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I am not asking how a suction cup can hang on a workpiece without falling down. I know explanation that atmosphere pressure is higher than inner pressure inside suction cups, so it hangs there, which is fine.

Let's imagine that there is a workpiece on the ground and I put a suction cup on it, then I'm lifting the suction cup. So my question is why the workpiece goes up with the suction cup? I can't see any reason for it, I'm just pulling the suction cup, not the workpiece.

If the suction cup and the workpiece were attached to each other physically, I would understand it. But in this case, there's no relationship between them, I just put the suction cup on it, that's all.

How does the suction cup transmit force I apply on it to the workpiece and lift it?

Thanks in advance. enter image description here

Jawel7
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3 Answers3

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Once the assembly workpiece and suction cup moves up with constant velocity, there is no difference comparing with both on the ground.

But it is true that initially an acceleration has to be applied to the suction cup to start the movement. We must verify the forces acting on the workpiece: its weight ($W$)downward, the normal force ($N$)from the suction cup, also downward, and the atmospheric pressure ($p$), upward. If there is acceleration upward:$$p - W - N = ma$$ We see that $N$ is the unknown of the equation. While it is greater than zero, there is contact between the objects. If the acceleration is big enough, they will be separated.

Claudio Saspinski
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  • I think for all this system, we should indicate this: there are two important areas for pressure. First one is surface area on workpiece covered by suction cup. Second one is outer surface area which atmospheric pressure will apply on from outside as directed downwards. Atmospheric pressure * workpiece surface area covered by suction cup(upward) > Atmospheric pressure * outer surface area of suction cub(downward). Otherwise, it doesn't function. What do you think? – Jawel7 Feb 04 '23 at 17:13
  • @Jawel7 depending on the form of the workpiece, there are regions with downward force, and others with upward force due to the atmospheric pressure. But, by hypothesis, due to the region of very low pressure on the top, the resultant force from the atmospheric pressure is upward. – Claudio Saspinski Feb 04 '23 at 18:19
  • Dear Claudio, let's analyze the workpiece separately in a free-body diagram. As it stays on ground, there's pressure difference force upwards, its weight and that's it. However, it stays on ground, right? It means its weight > pressure difference force, which is why it is not flying itself. Now, I'm pulling the cup up, and the workpiece goes up too. What has changed for the workpiece? I didn't pull it up, and if you analyze it in a free-body diagram, where is my pulling force on workpiece? So, I think there's another force on workpiece downward, which is another pressure difference force. – Jawel7 Feb 05 '23 at 09:44
  • This is due to pressure difference between inner and outer surface area of the cup, downward. However, this force should be less than the other pressure difference force exerted on the covered area of the workpiece. The thing is here that we are not pulling the workpiece up actually, but we are decreasing the downward force due to inner and outer surface of the cup. This is my only explanation. – Jawel7 Feb 05 '23 at 09:46
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Once the suction cup is pressed against the workpiece, it "sticks" to the workpiece because the air pressure inside the cup is less than the atmospheric pressure outside the cup. Now the cup and the workpiece are attached to each other. Consider the workpiece and the suction cup as a system. An external force pulling on the handle attached to the suction cup is an external force on the cup/workpiece system, and the entire systems moves due to the force.

Consider the motion of the workpiece alone with the workpiece initially at rest in the air with the suction cup attached. As the answer by @Claudio Saspinski explains the net force upward on the workpiece is $F_{\Delta p} - mg - N$. $F_{\Delta} p$ is the net force upward from the difference in the pressure on the bottom of the workpiece and the pressure on the top of the workpiece; this force is upwards due to the decrease in pressure on the top due the "vacuum" in the cup. $mg$ is the force of gravity downward on the workpiece of mass $g$. $N$ is the compressive force downward; see the answer by @AccidentalTaylorExpansion. For no motion, $F_{\Delta p} - mg - N = 0$. For acceleration $a$ upwards $F_{\Delta p} - mg - N = ma$ and $N$ is decreased.

See How do suction cups work? on this exchange.

