Force to propel a sliding gate up a 2,4 degree incline?

In summary, the Centurion D10 is a gate opener that is rated at 10kgf, but is able to move objects up a slope with a force of 30kgf.
  • #1
James Nilsen-Misra
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TL;DR Summary
What force is needed to propel a sliding (rolling) gate up a 2.456° degree incline?
The Gate has a mass of 380 Kg.
The wheels have a diameter of 120 mm, Steel UGroove wheels, runing on a 12mm diameter round bar track. (To reduce rolling resistance larger wheels have been included.)

(Wondering what strain this would give a Centurion motor, D5 Evo, rated to be able to work a gate of Max 500kg but its pulling force needs to be 17kg start up and not exceeding 10Kg during running.
Figure A
Screen Shot 2019-08-25 at 02.57.42.png


It slopes up from Left to Right (0.205911m Rise /4,8m run=4.28%)]
 
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  • #2
Total Force = <Operating-Force-If-Level> + (sin(2.5°) * <Weight-Of-Gate>)
= <Operating-Force-If-Level> + (0.0436 * 380kg)
= <Operating-Force-If-Level> + 16.56kG

Looks like you have exceeded the 10kg rating of the opener. :H
 
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  • #3
Tom.G said:
Total Force = <Operating-Force-If-Level> + (sin(2.5°) * <Weight-Of-Gate>)
= <Operating-Force-If-Level> + (0.0436 * 380kg)
= <Operating-Force-If-Level> + 16.56kG

Looks like you have exceeded the 10kg rating of the opener. :H
Thanks so much Tom.G,

So then to work out the maximum mass the gate could be, would it be thus:

10kg/0.04285216=233.36046537kg?

Total Force = <Operating-Force-If-Level> + (sin(2.456°) * <Weight-Of-Gate>)
= <Operating-Force-If-Level> + (0.04285216 * 233.36046537kg )
= <Operating-Force-If-Level> + 9.99999999kg
<Operating-Force-If-Level> + 10kg
 
  • #4
Screen Shot 2019-08-26 at 02.14.00.png


Made this comparison of the various models and their power output.
 
  • #6
Not sure what is going on here.

Your first post stated the Centurion D10 was rated:
...not exceeding 10Kg during running.
The calculation in your post #3, correct as far as it goes, ignores the rolling resistance of <Operating-Force-If-Level>. This calculation could be valid if the rolling resistance is small enough to be ignored (unlikely).

Then the table in your post #4 shows the D10 is rated at 30kg force, not 10kg.
 
  • #7
Tom.G said:
Not sure what is going on here.

Your first post stated the Centurion D10 was rated:"...not exceeding 10Kg during running."

The calculation in your post #3, correct as far as it goes, ignores the rolling resistance of <Operating-Force-If-Level>. This calculation could be valid if the rolling resistance is small enough to be ignored (unlikely).

Then the table in your post #4 shows the D10 is rated at 30kg force, not 10kg.

Sorry for the confusion. Well spotted, the "...not exceeding 10Kg during running" came in an email from a supplier, but when comparing the specs of different models in the published manuals, I didn't find any reference to 10kgf.

Yeh, was thinking that the force to overcome resistance would be just a pinch more than the force to counter the mass of the gate+the additional weight due to the incline...

How much additional force to set it in motion, and how much force to keep it in motion do you think is reasonable?

I came across Steven Holzner's article on "Calculating the Force Needed to Move an Object Up a Slope" Physics I For Dummies, 2nd Edition. But I've just scanned it and need a bit more time to see and apply it.

Coming back to the table I'm keen on using the ET Nice Drive 600m, well priced and their service centre is close by, which they recommend every 18 months.

I want the gate design mass to comply with the motor's output, a 41kgf initial push to set in motion, and 30kgf to overcome resistance and keep moving! With a little extra for bad weather conditions. :rolleyes:

Your help is greatly appreciated!
 
  • #8
Just a little bit of gravel or small stones under those wheels could add greatly to the forces needed. Because of unforeseeable complications like that you need a safety factor of x2 or x3.
 
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FAQ: Force to propel a sliding gate up a 2,4 degree incline?

1. What is the formula for calculating the force needed to propel a sliding gate up a 2.4 degree incline?

The formula for calculating the force needed to propel a sliding gate up a 2.4 degree incline is F = mgsinθ, where F is the force in Newtons, m is the mass of the gate in kilograms, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the incline in degrees.

2. How does the mass of the sliding gate affect the force required to propel it up the incline?

The force required to propel a sliding gate up a 2.4 degree incline is directly proportional to the mass of the gate. This means that the heavier the gate, the more force is needed to push it up the incline.

3. Does the angle of the incline affect the force needed to propel the sliding gate?

Yes, the force needed to propel a sliding gate up a 2.4 degree incline increases as the angle of the incline increases. This is because a steeper incline creates a greater component of the force acting against the direction of motion, making it harder to move the gate.

4. How can I decrease the force needed to propel the sliding gate up the 2.4 degree incline?

There are a few ways to decrease the force needed to propel a sliding gate up a 2.4 degree incline. One way is to decrease the mass of the gate by using lighter materials. Another way is to reduce the angle of the incline, making it less steep. Additionally, using lubrication or wheels on the gate can help reduce friction and make it easier to move.

5. Is there a maximum angle of incline that a sliding gate can be propelled up?

Yes, there is a maximum angle of incline that a sliding gate can be propelled up. This is determined by the coefficient of friction between the gate and the surface it is sliding on. If the angle of incline is too steep, the gate may not have enough friction to stay in place and could potentially slide back down. It is important to consider the weight and materials of the gate when determining the maximum angle of incline it can be propelled up.

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