I What is the minimum force required to lift an object?

[Yahya Sharif]

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. The force the scale pushes him up is also x N two forces in opposite directions. The force that lifts his body is the force the scale pushes him up which turned out to be the x N.

For a force to lift an object the force must be slightly greater than the object's weight or greater than 600 N . The force which lifted the human is x N which is far less than his weight 600 N.

*The scale shows mass 60 kg I converted it to force 600 N , force= mass* gravity acceleration gravity acceleration is 10 m/s^2

Does the x N which is less than 600 N lift the person? The x must be greater than 600 N. In fact the x is a very small force.

 

[russ_watters]

I think you can probably figure out the error in your reasoning g here - evidence tells you the answer so your premise has to be wrong...right?

What does f=ma tell us?

 

[Yahya Sharif]

What does f=ma tell us?

What the experiment tells?

 

[LastScattered1090]

What the experiment tells?

It sounds like maybe you need to go back to the more basic scenario. Consider when the person is just standing on the scale doing nothing. He is not exerting any additional downward force using his legs.

In this situation, the person is in equilibrium (motionless and unaccelerated) despite gravity pulling downward on him. So does the scale push upwards on him in this scenario? If so, by how much does it push upward on him?

 

[Yahya Sharif]

In this situation, the person is in equilibrium (motionless and unaccelerated) despite gravity pulling downward on him. So does the scale push upwards on him in this scenario? If so, by how much does it push upward on him?

When at equilibrium the body acts on the scale by the weight 600 N and the scale pushes up the body with an equal normal force 600 N then no acceleration.

When I push the scale by x I increase the force on the scale to be my weight 600 plus the force I pushed the scale with or 600+x N, the scale will instantly read it.

So I exert extra x , this extra x is a net force that accelerates my body upwards, then x N accelerates a 60 kg body against gravity even though the x is very smaller than the weight 600 N.

 

[LastScattered1090]

When at equilibrium the body acts on the scale by the weight 600 N and the scale pushes up the body with an equal normal force 600 N then no acceleration.

When I push the scale by x I increase the force on the scale to be my weight 600 plus the force I pushed the scale with or 600+x N, the scale will instantly read it.

So I exert extra x , this extra x is a net force that accelerates my body upwards, then x N accelerates a 60 kg body against gravity even though the x is very smaller than the weight 600 N.

Basically, you continue to confuse net force with total applied upward force in this scenario. You acknowledge that the scale pushes up with 600 N in the static scenario, so why are you having so much trouble acknowledging that it pushes upward with (600+x) N (which is a force unequivocally greater than the weight) in the dynamic scenario?

The net force is then
F_up - F_down = (600 + x) - 600 = x

The fact that this value is "much smaller than" the weight is irrelevant, because we've already subtracted out the weight, leaving the net (meaning the completely unopposed) upward force remaining.

An unopposed upward force of x Newtons will produce an acceleration of (x/60) m/s².

 

[Yahya Sharif]

The fact that this value is "much smaller than" the weight is irrelevant, because we've already subtracted out the weight, leaving the net (meaning the completely unopposed) upward force remaining.

An unopposed upward force of x Newtons will produce an acceleration of (x/60) m/s².

This violates the law of conservation of energy.
If you say I can use an x let's say 10 N to accelerate a 60 kg body one meter against gravity then the body can return back with force 600 N" the gravity weight " one meter, in such case I will do work " 10*1=10 Joules" far less than the work I get " 600*1=600 Joules"
The law is only conserved if I lift with force slightly greater than the weight. x=600 N , work1=600 Joules, mass return with weight 600 N work2=600 Joules work1=work2

 

[LastScattered1090]

This violates the law of conservation of energy.
If you say I can use an x let's say 10 N to accelerate a 60 kg body one meter against gravity then the body can return back with force 600 N" the gravity weight " one meter, in such case I will do work " 10*1=10 Joules" far less than the work I get " 600*1=600 Joules"
The law is only conserved if I lift with force slightly greater than the weight. x=600 N , work1=600 Joules, mass return with weight 600 N work2=600 Joules work1=work2

No, the Conservation of Energy is not being violated here. Over a vertical displacement of 1 metre, the gravitational force does -600 Nm of work, while the applied lifting force does +(600 + x) Nm of work. This means that the total work done of +x Nm is exactly the amount of kinetic energy you should have at the end of your leg movement.

