Doubt in resolving forces of a man in an elevator(normal force and weight)

In summary: X -9.8 the correct way to express the weight as a vector quantity?In summary, the correct way to express weight as a vector quantity is -mg, where g is the magnitude of the acceleration due to gravity. This is because weight is a downward force and using the convention of up as positive, it would be expressed as -mg. The force law is best expressed as a vector equation, with the sum of forces equaling mass times acceleration. To properly solve problems, it is important to first write down the force law vectorially and then resolve the signs accordingly.
  • #1
mutineer123
93
0
(Before I start, due to the use of vectors let me state I consider up to be positive and down to be negative)
Imagine a scenario, of a man in an elevator. And the elevator is accelerating upwards. Now in this case the Normal force > weight. So Normal Force - Weight =mass X acceleration. If you think this is right, why? Shouldn't it become Normal Force + Weight = mass X acceleration, because for weight(mg) the g will be -9.8 and hence -m X -9.8, will give you a positive value of
weight. I know its prolly wrong, because it would be insane to add Force and weight when their difference is what gives me the acceleration upwards, but then again why exactly is -9.8 wrong?
 
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  • #2
Mass is scalar.
How mass becomes -m?
 
  • #3
mutineer123 said:
So Normal Force - Weight =mass X acceleration. If you think this is right, why? Shouldn't it become Normal Force + Weight = mass X acceleration, because for weight(mg) the g will be -9.8 and hence -m X -9.8, will give you a positive value of
weight. I know its prolly wrong, because it would be insane to add Force and weight when their difference is what gives me the acceleration upwards, but then again why exactly is -9.8 wrong?
The best way to see it is as a vector equation:
ƩF = Normal Force + Weight = ma
Where 'normal force' and 'weight' are vectors.

Let the magnitude of the normal force be called N; the magnitude of the weight is mg, where g = 9.8 m/s^2. (FYI: g does not have a minus sign!) Now using your sign convention, we can rewrite that vector equation as:
ƩF = N -mg = ma
or N = mg + ma
 
  • #4
azizlwl said:
Mass is scalar.
How mass becomes -m?

it is -(m x -g) which will become +mg
 
  • #5
Doc Al said:
The best way to see it is as a vector equation:
ƩF = Normal Force + Weight = ma
Where 'normal force' and 'weight' are vectors.

Let the magnitude of the normal force be called N; the magnitude of the weight is mg, where g = 9.8 m/s^2. (FYI: g does not have a minus sign!) Now using your sign convention, we can rewrite that vector equation as:
ƩF = N -mg = ma
or N = mg + ma


(FYI: g does not have a minus sign!)
Why not?? It should, considering gravity is acting downwards!
 
  • #6
mutineer123 said:
(FYI: g does not have a minus sign!)
Why not?? It should, considering gravity is acting downwards!
The constant g is a magnitude, not a vector.
 
  • #7
mutineer123 said:
(FYI: g does not have a minus sign!)
Why not?? It should, considering gravity is acting downwards!
As A.T. explained, g is just the magnitude of the acceleration due to gravity--it does not have a sign. The full acceleration due to gravity with its appropriate sign would be -g. (Using up as positive.)

The weight of the man is a downward force of magnitude mg. So using your sign convention you'd express that as -mg.
 
  • #8
Hmm..Usually in these situations when many people try to help, simple things get confusing.
The OP chose the convention that up is positive. The forces are: normal-N, weight-W.
Normal force acts on the body up, so N is positive. Weight acts downwards, so W is negative. And as the mass is always positive scalar, g is negative.
So W=-9.81m.
So I think, as Doc Al already pointed out, the best way to do similar problems is to first write down the force law vectorialy. And there there is a sum of vectors.
Than as a second step you resolve the signs
 
  • #9
xAxis said:
Hmm..Usually in these situations when many people try to help, simple things get confusing.
The OP chose the convention that up is positive. The forces are: normal-N, weight-W.
Normal force acts on the body up, so N is positive. Weight acts downwards, so W is negative. And as the mass is always positive scalar, g is negative.
So W=-9.81m.
So I think, as Doc Al already pointed out, the best way to do similar problems is to first write down the force law vectorialy. And there there is a sum of vectors.
Than as a second step you resolve the signs

You are saying W=(-9.81m). Right? So now If i plug it into the equatio of N-W=ma won't I end up with N+W=ma ?? And maybe I know this already, but what exactly is the force law?
 
