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

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SUMMARY

The optimal slope for preventing sliding of a 10,000lb piece of equipment on a flat surface is calculated to be 11.54 degrees, based on a required horizontal force of 2,000lbs to initiate movement. This calculation utilizes the formula α = arcsin(2,000 / 10,000) and correlates with a coefficient of friction of 0.2, which translates to a 20% slope. The discussion emphasizes the importance of understanding the relationship between incline angle and friction, particularly noting that the force required to move the object can vary with the angle of incline.

PREREQUISITES
  • Understanding of basic physics concepts, particularly forces and friction.
  • Familiarity with trigonometric functions, specifically arcsine.
  • Knowledge of static and kinetic friction principles.
  • Ability to perform calculations involving weight and incline angles.
NEXT STEPS
  • Research the relationship between incline angles and static vs. kinetic friction.
  • Learn about the effects of different coefficients of friction on sliding resistance.
  • Explore advanced calculations for objects on inclined planes in physics.
  • Investigate real-world applications of incline calculations in engineering and equipment placement.
USEFUL FOR

Engineers, physics students, equipment operators, and anyone involved in the design or placement of heavy machinery on inclined surfaces will benefit from this discussion.

Lazorbeam
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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.
 
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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.
 

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