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Height of a Mass that receives a Force

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Homework Statement


A mass at rest of 70 Kg receives a force underneath of 360 N. Calculate the height of the mass when 2 seconds have passed.

Homework Equations


1) F = m.a
2) Y(t) = g/2*t^2 + Vi*t + Yi

The Attempt at a Solution


This is a problem i need to solve in order to start something in my thesis.

With force and mass we obtain the acceleration of the mass that is equal to 360/70 = 5.14 m/s² (formula 1). My problem is, since this is a vertical problem, the gravity is present and the formula 2 cannot be used with this acceleration.

Happening this and many other things I've tried, this might be a problem that has to be solved with Energy formulas. The problem of that is, my coursework of Physics in my Computer Science career has a really poor content of Work and Energy so i kinda need some tuition here.

The problem looks really simple to my brain, but not for the paper and the formulas.

Looking forward for some help,
Peter.
 

Answers and Replies

  • #2
BvU
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Hello Peter, welcome to PF :)

Looks as if your relevant equations are good enough to deal with this. One small expansion to ##\Sigma F = m\; a## is enough.
In other words : the force on the mass is 360 N - mg . Formula 1 gives the acceleration a.
And then use a in formula 2 to get Y. You may assume h = 0 at t = 0 (your Yi), and Vi is also 0 ("mass at rest").

Note: you will find that a is negative. Formulas stay true. So unless there is extra information, at time t=2 s the mass will be lower than at time t=0 s.
 
  • #3
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BvU, thank you for your answer. So, we are assuming there is no Normal force? The Normal force of the at rest mass should cancel the m.g. That means the "only" acting force in the mass is the 360 N one. With that said, a = 5.14 m/s².

After that, you are telling me to put "a" in formula 2. That doesn't seem possible to me, we are ignoring the gravity, i need something negative that tries to stop the object in the air.

Tell me if i said something wrong, thank you.
 
  • #4
haruspex
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So, we are assuming there is no Normal force?
The 360N is the normal force. Normal force just means a force at right angles to the surface of contact. In this case, the surface is horizontal, facing upwards, so the normal force is vertically up. One other force acts on the mass, and the net of the two forces produces the acceleration.
 
  • #5
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haruspex, thank you for your answer. The problem is, i'm the one that invented the problem, so, the 360 N force i mention is a "F" force that has nothing to do with action and reaction.
 
  • #6
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What does "is a "F" force that has nothing to do with action and reaction." mean?
You have something that applies a force of 360 N to the object, and you have earth applying a force of m*g to the object. Add both to get the total force on the object, that allows to find the acceleration.
 
  • #7
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mfb thank you for your answer. Let's assume for just one moment that there is no 360 N applied force. Where is the normal?
 
  • #8
haruspex
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mfb thank you for your answer. Let's assume for just one moment that there is no 360 N applied force. Where is the normal?
If there is no force applied by contact then the mass is in free fall. Your problem seems to be that you think there is some fundamental difference between an applied force and a normal force.
 
  • #9
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So by the second Newton's law, a standing mass in an horizontal plane where m.g is acting, there is no normal force generating a reaction?
 
  • #10
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If a mass is resting on a horizontal plane, then the horizontal plane applies a force of mg on the mass. This force can be called "normal force".
 
  • #11
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If a mass is resting on a horizontal plane, then the horizontal plane applies a force of mg on the mass. This force can be called "normal force".
Perfect. So with that said you now understand my concern, we have that situation, m.g acting to the center of earth and m.g acting upwards as a normal force. To that, shouldn't we add the 360 N applied force upwards?
 
  • #12
haruspex
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This force can be called "normal force".
I'd word that a bit differently, because I think this is behind Urmicoll's confusion. Urmicoll seems to be thinking of 'the normal force' as that force which arises when you put something on a horizontal surface, regardless of what else is going on. Rather, a normal force is that force applied by a surface which is sufficient to stop the object from penetrating the surface. So it is an example of an applied force, and any applied force that is perpendicular to the contact surface constitutes a normal force.
shouldn't we add the 360 N applied force upwards?
Urmicoll, it might help to be clearer about what you mean by then 'applying' a 360N force underneath. Describe a specific method.
 
  • #13
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I'd word that a bit differently, because I think this is behind Urmicoll's confusion. Urmicoll seems to be thinking of 'the normal force' as that force which arises when you put something on a horizontal surface, regardless of what else is going on. Rather, a normal force is that force applied by a surface which is sufficient to stop the object from penetrating the surface. So it is an example of an applied force, and any applied force that is perpendicular to the contact surface constitutes a normal force.

Urmicoll, it might help to be clearer about what you mean by then 'applying' a 360N force underneath. Describe a specific method.
Let's be more clearer then. The mass is a person, in each foot it has a cylinder that generates a force of 180 N upwards, i'm assuming that if each cylinder generates that force we can sum them and that gives us the 360 N upwards force.
 
  • #14
haruspex
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Let's be more clearer then. The mass is a person, in each foot it has a cylinder that generates a force of 180 N upwards, i'm assuming that if each cylinder generates that force we can sum them and that gives us the 360 N upwards force.
Fine. So the total upward force is 360N. The only other force acting on the person is gravity. If the person has a mass of more than about 36kg, acceleration will be downwards.
 
  • #15
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Fine. So the total upward force is 360N. The only other force acting on the person is gravity. If the person has a mass of more than about 36kg, acceleration will be downwards.
Perfect, so the conclusion is that a force that is major than m.g is needed in order to make the person move upwards?
 
