Person in Elevator with Scale

In summary: ... at rest, if you accelerate up at 3 m/s^2 you'd get a=+3 (which makes you heavier in reference to the scale) ... if you're sitting still in a chair in a car, and the car then takes off at a constant speed, you'll be pushed back into the chair, but you won't get heavier or lighter in reference to the scale (unless of course it's one of those things that measures in newtons and not kg) ... when you're moving at a constant speed in a car, it's just like sitting still, you're just moving at a constant speed. in reference to the car you're not moving, in reference to the floor you're moving at constant speed, so
  • #1
aceXstudent
14
0
Hi, this problem gave me some trouble and I'm not sure if I did this correctly.

The problem reads: A 75-kg person steps onto an elevator with a scale. What does the scale read when the elevator (a) is at rest, (b) climbing at 3.0m/s, (c) falling at 3.0m/s, (d) accelerating upward at 3.0m/s^2, and (e) falling downward at 3.0m/s^2?

This is the work I've done so far:

(a) 75kg

(b) Would it be correct to assume that a=0 thus F=0 (F=ma), which means Fscale=mg? Would the scale read 75kg then?

(c) *same assumption as above*

(d) If F=ma, then Fscale-mg=ma (up as positive). Thus m=Fscale/a+g. So then I just plug in the numbers (Fscale=75kgx9.8m/s^2)

(e) *same as above*

Err... or am I just completely out of the zone?
 
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  • #2
Parts a to c are fine.

For part d, I'm a bit confused how you ended up with Fscale = 75x9.8 although your original analysis was correct (i.e. Fscale = ma + mg).

For part e, of course I guess you realize the opposite happens and you will feel lighter (i.e. Fscale = mg - ma).
 
  • #3
weight is a force, like, weight is mass times an acceleration ... the accerlation is due to the little wonder we call gravity.
The problem here is that most scales are set up to be consistent with this, like they measure the object with the force broken down .. then again, if the scale reads lbs then it isn't this way ... you can figure it in terms of net acceleration, or net force, either way you're going to come out to the same answers.

accel due to gravity is 9.81 m / s^s , btw ...

edit: didn't mean to sound like a jerk with the accel of grav thing, I kept reading mg as milligrams (wow I'm quick) ... i never really got into writing g as gravity .. if 9.8 is what you're given as the value of it then use it ... geographically a_g differs
 
  • #4
The textbook given to us used 9.8m/s^2 (nine-point-eight-meters-per-second-square) for their gravity, and our instructor uses the system of the textbook.

As for the problem, the textbook wants me to give the reading in kg... I didn't understand how to do this either. As for part (d), I factored m out of the equation and divided it by (a+g) because the problem wanted me to solve in kg...
 
  • #5
(d) accelerating upward at 3.0m/s^2, and (e) falling downward at 3.0m/s^2?

mezarashi said:
Fscale = ma + mg

I like to factor out the "m" to make it easier.
Fscale = m(a + g)
Fscale = 75kg(a + 9.8)

No need for 2 questions, for part (e) when you are falling the acceleration would be negative because you are really dealing with a(sub y), or acceleration along only the y-axis.

edit: Fscale units will be in Newtons, so I would just divide by gravity to get back to kg.
 
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  • #6
I'm confused... the book wants it in kg, a scalar quantity instead of a force.

EDIT: I feel so dumb to not get that...
 
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  • #7
edited my post for that ;D
 
  • #8
maki said:
(d) accelerating upward at 3.0m/s^2, and (e) falling downward at 3.0m/s^2?
mezarashi said:
Fscale = ma + mg
I like to factor out the "m" to make it easier.
Fscale = m(a + g)
Fscale = 75kg(a + 9.8)
No need for 2 questions, for part (e) when you are falling the acceleration would be negative because you are really dealing with a(sub y), or acceleration along only the y-axis.
edit: Fscale units will be in Newtons, so I would just divide by gravity to get back to kg.


Well, for part (e) it would actually have to be Fscale=m(g-a) because the person's mass reading can't be -98kg. It would be 52kg.
 
  • #9
Well, for part (e) it would actually have to be Fscale=m(g-a) because the person's mass reading can't be -98kg. It would be 52kg.

Fscale = m ( a + g )
Fscale = 75 ( -3 + 9.8)
Fscale = 75 ( 6.8 )
Fscale = 510 Newtons
510 / 9.8 = 52.04 kg

where do you get a negative...
 
  • #10
A lot of books use 9.8 , some books use 10 ... I always use 9.81 because that's how it was in my first book . As you go around the world to different elevations the 'constant' acceleration of gravity changes, because the formula for finding Grav (which you'll probably learn in your class if you haven't yet) assumes the planet is a sphere, ... one small discrepency comes to mind immediately.. its called Everest ... so use 9.8 if that's what your teach/prof uses.

And sorry If i confused you on my scale talk, its just that lbs is a force unit, so a scale that reads lbs is reading the mass times grav, the ones that tell you actual mass are just set to cancel the 9.8 .. like, they measure the weight.. and the number bar that its put on is all scaled as if it was divided by the 9.8 ... since it says "What does the scale read?" .. and not "What weight does the scale read?" or "What mass... ?" then you have to assume on the type of scale it actually is :-/

if that confuses , then don't worry about it .. eventually you'll get it .. its dumb .. it really is
 
  • #11
This is all arising just from the inconsistent use of coordinates. If up is positive and down is negative, stick with it. If you stay that Fscale is in the up direction, it is positive, if a is in the downward direction, it is negative. Which direction is gravity in this convention?

So whichever approach you use would be okay as long as your consistent.
 
  • #12
i just base the coordinate system off rest-state. at rest, the gravity effect/constant is 9.8, if you then fall with accel 3 m/s^2 that's down from where you were at rest so a=-3 (which makes you lighter in reference to the scale)
 

1. What is the purpose of the person in the elevator with a scale?

The purpose of the person in the elevator with a scale is to gather data on the weight distribution of the elevator and its passengers. This can help determine the safe weight capacity of the elevator and ensure that it is not overloaded.

2. How does the person in the elevator with a scale collect data?

The person in the elevator with a scale will use a handheld or portable scale to weigh themselves and other passengers in the elevator. They may also take note of any additional items or objects being brought into the elevator.

3. What factors can affect the weight distribution in an elevator?

The weight distribution in an elevator can be affected by the number of people in the elevator, their individual weights, and the placement of their weight within the elevator. Additional factors such as objects being carried, uneven distribution of weight, and elevator movement can also impact the weight distribution.

4. How does weight distribution affect elevator safety?

An imbalanced weight distribution in an elevator can potentially lead to malfunctions or accidents, such as the elevator getting stuck between floors or falling. By collecting data on weight distribution, the elevator can be safely loaded with the appropriate weight limit to prevent these issues.

5. Are there any regulations or standards for elevator weight distribution?

Yes, there are regulations and standards for elevator weight distribution set by organizations such as the American Society of Mechanical Engineers (ASME). These regulations ensure the safety and proper functioning of elevators by setting weight limits and guidelines for weight distribution.

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