Calculating Static Friction To Eject Parts with Air

2″ Flat Super Air Nozzle

In today’s fast-paced world, companies are always looking for ways to do things faster, cheaper, more efficiently without sacrificing safety.

A cereal company had a high-speed system to check the quality of each box of cereal.  When a box did not meet the quality criteria for visual and/or weight, the box would be rejected.  The rejection system that they used was a quick blast of compressed air to remove the box from the conveyor line into a non-conforming bin.  For their first attempt, they tried to use a ¼” copper tube with a solenoid valve attached to a reservoir tank.  When a “bad” box was detected, the solenoid would be triggered, and compressed air would “shoot” the box off the rubber conveyor belt.  The ¼” copper tube can be an inexpensive, common, and easy-to-use device; but they found that the copper tube was very loud (above OSHA limits for noise exposure) and not very effective.  As a note, this company had a safety committee, and they wanted to keep all blowing devices below 80 dBA in this department.  The ¼” copper tube was around 100 dBA.  So, they contacted EXAIR to get our expertise on this type of application.

The cereal company gave me some additional details of the operation.  The box weighed 26 oz. (740 grams) with a dimension of 7.5″  wide by  11″ tall by  2 3/4″ deep (19 cm X 28 cm X 7 cm respectively).  The issue with the ¼” copper tube was the small target area compared to the area of the box.  With any slight variation in the timing sequence, the force would miss the center of mass of the box.  The box could then spin and remain on the conveyor belt.  This would cause stoppage and disruption in the system.  They asked if EXAIR had a better way to remove the defective boxes.

I recommended a model 1122, 2” Flat Super Air Nozzle.  The reason for this style of nozzle was for a variety of reasons.  First, we needed a larger area to “hit” the box.  This Flat Super Air Nozzle has a width of 2” versus the ¼” copper tube.  This increased the target area by 8 times.  So, any small variations in time, we could still hit the center of mass and remove the box.  The second reason was the force rating.  The model 1122 has a force of 22 oz. (624 grams) at 80 PSIG (5.5 bar).  This is slightly under the 26 oz. (740 grams) weight of the cereal box, but we are just sliding the box and not lifting it.  If we can overcome the static friction, then the box can be easily removed.  With Equation 1, we can calculate the required force.

Equation 1:

Fs = ms * W

Fs – Static Force (grams)

m– Static Friction

W – Weight (grams)

From the “Engineering Toolbox”, the static friction between rubber and cardboard is between 0.5 to 0.8.  If I take the worse case condition, I can calculate the static force between the belt and cereal box using Equation 1:

Fs = 0.8 * 740 grams

Fs = 592 grams

The model 1122 has a force of 22 oz. (624 grams), so plenty enough force to move the box from the rubber conveyor belt.

The third reason for this nozzle is the noise level.  The noise level of the model 1122 is 77 dBA, well below the safety requirement for this company.  Noise levels are very important in industries to protect operators from hearing loss, and the model 1122 was able to easily meet that requirement.  I added an additional reason for recommending the 2” Flat Super Air Nozzle; compressed air savings.  Companies sometimes overlook the cost when using compressed air for blow-off devices.  In this comparison, the ¼” copper tube will use 33 SCFM (934 SLPM) at 80 PSIG (5.5 bar) while the model 1122 will only use 21.8 SCFM (622 SLPM).  This is a 33% reduction in compressed air; saving them money.

At the intro, I mentioned that companies are looking to do things faster, cheaper, more effective without sacrificing safety.  For this company, we were able to increase production rates by removing every cereal box from the conveyor belt.  We also saved them money by reducing the compressed air requirement as well as keeping it safe by reducing noise.

If you have an application that needs products to be moved by air, you can contact an Application Engineer at EXAIR to help you with a solution.

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb

Friction Measurement

I had a customer wanting to reject a container off a conveyor belt.  The container held fruit, and when an optic detected a reject, they wanted to operate a solenoid to have a nozzle blow the container into the reject bin.  They had a range of containers that went from 6 oz. (170 grams) to 5 lbs (2,270 grams).  He wanted me to suggest one nozzle for all sizes, as they would automatically regulate the pressure for the full range of container sizes.  In looking at the largest size, this container will need the most force to remove.  The two factors that affects the force in this application is weight and friction.  When it comes to friction, it is generally an unknown for customers.  Here are a couple of things to help in determining the friction in your application.

Strawberry Delight
Strawberry Delight

Friction is a dimensionless number that represents the resistance created between two surfaces.  We have two types; static friction, ms, and kinetic friction, mk.  Static friction is the maximum amount of resistance before the object begins to slide.  Kinetic friction is the amount of resistance that is created when the object is sliding.  So, Static friction is always greater than kinetic friction, ms > mk.  For this application, we will have the air nozzle shoot horizontally to hit the target.  This is the most common and efficient way.

Let’s take a look our customer’s application.  We have a system to reject a non-conforming part with air.  The conveyor is a urethane belt.  The container is plastic.  We need to determine the correct nozzle to reject the 5 lb (2,270 gram) container.

Being that the conveyor belt is only 12” (30.5 cm) wide, we can determine that if we get the part moving, it will continue off the belt and into the reject bin.  The equation for the maximum amount of force required to move the container is Fs = ms * W(Equation 1).

Fs – Static Force – lbs (grams)

m– Static Friction

W – Weight lbs (grams)

One way to determine the amount of force is to use a spring scale.  The spring scale should have a maximum indicator to help tell you the maximum amount of force.  You will have to attach the scale to the container on the conveyor belt. Static friction is the resistance between two surfaces; so, you will have to use the same conditions as required for the operation.  Keep the scale parallel to the conveyor.  While slowly pulling on the scale, watch the dial.  Once the part begins to move, record the weight.  For the exercise above, it showed 1.82 lbs (826 grams) of force to move the 5 lb (2,270 gram) object.

Another way would be to determine the static friction, ms.  Static friction can be found by the angle at which an object starts to move.  By placing the container on a section of supported urethane conveyor belt and lifting one end of the conveyor belt until the object starts to slide, you can measure the angle or the height of the lift.  As an example, we take 3 foot (0.9 meter) of supported urethane conveyor belt and we lifted one end to a height of 1 foot (0.3 meters) before the 5 lb (2,270 gram) container moved.  To determine static friction, it is the tangent of the angle that you lifted, ms = tan(B) (Equation 2 below).  In this example, B = 20o.  Therefore Equation 2 gives us, ms = tan(20o) = 0.364.  If we plug this into Equation 1, we get the following:

Imperial Units                                                    SI Units

Fs = ms * W                                                         Fs = ms * W

= 0.364 * 5 lbs                                                    = 0.364 * 2,270 grams

= 1.82 lbs of force                                               = 826 grams of force
Now that we have the static force, we want to be slightly higher than that.  In looking at the force requirements that are in the EXAIR catalog, it shows that a model 1104 nozzle has a 1.9 lb (850 grams) of force.  This is at a 12” (30.5 cm) distance with a pressure of 80 psig (5.5 bar).  This nozzle will be able to slide the largest containers into a reject bin. With pressure manipulation, the customer can also use this same nozzle for the smaller containers.  If you have any applications that need products to be moved, you can always contact the application engineers at EXAIR to help you with a solution.

Variety of Nozzles
Variety of Nozzles

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb

 

Image courtesy of Chobist, Creative Commons License