What is the minimum force required to lift an object?

[Yahya Sharif]

Can you clarify, please.

You are saying that you exerted a force of less than 60 kg(force) and succeeded in lifting a human body with mass 60 kg(mass)? But you did not show us that experiment. Nor did you describe it.

I didn't do the experiment it is my predictions. I don't have the big scale to measure my weight.
The description is in post #1

 

[russ_watters]

I didn't do the experiment it is my predictions.

Why would you predict something that contradicts the experiment you did? What is the reasoning behind it?

Just to be clear: the experiment and it's description/reasoning are fine. But then you appear to be discarding your own experiment and predicting the human body works differently. That makes no sense. And is pretty obviously wrong (unless you just didn't explain it correctly, which is possible...).

 

[russ_watters]

Back to your OP:

A person stands on a scale. The scale reads his mass 60 kg . Now this human moves up his body short distance like someone tries to pick a fruit from a tree. The scale will start to increase by small forces x N in which the total read of the scale is 600+x N *. The force he exerts on the scale is x N.

This is very clearly wrong. The scale reads the force that is exerted on it (600+xN). That's its purpose. Your own videoed experiment correctly demonstrates how it works. At this point I'm having trouble believing you are being serious.

 

[Yahya Sharif]

What is this specific fixed maximum value x N as soon as the human body raises ? What is the value of this x N for the 60 kg body and why? Why is the x force is smaller in case of a "child" of less weight of 25 kg?
In the video as soon as I lifted the weight the maximum force was 1.29 kgf which is the force to lift the mass. So what about a human lifting his body in the OP, what is this maximum x as soon as he lifts his body?

 

[jbriggs444]

What is this specific fixed maximum value x N as soon as the human body raises ?

Minimum, not maximum. Any positive, non-zero value of $x$ will result in the human body rising. Which, technically, means that there is no minimum. Only a greatest lower bound: zero.

 

[Yahya Sharif]

Minimum, not maximum. Any positive, non-zero value of $x$ will result in the human body rising. Which, technically, means that there is no minimum. Only a greatest lower bound: zero.

This is the scenario :
I press with x=10 N " don't move" I increase to x=15 N " still do not move" when I reach an x>15 N I raise. This is the maximum which is a fixed number if I repeat the experiment.
If any force can lift me why I cannot move with the x=10 N?

 

[jbriggs444]

This is the scenario :
I press with x=10 N " don't move" I increase to x=15 N " still do not move" when I reach an x>15 N I raise. This is the maximum which is a fixed number if I repeat the experiment.
If any force can lift me why I cannot move with the x=10 N?

Please specify the experiment. Since you mass more than 1.5 kg, it cannot be the one that you seem to be describing.

If I (hypothetically) mass 60 kg and weigh 600 N then a force (600+x) N in magnitude is sufficient to lift me when x = 1. Also when x = 0.1. Also when x = 0.01. Also when x = 0.001. Also when x = $10^{-100}$.

Pick a positive number. Any positive number. Then 600 N plus that number of Newtons will suffice to lift me. It does not take an excess of 10 N. It does not take an excess of 15 N. Any excess will do.

 

[russ_watters]

This is the scenario :
I press with x=10 N " don't move" I increase to x=15 N " still do not move" when I reach an x>15 N I raise. This is the maximum which is a fixed number if I repeat the experiment.
If any force can lift me why I cannot move with the x=10 N?

That isn't correct, unless you're changing the equation without telling us. It's F = 160 + x, where 160 is your weight. If that's still the equation, clearly what you are saying is wrong.

Again, you demonstrated how this works. Why do you think it works differently for people?

 

[Yahya Sharif]

If I (hypothetically) mass 60 kg and weigh 600 N then a force (600+x) N in magnitude is sufficient to lift me when x = 1. Also when x = 0.1. Also when x = 0.01. Also when x = 0.001. Also when x = $10^{-100}$.

Pick a positive number. Any positive number. Then 600 N plus that number of Newtons will suffice to lift me. It does not take an excess of 10 N. It does not take an excess of 15 N. Any excess will do.