  • #16
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New problem guys, the person of the cylinder answered me with some new data. With 200 psi, this cylinders make a force of 446 Kg of force. In newtons, that should be 446/0.1019 = 4377 N. If we multiply that by 2 cylinders, we have 8754 N of force applied.

F = 8754; m.g = 686

With that said and what you told me,

(8754 - 686)/70 = a = 115.25 m/s².

With that acceleration... the person will be 225 m in the air in 2 seconds. That looks kind of absurd to me, is this correct?
 
  • #17
haruspex
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(8754 - 686)/70 = a = 115.25 m/s².
Right.
With that acceleration... the person will be 225 m in the air in 2 seconds. That looks kind of absurd to me,
It isn't absurd, but it is rather higher than it should be. What formula are you using to get that distance?
 
  • #18
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Right.

It isn't absurd, but it is rather higher than it should be. What formula are you using to get that distance?
I'm using the formula i posted as "2". Putting acceleration to it instead of g.

Y(2) = (115.25)*(2)²/2 + Vi.t + Yi ---> Vi and Yi are both 0 ---> 115.25*2 = 230.5 m.

My bad, is even more.
 
  • #19
haruspex
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I'm using the formula i posted as "2". Putting acceleration to it instead of g.

Y(2) = (115.25)*(2)²/2 + Vi.t + Yi ---> Vi and Yi are both 0 ---> 115.25*2 = 230.5 m.

My bad, is even more.
Oh yes - sorry. I had used 1 second instead of 2. It's perfectly reasonable. The acceleration will be 5g.
In practice, drag will bring the height down somewhat.
 
  • #20
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I already asked to the manufacturer of the cylinders, i have to wait until tomorrow. The thing i'm unsure is if the force he told me is present in the cylinder or in the stroke of the cylinder. If the force is in the stroke, that is the height, if it is not, it will be other of course. (The stroke is like 4 cm).
 

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  • #21
haruspex
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With that acceleration... the person will be 225 m in the air in 2 seconds. That looks kind of absurd to me, is this correct?
Sorry - wasn't thinking.
Your calculation is assuming the cylinders are applying that force all the way up. That's a physical impossibility.
What you need to look at is the total energy that the cylinders can deliver.
 
  • #22
BvU
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Well, you made quite some progress here, and now haru brings back the energy thing.
But I think it might be good if urmi gets a clear picture of the forces first:

You have 2 cylinders placed on the floor and on top of each cylinder there's a foot.

At rest, on the top surface of each cylinder you have 0.5 mg of the man down and 0.5 mg from the cylinder up: steady state, a=0.
On the floor you have m'g of (m' = 0.5 man + cylinder) down and m'g from the floor up: steady state, a=0. ignore cylinder mass if you want.

Putting 200 psi on the cylinders apparently causes a total upward force of 8754 N at the top level of the cylinders. down is still 2x 0.5mg, so as Urmi calculated in #16, an upward acceleration of 115 m/s2. That's 12 g and will crush the guy's joints and lumbar discs. Something needs to be mitigated there, but never mind.

This upward acceleration lasts for 4 cm only (the stroke). So for about (s = 0.5 a t2) 26 ms.

Now (assuming a very stiff person), you can do energy things or you can do momentum things.
Momentum thing says ##F = {dp\over dt} \approx {\Delta p \over \Delta t}## so ##\Delta p = 231## kgm/s, yielding (p = mv) a takeoff speed of 3.3 m/s at the end of the stroke. A jump of 1.1 m (energy thing: 0.5 mv2 = mgh) . But, as I said: in reality the guy will collapse with crushed joints and be some 4 cm shorter for the rest of his life in a wheelchair :)
 
  • #23
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@BvU: I think your velocity to height conversion went wrong.

12g over 4cm is not so bad - especially if your legs are bent a bit, the torso won't accelerate significantly (which also means you do not jump, on the other hand). If you jump down 1.25m, your feet decelerate from 5m/s to 0m/s in less than a centimeter, which corresponds to over 100g.
 
  • #24
BvU
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Oh, and with h = v0t - 0.5 a t2, you'll find that a very stiff object leaving the ground at 3.3 m/s will be back there in 0.67 seconds! (but I admit that's ignoring the height of the expanded cylinders...:) )
 
  • #25
BvU
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@mfb:
Yes, sorry: the 3.3 m/s yields half that height.

I play volleyball. Some folks jump pretty high (albeit a bit less than 1.25 m) and they don't end up in wheelchairs (at least not immediately), because they also flex knees and ankles when landing. So only the toes suffer 100 g, but they can handle it. Landing on flat feet is very bad indeed.

I agree that with bent knees (or better: on toetips) a thud of 4 cm is a mere unpleasantness. Since the OP wondered where in the sky he would be after 2 seconds, I took it for granted he (she?) wants to do the right thing to get as high as feasible.... which is healthwise the wrong thing to do.

@Urmi: note that mfb reaches you the way out on a plate! You want to accelerate and survive intact, make sure the push isn't over a short stretch. Think of the circus where two strong folks jump arm in arm on a teeterboard and launch a slender one more than three man-heights into the air ! (the slender one helps by starting crouched and stretching his legs while being launched). Check out this video . Note how the board flexes. And how one of them lands safely from much more that 1.25 m at the end.
 
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