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TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$ (usually there is even less volume), then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here Figure 2: Air exiting . (source: ijiset.com)

The dome is flexible, so while pushing it is easy to change its shape (thus force the the air out). Additionally the shape of the lip (at a tangent with the object makes it easy for the air to be driven out). However, after the external force is removed then because of the material elasticity, the suction cup is trying to return to the original shape.

enter image description here

Figure 3: Different stages. (source: Abetterchemtext)

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Figure 4: Low and high pressure regions. (source: wired)

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

Why some surfaces are better

The reason why smooth non porous surfaces tend to behave better with suction cups is that there is not much leakage (from the outside in).

enter image description here

Air coming in the suction chamber would increase the pressure inside the chamber, and therefore reduce the pressure differential $\Delta P$.

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$ (usually there is even less volume), then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape (thus force the the air out). Additionally the shape of the lip (at a tangent with the object makes it easy for the air to be driven out). However, after the external force is removed then because of the material elasticity, the suction cup is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

Why some surfaces are better

The reason why smooth non porous surfaces tend to behave better with suction cups is that there is not much leakage (from the outside in).

enter image description here

Air coming in the suction chamber would increase the pressure inside the chamber, and therefore reduce the pressure differential $\Delta P$.

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$ (usually there is even less volume), then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here Figure 2: Air exiting . (source: ijiset.com)

The dome is flexible, so while pushing it is easy to change its shape (thus force the the air out). Additionally the shape of the lip (at a tangent with the object makes it easy for the air to be driven out). However, after the external force is removed then because of the material elasticity, the suction cup is trying to return to the original shape.

enter image description here

Figure 3: Different stages. (source: Abetterchemtext)

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Figure 4: Low and high pressure regions. (source: wired)

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

Why some surfaces are better

The reason why smooth non porous surfaces tend to behave better with suction cups is that there is not much leakage (from the outside in).

enter image description here

Air coming in the suction chamber would increase the pressure inside the chamber, and therefore reduce the pressure differential $\Delta P$.

added 631 characters in body
Source Link
NMech
NMech
  • 24.3k
  • 3
  • 37
  • 76

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$ (usually there is even less volume), then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape (thus force the the air out). Additionally the shape of the lip (at a tangent with the object makes it easy for the air to be driven out). However, after the external force is removed then because of the material elasticity of, the material itsuction cup is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

Why some surfaces are better

The reason why smooth non porous surfaces tend to behave better with suction cups is that there is not much leakage (from the outside in).

enter image description here

Air coming in the suction chamber would increase the pressure inside the chamber, and therefore reduce the pressure differential $\Delta P$.

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$, then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape. However, because of the elasticity of the material it is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$ (usually there is even less volume), then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape (thus force the the air out). Additionally the shape of the lip (at a tangent with the object makes it easy for the air to be driven out). However, after the external force is removed then because of the material elasticity, the suction cup is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

Why some surfaces are better

The reason why smooth non porous surfaces tend to behave better with suction cups is that there is not much leakage (from the outside in).

enter image description here

Air coming in the suction chamber would increase the pressure inside the chamber, and therefore reduce the pressure differential $\Delta P$.

added 127 characters in body
Source Link
NMech
NMech
  • 24.3k
  • 3
  • 37
  • 76

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$, then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape. However, because of the elasticity of the material it is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$, then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape. However, because of the elasticity of the material it is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.

First of all some nomenclature (this is for the closed type, there is also the open type):

enter image description here

When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume $V_0$ of the suction cup, and then the volume reduces to $V_2 = \frac{V_0}{2}$, then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).

enter image description here

The dome is flexible, so while pushing it is easy to change its shape. However, because of the elasticity of the material it is trying to return to the original shape.

enter image description here

Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.

enter image description here

Then the "suction" force is equal to :

$$F = A\cdot dP $$

where :

  • A is the area of the suction cup (its projection)
  • dP is the difference in pressure ($P_{High}- P_{low}$)

numerical example

Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:

$$F = \frac{\pi d^2}{4} \Delta P $$ $$F \approx 2 \; [N]$$

Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.

added 109 characters in body
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NMech
NMech
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  • 37
  • 76
added 281 characters in body
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NMech
NMech
  • 24.3k
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added 653 characters in body
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NMech
  • 24.3k
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  • 76
Source Link
NMech
NMech
  • 24.3k
  • 3
  • 37
  • 76