Fun physics question

[zinfamous]

Its not homework, I am trying to convince my employer to purchase a different ramp and they are worried it will be too much force with the higher angle.

It's still homework, only you are getting paid to figure it out...

this. 😀

 

[Phoenix86]

Its a first approximation type problem.

So you need to find the force required to move the block up the ramp with only gravity acting as your resistance, no friction. Then find the distance needed to apply that force. Solve for the hypotenuse distance I guess.

The difference in potential energy is going to be mgh no matter what. I guess they want to see if you understand the effect of ramps on the normal force.

Sounds to me like they want to make sure it's not so fucking heavy they can't move it. 😉

 

[DrPizza]

You don't necessarily need the coefficient of friction. Just measure the force to move it across a level floor. The frictional force pushing it up the ramp will be just a tiny bit smaller, since the normal force will be reduced very slightly due to the very small angle. Assuming the force to push it across a level floor isn't too large, the difference is insignificant for what you want to do. Plus, some of the components that make up the total friction would be independent of the angle of the incline (e.g., if the bearings are in a very viscous lubricant, merely spinning the wheels, regardless of normal force, is going to have a frictional force.)

So, merely calculate the weight of the rack times the sine of whatever angle it is, and add that to the frictional force pushing it across a level surface. Actually, you don't even need to calculate the angle, since the sine of the angle is 8/sqrt(122^2+8^2) for one of the ramps, and 12/sqrt(156^2+12^2) for the other ramp.

And, fwiw, I believe the frictional force would drop to no less than the cosine of the angle times the frictional force over a horizontal surface. That is, 156/sqrt(156^2+12^2) times the frictional force on a level surface. Or, 99.7% of the original frictional force. The other works out to 99.8%.

For those thinking coefficient of friction really mattered though, you're probably thinking static coefficient of friction. You're not doing work against the static friction though - that's what's making the wheels turn rather than slide. E.g., you can walk easily across a floor vs. walking across ice because of the static coefficient of friction, not despite it.

 

[DrPizza]

Oh, I never answered the question though -

On the 8" ramp, the additional force to move the rack up the slope is 8/sqrt(122^2+8^2) times the weight
=6.5433% of the weight.

On the 12" ramp, the additional force to move the rack up the slope is 12/sqrt(156^2+12^2) times the weight
=7.6696% of the weight.

*edit: I blame any errors on a lack of paper, pencil, and using the google calculator.

 

[moonbogg]

I used 5deg angles to be conservative. I neglected friction because you are using wheels.

about 300lbs of PUSH or PULL power is what you need. I would use something with about 1.5x to 2x that in order to not stress the system you are using to move the thing. So if you got something that can push or pull 500 to 600lbs, you are good to go with a decent margin.

 

[Phoenix86]

Oh, I never answered the question though -

On the 8" ramp, the additional force to move the rack up the slope is 8/sqrt(122^2+8^2) times the weight
=6.5433% of the weight.

On the 12" ramp, the additional force to move the rack up the slope is 12/sqrt(156^2+12^2) times the weight
=7.6696% of the weight.

*edit: I blame any errors on a lack of paper, pencil, and using the google calculator.

edit: can't you just compare the difference in angles since it's a straight multiplier? So about 17.5% more force.

6.5433%x1.175=7.688

Love all the approaches though. 😛

 

[Phoenix86]

Actually, after considering things.

Get a lift. Even generic freight is not rolled up and down ramps, why the hell would you roll expensive/heavy server racks.

 

[DrPizza]

6.5433%x1.175=7.688

Love all the approaches though. 😛

Good thing it's not at an angle greater than about 57 degrees, otherwise your approach required a force greater than its weight. 😛 Though, without putting much thought into it, I believe your approach works for very small angles because the derivative of sin(x) is very close to 1 for such small angles.

 

[TheSiege]

I am amazed at the spectrum of people just in this thread, from obvious trolls to the incredibly intelligent and helpful. Thanks again, I appreciate the help.