
Lazorbeam
Messages
3
Reaction score
0
Hey guys,

Just want to make sure I'm doing everything right here with some simple calcs. My physics days are long gone and it's been a while since I looked at any of this stuff.

A 10,000lb piece of equipment is placed on a flat surface. 2,000lbs horizontal pounds-force is required to move it (brakes on).

Assuming the type of surface is the same, what slope/incline/decline does the equipment start sliding?

The answer I have is quite simply;

α = arcsin (2,000 / 10,000)
α = 11.54 degrees

Or, even simpler, since the coefficient of friction (more like sliding resistance) is 0.2, the slope is 20% which converts to 11.54 degrees.

Please confirm that I have not yet gone senile. Thank you.
 
Last edited:
Physics news on Phys.org
I moved the thread to our homework section as this is very homework-like.

Lazorbeam said:
α = arcsin (2,000 / 10,000)
α = 11.54 degrees
If the 2000 N are independent of the incline, that is right.
Usually, that force depends on the angle (but the coefficient of friction is constant).
 
Lazorbeam said:
Please confirm that I have not yet gone senile. Thank you.
No way for us to confirm that.

I think that if you work this all out for an object on an incline with friction, μ = tan(θ). (It's only a slightly different result for your example.) There is a component of the weight parallel to the incline surface, when you tilt the incline.

The details are a bit different for static vs. kinetic friction, but that's the general result.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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...

Similar threads

Replies
1
Views
3K
Back
Top