  • #10
Doc Al said:
As A.T. explained, g is just the magnitude of the acceleration due to gravity--it does not have a sign. The full acceleration due to gravity with its appropriate sign would be -g. (Using up as positive.)

The weight of the man is a downward force of magnitude mg. So using your sign convention you'd express that as -mg.

he full acceleration due to gravity with its appropriate sign would be -g. So why are we not using the this? If it were then we would get (-W), and if we plug this value into the equation
N-W= ma we would get N-(-W)=ma which would become N+W=ma. My method is to first find why the elevator is accelerating and hence get an appropriate equation(which I have done:N-W= ma). My next step is to plug in those values(with their vectorial signs), here is where I am stuck, as it become N + W.
 
  • #11
mutineer123 said:
(Before I start, due to the use of vectors let me state I consider up to be positive and down to be negative)
Imagine a scenario, of a man in an elevator. And the elevator is accelerating upwards. Now in this case the Normal force > weight. So Normal Force - Weight =mass X acceleration. If you think this is right, why? Shouldn't it become Normal Force + Weight = mass X acceleration, because for weight(mg) the g will be -9.8 and hence -m X -9.8, will give you a positive value of
weight. I know its prolly wrong, because it would be insane to add Force and weight when their difference is what gives me the acceleration upwards, but then again why exactly is -9.8 wrong?

Start with Newton's second law of motion: the sum of vertical forces is equal to mass times the acceleration. Note that this is a vector sum, which means that for a 1D problem, you have to take the sign of the force (up being positive and down being negative) into account.

One of the points that others have been trying to make to you in this thread is that most people define the constant g as g = +9.81 m/s2, in which case the acceleration due to gravity is -g, and the weight is -mg. Keeping this in mind:

ƩF = ma

Fnorm + Fgrav = ma

N - mg = ma

N = ma + mg

The *magnitude* of the normal force is equal to the *magnitude* of the weight plus the magnitude of the net force.

EDIT: If you want to use the notation "W" for Fgrav, and N for Fnorm, that's fine. I myself switched to N halfway through to save typing.

EDIT: I typed this before seeing that Doc Al typed nearly the exact same thing.
 
  • #12
mutineer123 said:
You are saying W=(-9.81m). Right? So now If i plug it into the equatio of N-W=ma won't I end up with N+W=ma ?? And maybe I know this already, but what exactly is the force law?

F1+F2=ma is the standard equation if we do not know the direction of the vector.
You shouldn't start with N-W=ma
If W is in direction of N, which means added force, your equation still subracting.
It means your equation N-W=ma to be standard form is wrong.
Substituting N and F2=-W for the case on acceleration upward.
N+(-W)=ma
 
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  • #13
mutineer123 said:
he full acceleration due to gravity with its appropriate sign would be -g. So why are we not using the this? If it were then we would get (-W), and if we plug this value into the equation
N-W= ma we would get N-(-W)=ma which would become N+W=ma. My method is to first find why the elevator is accelerating and hence get an appropriate equation(which I have done:N-W= ma). My next step is to plug in those values(with their vectorial signs), here is where I am stuck, as it become N + W.

How about you just use forces as vectors. The net force, ma, is up because he is acclerating up. This is a positive number. Now what were the actual forces involved in accelerating the man up. Well, the Earth is always pulling down on him with a force of mg. So mg is down. The normal force from the elevator floor on the man is up, so N is up positive. If you have one vector up, N, and another vector down, mg, what is the resultant vector?

Its ma which is up. If you add the forces acting on the man you get N + (-mg) = ma If he was accelerating down you would get N + (-mg) = -ma. If he is not accelerating you get N + (-mg) = 0 oh and mg = W

You are adding the Force vectors that are pointing in opposing directions on the man and getting a resultant that also has magnitude and direction.

If you write N + W you are basically stating that both the normal force and weight are acting up? that's not what is happening.

You might be best served by drawing a free body diagram and just work with forces as vectors.

I apologize to others that have basically stated the same thing.
 
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  • #14
A.T. said:
The constant g is a magnitude, not a vector.

infact g is a vector since its acceleration to gravity.
 