But experiments do not say this. If you stand on the scale with weight 60 kg and you apply a smaller force let's say x=10 N the scale will read 600+10 N but you still do not rise. You will need to maximize your force to x N to lift your body.

 

[jbriggs444]

But experiments do not say this. If you stand on the scale with weight 60 kg and you apply a smaller force let's say x=10 N the scale will read 600+10 N but you still do not rise. You will need to maximize your force to x N to lift your body.

To be clear, you are standing on the scale and you push down on a nearby counter with a downward force of 10 N, you are concerned that you do not then rise up off of the scale?

Have you examined the scale in this circumstance to see whether the updated scale reading is now 590 N? The total supporting force would then be unchanged at... 600 N.

If you actually performed this experiment instead of thinking about it, you would discover that at x = 600 N, the scale reading would have gone to zero and you would then be lifting your feet free of the scale.

Edit: The actual behavior will depend crucially on whether the scale in question is a constant force device (one that will deliver a constant force of 600 N regardless of its vertical posision) or a constant position device (one that will deliver whatever force is required to maintain the position of the object resting upon it and will report that delivered force on a dial or digital readout).

A standard bathroom scale is, for most practical purposes, a constant position device.

Edit 2: Or, maybe this is what you are getting at...You are standing on a scale that reads 600 N. You kneel down and push with your hands, applying a downward force of 10 N with your hands on the top surface of the scale. You now expect the scale to read 610 N?

 

[russ_watters]

Edit 2: Or, maybe this is what you are getting at...

My interpretation of the scenario is jumping off the scale. The equation holds as originally described for that scenario. But the OP can clarify...

 

[Yahya Sharif]

Edit 2: Or, maybe this is what you are getting at...You are standing on a scale that reads 600 N. You kneel down and push with your hands, applying a downward force of 10 N with your hands on the top surface of the scale. You now expect the scale to read 610 N?

I do not kneel. I stand on the scale. my weight is 60 kg. I lift my body up short distance by feet and calves' muscles like someone trying to pick a fruit from a tree. If I press by a small force like 2 N I will not rise. The scale will read 600+2=602 N. I will need to maximize my force to some specific force x N to lift my body.
I start pressing the scale, the x continues to increase, as soon as my body raises I reach a maximum of x N then by definition the force to lift the body is the x N. This x has a specific value for the 60 kg If I repeat the experiment. If the person is " a child" of 25 kg the force x will be less.
In the video it is the same thing first I measured the weight which is 1.29 kg then I started to lift the weight, the force increases. As soon as the mass raises, the force reaches a maximum of 1.29 kgf that means by definition the force I used to lift the object equals the weight 1.29 kgf.
The force I used to lift the object in the video equals to its weight. The force I used to lift my body in the experiment x N is less than weight.
Humans and animals use force to lift or move their bodies smaller than the force they use to lift or move other objects with the same mass. So human movements such as : walking, running, jumping, dancing, etc, are done relatively with little effort.
So what is this maximum specific force x N that lifted the human?
P.S
The maximum force x is just a few Newtons compared to the weight 600 N. So the force to lift the 60 kg human body is just a few Newtons.

 

[Steve4Physics]

So what is this maximum specific force x N that lifted the human?

You mean minimum. And there is no limit.

There is some confusion here. I wonder if it's related to this...

If there is only a small change in force, internal friction in the scales may prevent the reading from changing. You will get a false reading.

But if you had perfect scales, then as soon as as you start to accelerate upwards (e.g. by straightening bent legs), the reading would increase.

If the acceleration is very small the increase will also be very small. E.g. a reading of 60.000kgf might increase to 60.001kgf.

 

[jbriggs444]

I do not kneel. I stand on the scale. my weight is 60 kg. I lift my body up short distance by feet and calves' muscles like someone trying to pick a fruit from a tree. If I press by a small force like 2 N I will not rise.

Wrong.

 

[Delta2]

Your CoM might rise without your feet leaving the ground. If your CoM is changing position then there is some net external force on your body.

