Calculating the weight when faced with multiple accelerations

In summary: Ah,now I see well 2 forces,the one pulling him up and the one pulling him down (the question states friction and air resistance shouldn't be taken into...
  • #1
arhzz
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Homework Statement
A man tries to convince his wife with a trick that he has already lost a lot. He stands on a scale in an elevator. His mass is 95 kg. As soon as the elevator starts going down, he shows his wife the display.
a) The elevator accelerates at 0.2 g when leaving the floor. The scale is showing a mass of (blank) at this moment.That corresponds to a force of (blank)

b)Bad luck, but the wife also looks at the display when the elevator descends at a constant speed of 2.0 m / s. She sees a mass of (blank)
Relevant Equations
F = m*g
Hello! Now what I've done is first calculate the acceleration. a = 1,962 m/s^2. Than I tried looking at the solution simply through the Newtons laws (or axioms I don't know how to say it in english). Since the elevator is moving through the air the gravitation constant has to be taken into account.Than I've used this property; $$
\frac{m1}{m2} = \frac{a2}{a1}$$ Now I have one mass and one accelartion(the one I calculated) we are looking for m2 so we only need a2,I took a2 to be g since the way I see it we have 2 Forces,one where the elevator accelerates and the second the one where the elevator is getting pulled down by F = m*g. So after that I did this. $$ m2 = \frac{m1a1}{a2} kg$$ m2 comes out to be 19kg.Now this doesn't really seem realistic to me.I've looked at some other formulas and I've stumbled upon this one $$ a = \frac{m2}{m1+m2} * g $$ Now this should apply to my problem,I just don't know how to get an expression for m2,the algebra doesn't add up.My intuition tells me that somehow I should get to a form where m2 = m1 -19 = 76kg,and than we can easily calcualte the Force (with 76kg should be 745,56N).

Any insights?for b) I haven't tried anything yet,kind of stuck on a.Thank you!
 
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  • #2
I can't follow what you are doing here. We have only one mass: the mass of the man is ##m = 95kg##.

There are not two masses in the question.
 
  • #3
Okay maybe my statement is bad.But the main question is. We have a mass of 95kg. If we are moving upward with an acceleration of 0,2 *g, what will the mass be? So we need to calculate the second mass.
 
  • #4
arhzz said:
Okay maybe my statement is bad.But the main question is. We have a mass of 95kg. If we are moving upward with an acceleration of 0,2 *g, what will the mass be? So we need to calculate the second mass.
The question should ask for the weight, not the mass. Or the mass-equivalent. Scales don't measure mass, they measure force or mass-equivalent.

The question as stated is causing confusion.

I suggest you calculate the force and use the formula ##m_eg = F## (where ##F## is the force on a scale and ##m_e## is the mass-equivalent).
 
  • #5
PeroK said:
The question should ask for the weight, not the mass. Or the mass-equivalent. Scales don't measure mass, they measure force or mass-equivalent.

The question as stated is causing confusion.

I suggest you calculate the force and use the formula ##m_eg = F## (where ##F## is the force on a scale and ##m_e## is the mass-equivalent).
I have changed the mass into weight in the title,to try clear up the confusion.But what exactly is the mass equivalent? The F=mg comes familiar but not the term "mass equivalent" ?
 
  • #6
arhzz said:
I have changed the mass into weight in the title,to try clear up the confusion.But what exactly is the mass equivalent? The F=mg comes familiar but not the term "mass equivalent" ?
If the scales read ##kg##, then that is the mass that would be needed with normal gravity to show that reading. The best term for this is probably "mass-equivalent". I don't know if there is a better term. But, "mass" is wrong.
 
  • #7
PeroK said:
If the scales read ##kg##, then that is the mass that would be needed with normal gravity to show that reading. The best term for this is probably "mass-equivalent". I don't know if there is a better term. But, "mass" is wrong.
Okay so If I understood this correctly.We need to calculate the force F with m 95kg and the acceleration 0,2 *g.Than we need to find for what mass equivalent when dealing with simply g and the same force we get the reading of 95kg?
 
  • #8
arhzz said:
Okay so If I understood this correctly.We need to calculate the force F with m 95kg and the acceleration 0,2 *g.Than we need to find for what mass equivalent when dealing with simply g and the same force we get the reading of 95kg?
First, you need to analyse the forces on the man. Have you heard of a free-body diagram?
 
  • #9
PeroK said:
First, you need to analyse the forces on the man. Have you heard of a free-body diagram?
No,not really.
 