I emphasized the most important words in your reply above: "I lift". Yes, you (or whatever is pushing upward, i.e. the scale in this case) has to lift with a force greater than the weight. But this does not mean that the net upward force must be greater than the weight. That's your fundamental misconception here. The net force and the applied lifting force are not the same thing. In order for an object to accelerate in a certain direction, the net force in that direction must be non-zero. Period. That's Newton's Second Law. When considering vertical forces in a gravitational field, there is always a downward force of mg. As a consequence, the only way to achieve a non-zero net upward force is that any applied upward forces must total to greater than mg, so that the balance (net) is non-zero upward.

In summary:

(total applied upward force) > weight
and
(net upward force) > 0

are two different ways of stating the same requirement here. Don't confuse them.

 

[Yahya Sharif]

This means that the total work done of +x Nm is exactly the amount of kinetic energy you should have at the end of your leg movement.

If my kinetic energy at the top is +x Joules, how it is possible that I store back a potential energy of -600 Joules? It is the same issue of violating the law of conservation of energy.

 

[hmmm27]

If it takes 600N to hold a weight up without rising or falling, what happens if you apply 700N instead of 600 ? What if you apply 500N instead of 600N ?

If there's a 100N weight on top of the table, how many N is it going to require to lift it vertically off the table ? 1N or 101N ?

 

[Yahya Sharif]

If it takes 600N to hold a weight up without rising or falling, what happens if you apply 700N instead of 600 ? What if you apply 500N instead of 600N ?

If you you apply 700 N it will rise, if you apply 500 N it will remain.

If there's a 100N weight on top of the table, how many N is it going to require to lift it vertically off the table ? 1N or 101N ?

You need a force greater than 100 N.

 

[hmmm27]

If you you apply 700 N it will rise, if you apply 500 N it will remain.

Your physics math needs work.

You need a force greater than 100 N.

But, by your apparent logic, the object is already applying a force of 100N to the table, so you only need pull/push it up with a small amount, say 1N

 

[LastScattered1090]

If my kinetic energy at the top is +x Joules, how it is possible that I store back a potential energy of -600 Joules? It is the same issue of violating the law of conservation of energy.

First of all, this line of reasoning is irrelevant to the question you asked about forces. The question about forces has been asked and answered. If you take the time to consider my answer, you'll see it must be correct. Thus, any counterarguments you come up with against it (based on energy arguments or whatever) must be flawed too.

Second, the x Nm of energy didn't just come out of nowhere. Your body converted internally-stored chemical energy into mechanical energy in order to get your muscles to move in the first place.

The 1 m of displacement is just the amount of vertical rise that we are assuming happens while the upward force is being applied (i.e. while you are pushing off from the ground). But once you've finished propelling yourself, remember what I said above? You now have kinetic energy! Which means you'll continue to move upwards. Beyond the initial height of 1 m. If you continue to move upwards, gravity will do even more negative work, resulting in even more increase in potential energy.

 

[Yahya Sharif]

But, by your apparent logic, the object is already applying a force of 100N to the table, so you only need pull/push it up with a small amount, say 1N

It will not move by applying 1 N force because I can do a simple experiment by hanging the object on a scale and lift the scale slowly measuring the minimum force to lift it with, which will turn out to be greater than 100 N. Both are separate experiments.

 

[Yahya Sharif]

First of all, this line of reasoning is irrelevant to the question you asked about forces. The question about forces has been asked and answered. If you take the time to consider my answer, you'll see it must be correct.

Energy is also an issue. If a smaller force x N used to lift a bigger 60 kg body I will exert small input energy to get greater output potential energy which violates the law of conservation of energy as in the previous posts.

 

[Ibix]

Energy is also an issue. If a smaller force x N used to lift a bigger 60 kg body I will exert small input energy to get greater output potential energy which violates the law of conservation of energy as in the previous posts.

If you apply 101N through to lift your mass 1m, then 100J will be added to the potential energy (because the gravitational force does -100J of work on the mass) and 1J to kinetic energy. There is no energy conservation problem here.

 

[LastScattered1090]

If you apply 101N through to lift your mass 1m, then 100J will be added to the potential energy (because the gravitational force does -100J of work on the mass) and 1J to kinetic energy. There is no energy conservation problem here.