  • #15
Puneeth423 said:
infact g is a vector since its acceleration to gravity.
The constant g is the magnitude of that vector.

That is how g is generally defined in those formulas. It doesn't make sense to have the direction, which depends on an arbitrary axis orientation choice, included in a constant. Because then it is not a constant anymore, but depends on your axis orientation.
 
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  • #16
mutineer123 said:
(Before I start, due to the use of vectors let me state I consider up to be positive and down to be negative)
Imagine a scenario, of a man in an elevator. And the elevator is accelerating upwards. Now in this case the Normal force > weight. So Normal Force - Weight =mass X acceleration. If you think this is right, why? Shouldn't it become Normal Force + Weight = mass X acceleration, because for weight(mg) the g will be -9.8 and hence -m X -9.8, will give you a positive value of
weight. I know its prolly wrong, because it would be insane to add Force and weight when their difference is what gives me the acceleration upwards, but then again why exactly is -9.8 wrong?

Basic mistake you are doing is you are considering the direction of weight as negative two times. You are taking weight as negative in equation(N-W=ma) because the acceleration(g) in weight formulae(mg) is negative. When you take g already negative in representing weight how can you take again g as negative. Remember weight and g are same except that they change in magnitude due to multiplication of mass m, which have magnitude but no direction.
 
  • #17
A.T. said:
The constant g is the magnitude of that vector.

That is how g is generally defined in those formulas. It doesn't make sense to have the direction, which depends on an arbitrary axis orientation choice, included in a constant. Because then it is not a constant anymore, but depends on your axis orientation.

I realize we are all starting from different standpoints but there is no way that the direction of the Weight force (mg) is in doubt and, as m is definitely a scalar, then g must involve direction - i.e. it must be a vector. Also, g is also an acceleration, which, by definition, is a Vector.

The only way an individual can sort this out in their minds is to draw a diagram of the forces involved and rigorously think in which direction they act. The equation of motion falls out of it. But you need to remember that the elevator is NOT in equilibrium if it is accelerating so you cannot write an equation based on that assumption.

This is precisely what you need to do and you need to be strictly consistent, once you have chosen the reference frame. If you are, then your answer will be correct. You can confirm what I say by doing the sums using 'up=positive' and 'down=positive'. You will get the same answer if you are rigorous.
 
  • #18
sophiecentaur said:
I realize we are all starting from different standpoints but there is no way that the direction of the Weight force (mg) is in doubt and, as m is definitely a scalar, then g must involve direction - i.e. it must be a vector. Also, g is also an acceleration, which, by definition, is a Vector.
The standard convention is to have g stand for the magnitude of the acceleration due to gravity. It has no direction; it's a constant, not a vector.

The acceleration of a falling object is a vector with magnitude g and direction down.

The weight of an object is a vector with magnitude mg and direction down.
 
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  • #19
Doc Al said:
The standard convention is to have g stand for the magnitude of the acceleration due to gravity. It has no direction; it's a constant, not a vector.

The acceleration of a falling object is a vector with magnitude g and direction down.

The weight of an object is a vector with magnitude mg and direction down.

I can't disagree totally but somewhere along the line you need a direction / vector. If you are doing a ballistics problem, would you not put an arrow on the 'g' acting on the projectile? In the SUVAT approach, wouldn't the value of acceleration be entered as g or -g depending on which way up you happened to be working? The fact is that you have to be aware of the direction if you want the right answer. 9.81ms-1 is not enough.
Perhaps I'm just saying that it could be confusing to treat g as positive, automatically and not be constantly aware of sign. But, if the 'm' in 'ms-1' is only distance and not displacement, by definition, then fair enough.
 
  • #20
mutineer123 said:
he full acceleration due to gravity with its appropriate sign would be -g.
Sure. The acceleration of an object in free fall would be, using your sign convention, -g.
So why are we not using the this?
There's nothing in free fall in this problem. The only appearance of g in your problem is to give you the magnitude of the weight, mg.
If it were then we would get (-W), and if we plug this value into the equation
N-W= ma
we would get N-(-W)=ma which would become N+W=ma. My method is to first find why the elevator is accelerating and hence get an appropriate equation(which I have done:N-W= ma). My next step is to plug in those values(with their vectorial signs), here is where I am stuck, as it become N + W.
As long as you insist on using the equation N-W = ma you'd better learn what the symbols stand for. In that equation, W stands for the magnitude of the weight. The downward direction is already incorporated in the minus sign.

