Equipment sliding on an incline with 2000 lb friction force

[Lazorbeam]

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.

 

[mfb]

I moved the thread to our homework section as this is very homework-like.

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).

 

[SammyS]

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.