  • #12
arhzz said:
Ah,now I see well 2 forces,the one pulling him up and the one pulling him down (the question states friction and air resistance shouldn't be taken into account)
Okay, but that's not very descriptive. What about:

A downward gravitational force of ##mg## and an upward normal force of ##N## from the scales.

What is the net force? And how does that relate to acceleration?
 
  • #13
PeroK said:
Okay, but that's not very descriptive. What about:

A downward gravitational force of ##mg## and an upward normal force of ##N## from the scales.

What is the net force? And how does that relate to acceleration?
Well my understanding of the net force is that it is the sum of all forces that act upon an object.Since the direction of the gravitational force is pointing down (its pulling us to the ground) it should be -Fd and the upward force is pulling so +Fup. Now the get the exact value of the netforce I would need to calculate both of the forces and than add them up(in these case subract them).The way net force relates to accelration is the usual a = F/m if I am not mistaken, or were you referring to something else?
 
  • #14
arhzz said:
Well my understanding of the net force is that it is the sum of all forces that act upon an object.Since the direction of the gravitational force is pointing down (its pulling us to the ground) it should be -Fd and the upward force is pulling so +Fup. Now the get the exact value of the netforce I would need to calculate both of the forces and than add them up(in these case subract them).The way net force relates to accelration is the usual a = F/m if I am not mistaken, or were you referring to something else?
What does that all mean for this question?
 
  • #15
Well it means I should start implementing this by using math.Basically get the netforce,and than out of the net force we should be able to get the mass.I'll try to solve it and let you know how it went.

Thank you!
 
  • #16
Okay this is what I've got. So for the upward force (Fp) I have this;
Fp = mg = 931,95 N

For the downward force(Fd) Note: a is 0,2*g
-Fd = ma = -186,39 N

Now the net force (F)
F = Fp - Fn = 745,56

And that is also the righr answer,according to the solutions. Now here is the part I don't really understand. As we've discussed before about the relation to acceleration and net force I've done this to get the mass.
$$m =\frac{F}{a} $$ where a is again 0,2*g and I get the mass of 380kg which is obvioulsy wrong.I've than tried putting a as simply g and not 0,2*g and it worked,I got the right result (76kg).Now thing is I don't understand how.In particullar, if we are looking at the weight that appers when moving downwards,wouldnt it make sense to use the acceleration that happens with that movement? Because we are moving down but with an accelration of 0,2*g and not g. What am I missing here?
 
  • #17
arhzz said:
Okay this is what I've got. So for the upward force (Fp) I have this;
Fp = mg = 931,95 N

For the downward force(Fd) Note: a is 0,2*g
-Fd = ma = -186,39 N

Now the net force (F)
F = Fp - Fn = 745,56

And that is also the righr answer,according to the solutions. Now here is the part I don't really understand. As we've discussed before about the relation to acceleration and net force I've done this to get the mass.
$$m =\frac{F}{a} $$ where a is again 0,2*g and I get the mass of 380kg which is obvioulsy wrong.I've than tried putting a as simply g and not 0,2*g and it worked,I got the right result (76kg).Now thing is I don't understand how.In particullar, if we are looking at the weight that appers when moving downwards,wouldnt it make sense to use the acceleration that happens with that movement? Because we are moving down but with an accelration of 0,2*g and not g. What am I missing here?
I think you know the answer: the reading on the scales must be ##0.8 \times 95 kg = 76 kg##.

But, your physics is not right and you're not able to show this formally. There's a lot you need to sort out.

First, I'd take the direction of motion (downwards) as positive, so we have: $$F_{net} = F_g - N = mg - N$$ where ##N## is the normal force between the man and the scales.

Next, you know that the net force is related to the acceration by: $$F_{net} = ma = 0.2mg$$ And if you put these two equations together you can find the normal force, ##N##.
 
  • #18
Okay,I know the answers that isn't the problem,I just don't understand how to get to those answers,or the steps that are made to get to them.Now for the normal force I did your formula (calculated it) and I got the same exact result.Maybe my variable names or simply the way I put my answer may be off putting(I did it over my phone so barely any LaTeX) but I 100% understood your guidance and know what I needed to do to get to the normal force.The thing that is confusing me is the following;
How do we get to the 76kg? How did you figure out that you needed to multiply m by 0,8?? Do we even need the normal force to calculate the mass? If yes than how would I do it?
 