John Darby
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  • I see your point to think of it as an entire system, however do we know physical reason? When we analyze the workpiece seperately, we will see gravitational force and another force caused by difference between atmospheric pressure and inner pressure inside the cup. Where is our force applied on the cup? How did it get transmitted to the workpiece? – Jawel7 Feb 04 '23 at 16:02
  • Please see my expanded answer. – John Darby Feb 04 '23 at 17:18
  • But you are ignoring the force on top of suction cup, exerted by ambient pressure. Okay there is a force due to pressure difference between top and bottom surface area of workpiece. But what about another force exerted by ambient pressure on top of suction cup? – Jawel7 Feb 04 '23 at 17:26
  • For this system to function, I think the workpiece area covered by suction cup must be larger than outer surface area of suction cup. Do you agree? – Jawel7 Feb 04 '23 at 17:27
  • The force from the atmospheric pressure on top of the cup is a force on the cup, not a force on the workpiece. The force on the workpiece for the area covered by the cup is due to the pressure in the cup which is lower than atmospheric pressure. – John Darby Feb 04 '23 at 17:34
  • Dear John, you are still ignoring.. When you put the cup on the workpiece and squeeze and release it, just analyze the workpiece. There's weight force downward, there's low pressure area force upwards. However, unless you lift the cup, the workpiece will stay on ground. When you lift the cup, the workpiece will go up as well. The question is what has changed for the cup? According to your assumption, everything must be the same. Its weight is the same, and low pressure area force is the same as well. But how is it going up then? It means that there should be another force here. – Jawel7 Feb 05 '23 at 01:47
  • Additionally, it's a huge mistake to think that "the force from the atmospheric pressure on top of the cup is a force on the cup, not on the workpiece.". How is it possible? This force applies on the cup, and the cup will transmit this force to the workpiece. Then the workpiece will react to the cup to the same extend. In this way the cup will stay on the workpiece instead of going down by crushing the workpiece.. – Jawel7 Feb 05 '23 at 01:49
  • The other force on workpiece is normal force that changes when you lift up. See answer by @Claudio Saspinski. Forces on the cup are: the cup weight down, from the atmospheric pressure down, from the lower pressure in the cup up, the normal force from the workpiece on the cup up, and a pulling force from the handle on the cup if you pull the cup up. When you state "the atmospheric pressure on the top of the cup is transmitted to the workspace" this is not correct. If this were true, there would be no suction which would render the suction cup useless. Pressure in cup < pressure outside cup. – John Darby Feb 05 '23 at 03:55
  • What I don't understand here is how my pulling force is transmitted to the workpiece while I applied it on the cup? Additionally, I'm not saying the ambient pressure is directly transmitted to the workpiece from top of the cup. Since the outer surface of the cup is less than the workpiece area covered by the cup, the net force due to ambient pressure will be upward. It makes our pulling up easier, this is my understanding. Otherwise, it's so hard to understand that the workpiece goes up with my pulling force, but I'm not applying any force on workpiece, I'm just pulling the cup. – Jawel7 Feb 05 '23 at 09:17
  • When you pull up on the cup you lower the normal force on the workpiece. Here is an analogy: when you jump off the ground you increase the normal force on you from the ground; that normal up exceeds force of gravity down and you accelerate up (leave the ground). – John Darby Feb 05 '23 at 13:27
  • Okay, I see. Then my question is what you mean by "normal force on the workpiece" ? What creates that normal force on the workpiece? – Jawel7 Feb 05 '23 at 13:28
  • Consider no motion, cup stationary. There must be normal force up on cup from workpiece to balance the net down force from cup's weight and difference in pressure outside/inside cup. By Newton's third law there is also normal force down on workpiece from cup equal in magnitude to the normal force up on cup from workpiece. When you accelerate the cup up by applying a pulling force, looking at the workpiece alone the magnitude of the normal force down on the workpiece must decrease to allow acceleration up as @Claudio Saspinski shows. By Newton #3 normal force up on cup also decreases. – John Darby Feb 05 '23 at 18:26
  • The normal force is the force that surfaces exert to prevent solid objects from passing through each other. Normal force is a contact force. If two surfaces are not in contact, they can't exert a normal force on each other. It is due to Coulombic repulsion between atoms as the objects are pushed together, – John Darby Feb 05 '23 at 18:31
  • Let us continue this discussion in chat. – John Darby Feb 06 '23 at 13:23
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Because there is air on one side and no air on the other side, there is a pressure difference which will result in an unbalanced force on the plate. Because of Newtons' law the plate either has to accelerate or acquire another force from somewhere else to balance it. This force comes from the joint between the plate and the cup. The plate tries to accelerate, but the cup is in the way so a normal force develops. This normal force causes a lot of friction because friction is proportional to the normal force. This large compression force, combined with a frictional force that prevents slipping, results in the system acting like a single object.

enter image description here

AccidentalTaylorExpansion
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  • Just after I put the cup on the workpiece, the net force acting on the workpiece is downward, since its weight is more than the upward force caused by pressure difference on the upper and lower surface of the workpiece. When I apply force on the cup to lift it, what has changed for the workpiece? Why is it going up as well? I don't understand this. – Jawel7 Feb 04 '23 at 16:09
  • @Jawel7 The blue arrows that I drew can be very large. They can easily be much larger than the weight of the cup. the blue arrow is just the total force that is working on the cup or plate. When the cup is resting on the plate with no air removed it exerts $N=m_\text{cup}g$ on the plate (In my drawing the plate is on top but I think your example has the cup on top so that is what I use here). After removing the air the cup exerts $N=m_\text{cup}g+\text{pressure forces}$ on the plate. After lifting the cup and plate together, the cup exerts $N=\text{pressure forces}-m_\text{plate}g$ on the – AccidentalTaylorExpansion Feb 05 '23 at 14:47
  • ... plate. If $m_\text{plate}g>\text{pressure forces}$ the plate will drop of course. We can reason each of these forces by drawing a free body diagram. So what has changed during the lifting? Before lifting, the pressure forces were pressing the two objects together with excess. After lifting, the pressure forces are still pressing the objects together but now just a little less hard. – AccidentalTaylorExpansion Feb 05 '23 at 14:53