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

Doc Al

Quote by sophiecentaur

I realise 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.

sophiecentaur

Quote by Doc Al

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.

Doc Al

Quote by mutineer123

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.

Doc Al

Quote by sophiecentaur

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.

A.T.

Quote by sophiecentaur

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.

Quote by sophiecentaur

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.

sophiecentaur

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'.

xAxis

Quote by mutineer123

Quote by xAxis

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.

pgardn

Jesus...

I cant imagine whats 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.

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sophiecentaur Recognitions: Gold Member Science Advisor	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'.

pgardn	Jesus... I cant imagine whats 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.