Exactly, and this is something which I already said way back in post #8. I think this thread might not be going anywhere..

 

[Ibix]

Exactly, and this is something which I already said way back in post #8. I think this thread might not be going anywhere..

Let's hope the silence means OP is thinking about what's been written, rather than arguing that well-tested theories don't work for simple problems.

 

[Yahya Sharif]

How a person of 60 kg can jump high and fast against gravity with his only leg muscles while he will barely move a rock of 60 kg with his all body muscles? He uses small force x N to lift his body and a small force to jump.

 

[malawi_glenn]

How a person of 60 kg can jump high and fast against gravity with his only leg muscles while he will barely move a rock of 60 kg with his all body muscles?

https://www.physicsforums.com/insights/how-can-we-jump-when-the-ground-does-no-work/

Why he can not move a rock of 60 kg? Because he does not hit the gym as often as he hits the library!

 

[Delta2]

How a person of 60 kg can jump high and fast against gravity with his only leg muscles while he will barely move a rock of 60 kg with his all body muscles? He uses small force x N to lift his body and a small force to jump.

This happens because we lift a weight with the muscles of our hands + arms + muscles of our middle which are not so strong as the muscles of our legs.

When you want to do big damage to something you hit it with your leg not your hand or middle !

 

[malawi_glenn]

This happens because we lift a weight with the muscles of our hands + arms + muscles of our middle which are not so strong as the muscles of our legs.

Hence the saying "lift with your legs"

When you want to do big damage to something you hit it with your leg not your hand or middle !

Never skip legday bro

 

[Yahya Sharif]

This happens because we lift a weight with the muscles of our hands + arms + muscles of our middle which are not so strong as the muscles of our legs.

He cannot lift the rock with all muscles including his legs. Think of someone lying on the floor pushing a rock by his hands and pushing a wall by his feet to move the rock. He will hardly be able to overcome its friction let alone lifting it.
The difference is also big. What force I need to throw the rock in air like my body?! perhaps I will need hundreds times of my body force to through a rock. Jumping also needs smaller energy I can jump many times.
But think of the energy needed to throw a rock of 60 kg in air several times! Is that possible with hundreds times energy of my body?!

 

[malawi_glenn]

He cannot lift the rock with all muscles including his legs.

You do realize that he does not just have to lift the stone, but also his entire body, with those skinny legs?

This is why squats with just your body weight is a good beginner exercise before you start with a bar on your neck! If you can't even squat your body weight, no need to buy an expensive gym membership!

Seriously, what is it that you really have a hard time to understand? How a person can jump or what?

 

[Delta2]

@Yahya Sharif you got some sort of intuitive reasoning however it has some serious flaws which me and other members have highlighted but you keep insisting. I think no matter how I insist too you will not accept your flaws. So goodbye from me.

 

[russ_watters]

How a person of 60 kg can jump high and fast against gravity with his only leg muscles while he will barely move a rock of 60 kg with his all body muscles? He uses small force x N to lift his body and a small force to jump.

Because 60+60=120 and 120 is a lot bigger than 60.

 

[malawi_glenn]

This violates the law of conservation of energy.

We do have chemical energy stored in our bodies. In the muscles we have glycogen, also we have fat stored - but not me because I am ripped to the bonez working out at the gym 2 h everyday before I hit my math books for some brain gainz. Food is a vital component for your health and brain power. Eat lots of whole foods like fruit and salmon, stay away from starbucks and pizza hut - unless it is legday, then everything is allowed

 

[Steve4Physics]

He cannot lift the rock with all muscles including his legs. Think of someone lying on the floor pushing a rock by his hands and pushing a wall by his feet to move the rock. He will hardly be able to overcome its friction let alone lifting it.

You are not comparing like-with-like, The strong muscles in legs can easily move heavy objects, e.g. watch this:

 

[nasu]

Where did russ_watters said this? Aren't you misquoting him?

 

[Yahya Sharif]

Because 60+60=120 and 120 is a lot bigger than 60.

The difference is big. What force I need to throw a rock of 60 kg in the air? what force of my legs I need to throw my body 60 kg jumping? The force I need to throw the rock is hundreds times the force I need to jump even though the body and the rock are of the same mass.