Calculate Load carrying capacity of vertically mounted suction cup

[Acenish]

TL;DR Summary

Calculate Load carrying capacity of vertically mounted suction cup

I'm looking for means of calculating load carrying capacity of a vertically mounted suction cup.
Suction Cup effective diameter = 90mm
Mounted on a vertical glass surface
Pressure inside the suction cup = 0.75bar
Pressure difference = 0.25bar
Weight is suspended from a rod attached in the centre of the suction cup.
Rod length = 150mm

Need to calculate how much weight can be suspended before the suction cup pulls off the glass surface

Assume, coefficient of friction between suction cup and glass as 0.2

 

[Baluncore]

Welcome to PF.
Is this a homework problem?
Have you tried to solve the problem yourself?

 

[Acenish]

Hi
No its not a homework problem, rather a curiosity in suction cup design and load capacity.
This arose while using suction cup phone mounts or clothes line as well as with respect to nerf arrows.

Below is a diagram of the suction cup with a suspended load.

Applying force equilibrium, we get
Fsuction = N (normal force)
Ffric = W (load)
now, Ffric = µ x N = µ x Fsuction
& Fsuction = ΔP x A
so, Ffric = µ x ΔP x A = W

Assume, µ = 0.1
Patm =1 bar (0.1N/mm²)
Pi = 0.8bar (0.08N/mm²)
A = π x d² /4 = 6361.73mm² (d = 90mm)

thus, W = 0.1 x (0.1 - 0.08) x 6361.73 = 12.72N = 1.2kg (approx)

However, what has me puzzled is that the Load W will also have a moment acting on the suction cup and this moment if considered in the centre of the cup, will be balanced by the Fsuction till it fails under a 'peeling-off' action (in this case the top end will peel as W is acting downwards.

This gives, W x 150mm = Fsuction x 45 (radius of cup)....... considering moment in the centre.
=> W = (ΔP x π x d² /4) x 45 / 150 = 38.17 N = 3.8kg (approx)

Was hoping to get feedback on whether either of the approaches are correct.
Thanks

 

[Baluncore]

Was hoping to get feedback on whether either of the approaches are correct.

The cup could unstick from the glass, or it could slide down the glass.

Area of cup = Pi*(0.045)^2 = 6.36e-3
Differential pressure = 0.25 bar * 100 kPa/bar = 25 kPa
Cup force against wall = area * pressure = 159 N.

At what vertical load will the cup slide down the wall?
Note that it is independent of the rod lever arm length.
Apply friction; 159 N * 0.2 = 31.8 N.
31.8 / 9.8 = 3.24 kg.

If it did not slide, the cup would break off the surface at;
Assume the length of the rod is measured from the glass.
Fvert = 159 N * (45 mm / 150 mm) = 47.7 N.
47.7 / 9.8 = 4.867 kg.

So the cup will slide at 3.24 kg.

 

[Baluncore]

Assume, coefficient of friction between suction cup and glass as 0.2

Assume, µ = 0.1

 

[Acenish]

Thanks for response.
Values are arbitrary and I am interested in the approach.

Glad to know, I wasn't off the mark!