# Height of a Mass that receives a Force

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1. Dec 11, 2014

### Urmicoll

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Dec 11, 2014

### BvU

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### haruspex

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### Staff: Mentor

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### haruspex

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### Staff: Mentor

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### haruspex

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.

13. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### haruspex

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### Urmicoll

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. Dec 11, 2014

### haruspex

Right.
It isn't absurd, but it is rather higher than it should be. What formula are you using to get that distance?

18. Dec 11, 2014

### Urmicoll

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.

19. Dec 11, 2014

### haruspex

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. Dec 11, 2014

### Urmicoll

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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