 

[russ_watters]

I do not kneel. I stand on the scale. my weight is 60 kg. I lift my body up short distance by feet and calves' muscles like someone trying to pick a fruit from a tree. If I press by a small force like 2 N I will not rise. The scale will read 600+2=602 N. I will need to maximize my force to some specific force x N to lift my body.

That scenario is now clear, thanks for that at least. As others said, your understanding of the forces is wrong. Can you explain *why* you believe the physics works differently for humans than for the weight you tested?

Ultimately it's up to you if you want to learn/accept the reality or not. We can help, but there is only so much we can do if you won't explain your reasoning. This won't be productive if you just keep repeating the wrong thing over and over without explanation.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[russ_watters]

Let's give the OP one more chance...

 

[Yahya Sharif]

The human will press the scale with x N. the scale will push with the same force x N then the force that lift the human is x N which is the force upwards by the scale.
A human legs undergoes a very small force upwards by the scale " x N" that results in very small pressure on the legs when the human stands on feet.
Observations:
1)a A human of 60 kg stands on his feet.
1)b The human is lying on the ground and legs are up and then putting a rock of 60 kg on the human's feet.
2)a The human is just lying on the ground.
2)b The human is lying on the ground on his belly a rock of 60 kg is put on his back, the rock has the human length and width.
In 1)a scenario the pressure on legs will be small that the human can stand for hours and this because the force upwards on the legs by the ground or a scale is a small force x N. In 1)b scenario the pressure will be extremely big and the legs cannot bare this 60 kg rock. This is also how a human knee can bare massive body 60 kg for years because the force on the legs is small x N.
In 2)a scenario the back will undergo small pressure that the human will be comfortable lying for hours. In 2)b scenario the rock will make very big pressure that it might break the chest. In all scenarios the human mass and the rock mass are equal 60 kg.

So the force that lifted the human In the experiment in the OP is a smaller force x N that appears as a smaller pressure on legs when standing or on the back when lying.

 

[russ_watters]

Okay, so your misunderstanding stems from the idea that because you do not have to exert much effort to stand up straight that means the force being applied to the ground is small. That is not the case. It just happens that our bodies are well configured for standing straight. In that condition we are essentially static objects. In other positions we are much less efficient at holding still. To see what I mean try this:

1. Start from a normal standing position.

2. Now bend your knees while keeping your torso vertical to lower your body a few centimeters until your knees are bent at an angle of 90°.

3. Hold this position until you understand the force that is actually required to be applied to the ground to hold you up. It should only take a few seconds.

 

[Ibix]

Another simple experiment you can do is to hold a weight of a few kilograms close to your chest, then extend your arms horizontally in front of you. I have a 10kg weight to hand. Close against my chest I can hold it for minutes with no trouble (I got bored before I got tired), but held out straight-armed in front of me I could only manage for a few seconds.

The force that needed to be exerted on the weight didn't change. The leverage my muscles had on my skeleton (and hence how hard they had to work) changed a lot.

Body mechanics is not simple introductory-level physics, and it's made complicated because you are only really aware of sensations out of the ordinary. For example, how do you think it would feel to have a 1kg weight pressing on 1cm² of your skin? Actually you already have that - atmospheric pressure - and you aren't aware of it. Another 1kg you would feel, and you would feel it differently if it were a single weight pressing on one place or if you doubled atmospheric pressure.

 

[Yahya Sharif]

Another simple experiment you can do is to hold a weight of a few kilograms close to your chest, then extend your arms horizontally in front of you. I have a 10kg weight to hand. Close against my chest I can hold it for minutes with no trouble (I got bored before I got tired), but held out straight-armed in front of me I could only manage for a few seconds.

The force that needed to be exerted on the weight didn't change. The leverage my muscles had on my skeleton (and hence how hard they had to work) changed a lot.

Body mechanics is not simple introductory-level physics, and it's made complicated because you are only really aware of sensations out of the ordinary. For example, how do you think it would feel to have a 1kg weight pressing on 1cm² of your skin? Actually you already have that - atmospheric pressure - and you aren't aware of it. Another 1kg you would feel, and you would feel it differently if it were a single weight pressing on one place or if you doubled atmospheric pressure.