Much better, in my opinion, is to learn the vector form of Newton's 2nd law as I describe in my first post. Then you'd really understand how things work.
 
  • #21
sophiecentaur said:
I can't disagree totally but somewhere along the line you need a direction / vector.
Of course.
If you are doing a ballistics problem, would you not put an arrow on the 'g' acting on the projectile?
Of course not. g is a constant. The acceleration, a, would be written as [itex]\vec{a}[/itex]. The magnitude of that vector would be g. Using the symbol g to represent the acceleration vector would be incredibly confusing.
In the SUVAT approach, wouldn't the value of acceleration be entered as g or -g depending on which way up you happened to be working?
Of course.
The fact is that you have to be aware of the direction if you want the right answer. 9.81ms-1 is not enough.
Of course.
Perhaps I'm just saying that it could be confusing to treat g as positive, automatically and not be constantly aware of sign. But, if the 'm' in 'ms-1' is only distance and not displacement, by definition, then fair enough.
One certainly needs to be aware of the direction of any vector quantity. Some books--like Halliday and Resnick--take great care to explicitly warn the student to not substitute -9.8 m/s^2 for g.
 
  • #22
sophiecentaur said:
there is no way that the direction of the Weight force (mg) is in doubt
mg is the magnitude of the weight force. It doesn't have a direction. To get the weight force vector according to your specific coordinate system, you have to multiply your specific downwards unit vector with mg.

sophiecentaur said:
Also, g is also an acceleration, which, by definition, is a Vector.
The general constant g=9.81m/s2 is by definition the magnitude of an acceleration vector.
 
  • #23
OK. We've gone over this enough. I think we are arguing more about notation than anything else. Yours seems to be more formal and 'correct'.
 
  • #24
mutineer123 said:
xAxis said:
Hmm..Usually in these situations when many people try to help, simple things get confusing.
The OP chose the convention that up is positive. The forces are: normal-N, weight-W.
Normal force acts on the body up, so N is positive. Weight acts downwards, so W is negative. And as the mass is always positive scalar, g is negative.
So W=-9.81m.
So I think, as Doc Al already pointed out, the best way to do similar problems is to first write down the force law vectorialy. And there there is a sum of vectors.
Than as a second step you resolve the signs
You are saying W=(-9.81m). Right? So now If i plug it into the equatio of N-W=ma won't I end up with N+W=ma ?? And maybe I know this already, but what exactly is the force law?
That's exactly what has been pointed out. The force law says that when more forces act on the body, it moves as if only one force acts on it. And that force is their resultant. That's why the equation of motion is ƩFi = ma. You don't subtract forces as you did in N-W=ma. You can check that this way, you will get the same result with any sign convention.
 
  • #25
Jesus...

I can't imagine what's going to happen when this poor person tries to get an equation for an attwood machine... or a person standing in a device attached to a pulley with the person holding one rope of the pulley.

Free body diagram and then learning how to add vectors after establishing which direction is positive and negative suggested again.
 

1. What is "normal force" in the context of resolving forces of a man in an elevator?

The normal force is the force exerted by a surface on an object that is in contact with it. In the context of an elevator, it refers to the force that the elevator floor exerts on the person standing on it.

2. How does the normal force change when a person is standing on a moving elevator?

The normal force changes based on the acceleration of the elevator. If the elevator is moving with a constant velocity, the normal force will be equal to the person's weight. However, if the elevator is accelerating or decelerating, the normal force will change accordingly.

3. How is the normal force related to the weight of a person in an elevator?

The normal force is equal in magnitude and opposite in direction to the weight of the person. This means that if the person's weight increases, the normal force will also increase to balance it out.

4. What happens to the normal force when a person is in a free-falling elevator?

In a free-falling elevator, both the person and the elevator are accelerating towards the ground at the same rate. This means that the normal force will be zero, as there is no contact between the person and the elevator floor.

5. How does the normal force affect the overall force exerted on a person in an elevator?

The normal force is just one of the forces acting on a person in an elevator. It is important to consider all other forces, such as weight, friction, and air resistance, to determine the overall force exerted on the person. The normal force helps to counteract the weight of the person, allowing them to remain in contact with the elevator floor.

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