  • #19
arhzz said:
So for the upward force (Fp) I have this;
Fp = mg = 931,95 N
The first thing to understand is how a set of bathroom scales works. E.g. if it contained a beam balance and worked by shifting a fixed mass along the beam until the torque balanced the weight placed on top would it show different numbers as the elevator accelerates?
How do they really work? What do they directly measure? What does that tell you about Fp?
 
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  • #20
arhzz said:
How do we get to the 76kg? How did you figure out that you needed to multiply m by 0,8?? Do we even need the normal force to calculate the mass? If yes than how would I do it?
Suppose the scales normally show a reading of ##m \ kg##. That's represents a force of ##mg##. If the whole system accelerates downwards at ##0.1g##, then 10% of the gravitational force is used for acceleration, leaving 90% of the force on the scales, so a reading of ##0.9m \ kg##. If the acceleration increases, then the force and reading will reduce proportionally. If the acceleration is ##g## then there is no force or reading on the scales.

So, you can write down the answer to this problem without ant real calculation, just by a physical understanding of what's happening.

That said, it's always a good exercise to apply formal methods to a problem to which you know the answer. Do a free-body diagram, write down the force equation, solve for any unknowns.

The final piece in the jigsaw is that if the force on the scales is ##N##, then the reading is ##m_e = N/g##.

If you using the scales normally, then ##N = mg## and ##m_e = m##. I.e. the reading on the scales (mass equivalent, or whatever you want to call it) is ##m##. But, if the force is more or less than ##mg##, then the reading is more or less than ##m##.

Note that it may be slightly pedantic not to call the reading on the scale a "mass", but it's always worth being precise in physics.
 
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  • #21
PeroK said:
Suppose the scales normally show a reading of ##m \ kg##. That's represents a force of ##mg##. If the whole system accelerates downwards at ##0.1g##, then 10% of the gravitational force is used for acceleration, leaving 90% of the force on the scales, so a reading of ##0.9m \ kg##. If the acceleration increases, then the force and reading will reduce proportionally. If the acceleration is ##g## then there is no force or reading on the scales.

So, you can write down the answer to this problem without ant real calculation, just by a physical understanding of what's happening.

That said, it's always a good exercise to apply formal methods to a problem to which you know the answer. Do a free-body diagram, write down the force equation, solve for any unknowns.

The final piece in the jigsaw is that if the force on the scales is ##N##, then the reading is ##m_e = N/g##.

If you using the scales normally, then ##N = mg## and ##m_e = m##. I.e. the reading on the scales (mass equivalent, or whatever you want to call it) is ##m##. But, if the force is more or less than ##mg##, then the reading is more or less than ##m##.

Note that it may be slightly pedantic not to call the reading on the scale a "mass", but it's always worth being precise in physics.

Ahhh ,okay now it all makes sence.The fact that we are using only 20% of the gravitational force is the part that made all the confusion to me.Now it makes perfect sense.This was actually what I was trying to achieve,since the first time around i solved this simply by trying out all of the methods and I figured out simply by the way the question is phrased what the answers must be and I got the full point score on the exercise but I wasnt happy that I understood how to get there without doing some "educated guessing".

Thank you!
 

Related to Calculating the weight when faced with multiple accelerations

What is the formula for calculating weight when faced with multiple accelerations?

The formula for calculating weight when faced with multiple accelerations is W = m x (a1 + a2 + a3 + ...), where W is the weight, m is the mass, and a1, a2, a3, etc. are the accelerations acting on the object.

How do I determine the direction of the weight when faced with multiple accelerations?

The direction of the weight when faced with multiple accelerations is always towards the center of the Earth. This is because weight is a force that is caused by the gravitational pull of the Earth on an object.

Can I use the same formula to calculate weight for objects with different masses?

Yes, you can use the same formula to calculate weight for objects with different masses. The only difference will be the value of the mass (m) in the formula.

What happens to the weight when there are multiple accelerations acting in opposite directions?

When there are multiple accelerations acting in opposite directions, the weight will be the difference between the two accelerations. For example, if there is an acceleration of 5 m/s² upwards and an acceleration of 3 m/s² downwards, the weight will be 2 m/s² upwards.

Can I use this formula to calculate weight in other planets or objects with different gravitational pulls?

Yes, you can use this formula to calculate weight in other planets or objects with different gravitational pulls. However, you will need to use the gravitational pull of that specific planet or object in place of the acceleration due to gravity (g) in the formula. For example, if you want to calculate the weight of an object on Mars, you will use the acceleration due to gravity on Mars (3.71 m/s²) instead of the acceleration due to gravity on Earth (9.81 m/s²).

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