Let's say for the the experiment in the OP the foot is 20 cm or 0.2 meters long , Now let's calculate for a 0.2 m lever:
First the lever will be class 2 :
The weight for 60 kg will be 60*10=600 Newtons.
Class 2 is the fulcrum at the toes , and in this case both the weight of my body and the force of my calves' muscles I lift my body with will be at the heel:
F: force of my weight
f: force of muscles strength
L: the distance of the weight from the heel to the toes.
l: distance of the muscles force from the heel to the toes.
f * l=FL
F=600 and L=l =0.2

f*0.2=600*0.2
f=600 Newton
The force needed to lift my body in the experiment does not change which is 600 N. I must exert a 600 N force to lift my body but I actually lift by weak calves' muscles and feet's muscles with the small force x N.

 

[hmmm27]

DRAW A DIAGRAM

of the simplest example you can think of that illustrates your problem with physics.

Label all forces.

 

[Yahya Sharif]

Why putting a rock of 60 kg on my back while I am lying on belly will give immediate extreme pressure that might break the chest but lying with human body mass 60 kg will cause small pressure that a human can lie for days ? The body and the rock are of the same weight. The mass of the rock has the same human shape so that the pressure by the rock on the back will be equivalent to the pressure by the human body on the back.

 

[jbriggs444]

Let's say for the the experiment in the OP the foot is 20 cm or 0.2 meters long , Now let's calculate for a 0.2 m lever:

So it seems that you will be working to calculate the force of the calf muscle as concentrated in the Achilles tendon.

First the lever will be class 2 :
The weight for 60 kg will be 60*10=600 Newtons.

Right. 600 Newtons upward force from the ground on the foot centered near the toes. Hypothetically 20 cm forward from the ankle joint.

Class 2 is the fulcrum at the toes , and in this case both the weight of my body and the force of my calves' muscles I lift my body with will be at the heel:

Right so you envision the toes as the fulcrum for this lever. Not the best spot to use to illustrate the principle, but let us see where you are going with this.

F: force of my weight
f: force of muscles strength
L: the distance of the weight from the heel to the toes.
l: distance of the muscles force from the heel to the toes.

Ok. Achilles tendon attaches to foot at distance lower case "l" from the toes.
Meanwhile, the weight of the body attaches to the foot at a distance upper case "L" from the toes.

f * l=FL
F=600 and L=l =0.2

Wait a minute. You are saying that the attachment point for the Achilles tendon and the center of the ankle joint are co-located. Please consult a drawing of a real foot.

Personally, I would have treated this as a lever with the fulcrum at the ankle. But then I'd have first realized that there is no need to treat the foot as a lever at all. We have a body with a center of mass and we have an external force. We don't need any levers to describe the resulting motion.

 

[Ibix]

Why putting a rock of 60 kg on my back while I am lying on belly will give immediate extreme pressure that might break the chest but lying on back with human body mass 60 kg will cause small pressure that a human can lie for days even though the body and the rock are of the same mass?

I think you are over estimating how much pressure a 60kg rock 2m long and say 50cm wide will exert. It's 600Pa, which is about the same as water pressure at a depth of 6cm - easily survivable. The rock would be between one and two centimeters thick, depending on what rock you have in mind.

Of course the top surface of a supine human is not flat and some points will suffer a higher pressure, so I'd expect it to be uncomfortable (rather like lying face down on flat concrete is, in fact) but hardly fatal.

The force needed to lift my body in the experiment does not change which is 600 N. I must exert a 600 N force to lift my body but I actually lift by weak calves' muscles and feet's muscles with the small force x N.

You are missing the point of both @russ_watters' comment and mine. The point is that your sensation of how hard or painful it is to do something is a terrible guide to how much force is needed. Depending on how you arrange your skeleton, the amount of force you need to apply to hold a fixed mass stationary can vary wildly. That isn't because humans are an exception to the laws of physics, it's because we can change our configuration so that the force we need to remain in a given configuration varies.

 

[Yahya Sharif]

View attachment 305068

Personally, I would have treated this as a lever with the fulcrum at the ankle. But then I'd have first realized that there is no need to treat the foot as a lever at all. We have a body with a center of mass and we have an external force. We don't need any levers to describe the resulting motion.

I am not a native speaker I really meant by the heel the joint of the ankle. I can make my body perfectly vertical when I balance my body holding something then the idea of a lever will work. The force must lift the human is still 600 N.

 

[Yahya Sharif]

I think you are over estimating how much pressure a 60kg rock 2m long and say 50cm wide will exert. It's 600Pa, which is about the same as water pressure at a depth of 6cm - easily survivable.

Of course the top surface of a supine human is not flat and some points will suffer a higher pressure, so I'd expect it to be uncomfortable (rather like lying face down on flat concrete is, in fact) but hardly fatal.

There is a difference between the pressure under water and the pressure of a mass on body.
At 10 meters under water the pressure on the body is 1 atmosphere which is equivalent to 101325 pascal the human will survive.

https://www.google.com/search?q=atm....69i57j0l5.12585j0j7&sourceid=chrome&ie=UTF-8

But 101325 pascal or N/m^2 is a rock of 10132.5 kg on the human body of 2 m length and 50 cm width so a mass of 10 tonnes is on the human body will crush all human body bones.

You are missing the point of both @russ_watters' comment and mine. The point is that your sensation of how hard or painful it is to do something is a terrible guide to how much force is needed.

Damage. Damage measures the severity of pressure on the human. Compare putting a protrusion of 18 cm width 18 cm length of a rock of 60 kg on belly vs lying on a concrete protrusion of 18 cm width 18 cm length on belly and my rest of the body is on air.
In case of human you will feel some pain but in the case of a rock the rock will damage the belly completely.

 

[malawi_glenn]

But 101325 N/m^2 is a rock of 10132.5 kg on the human body of 2 m length and 50 cm width so a mass of 10 tonnes is on the human body will crush all human body bones.

You can not make that rock have 1 m² contact on the human so the pressure will be much larger.

Humans can be buried in sand which are tiny tiny rocks. But the weight of the sand is more evenly distributed along the human body thus increasing the area of contact.

Damage measures the severity of pressure on the human. Compare putting a protrusion of 18 cm width 18 cm length of a rock of 60 kg on belly vs lying on a concrete protrusion of 18 cm width 18 cm length on belly and my rest of the body is on air.
In case of human you will feel some pain but in the case of a rock the rock will damage the belly completely.

Eh, no it won't crush your belly...

 

[Ibix]

do not know

And this is the problem. You are guessing instead of calculating. Of course, you cannot calculate because you have no model except your own personal incredulity that human bodies obey the laws of physics.

there is a different between the pressure under water and the pressure of a mass on body.

There is. It's that water pressure is isotropic (more or less) and the rock isn't. Nevertheless, it's worth comparing your rock to water. If I put a child's paddling pool on you and filled it to a depth of 6cm, do you think it would hurt? If I put a 60kg 2m×0.5m rock on you do you think it would hurt? You've already answered yes to the second question, and you now know the paddling pool weighs the same. But what's your instinctive response to the paddling pool? I bet it's not "that would destroy me" - but if your instincts are a consistent physical model then they should tell you the paddling pool with 6cm of water will crush you the same as the rock.

Of course, neither will crush you.

At 10 meters under water the pressure on the body is 1 atmosphere which equivalent to 101325 pascal

And I expect a "paddling pool" ten meters high filled with water and placed on your chest would crush you, yes.

Compare putting a protrusion of 18 cm with 18 cm length of a rock of 60 kg on belly vs lying on a concrete protrusion of 18 cm width 18 cm length on belly and my rest of the body is on air.
In case of human you will feel some pain but in the case of a rock the rock will damage the belly completely.

Nonsense. Go to YouTube and search for "weightlifting fails". The site is awash with videos of idiots benchpressing too much without a spotter and ending up trapped under their barbells' bars. That easily provides pressures well above your 18×18 rock example, even if you neglect the initial extra pressure from stopping the weights' fall. It doesn't look comfortable, and I bet they have impressive bruising, but they're not cut in half.