Apparent weight on an elevator. Magnitude of its acceleration?

In summary, the person is standing on a bathroom scale in an elevator, with a weight of 160 lb but a reading of 123 lb. The elevator is accelerating downwards, causing the person's apparent weight to decrease. The magnitude of the acceleration can be calculated using the equation 740.625=55.792a, resulting in an acceleration of 13.27475 m/s2. However, this is not entirely accurate as the direction of acceleration must also be taken into account. By dividing the equation 9.8/160=x/123, we can find that the acceleration in the same direction as gravity is 7.234 m/s2.
  • #1
WPCareyDevil
30
0

Homework Statement


You are standing on a bathroom scale inside an elevator. Your weight is 160 lb, but the reading of the scale is 123 lb.
(a) What is the magnitude and direction of the acceleration of the elevator?
? m/s2
(b) Can you tell whether the elevator is speeding up or slowing down?
do not know
slowing down
speeding up


Homework Equations


weight=mg
ma=ma


The Attempt at a Solution



160lbs = 75.574kg

123lbs: 55.792kg

g= 9.8ms

obviously, the elevator is accelerating downwards, reducing his apparent weight.

his weight when on the ground is 740.625N.

So, 740.625=55.792a
a=13.27475. The question is asking for magnitude, but there is no x component so sqrt(a^2) would still be 13.27475

Obviously, this isn't correct. Can someone point me in the right direction?
 
Physics news on Phys.org
  • #2
WPCareyDevil said:

Homework Statement


You are standing on a bathroom scale inside an elevator. Your weight is 160 lb, but the reading of the scale is 123 lb.
(a) What is the magnitude and direction of the acceleration of the elevator?
? m/s2
(b) Can you tell whether the elevator is speeding up or slowing down?
do not know
slowing down
speeding up


Homework Equations


weight=mg
ma=ma


The Attempt at a Solution



160lbs = 75.574kg

123lbs: 55.792kg

g= 9.8ms

obviously, the elevator is accelerating downwards, reducing his apparent weight.

his weight when on the ground is 740.625N.

So, 740.625=55.792a
a=13.27475. The question is asking for magnitude, but there is no x component so sqrt(a^2) would still be 13.27475

Obviously, this isn't correct. Can someone point me in the right direction?

If it was slowing down while going up it could read the same. All you know is acceleration is less down on the scales.
 
  • #3
Ahh, you're right. That makes sense - I didnt think of that.


I tried using -13.27, and subtracting 9.8. What am I missing with the calculation?

Thank you for the tip!
 
  • #4
WPCareyDevil said:
Ahh, you're right. That makes sense - I didnt think of that.I tried using -13.27, and subtracting 9.8. What am I missing with the calculation?

Thank you for the tip!

You can divide by like equations.

9.8/160 = x/123

I know the units are silly but they cancel as long as your acceleration is to be in m/s2

The difference then with 9.8 must be the acceleration in the same direction as g.
 
  • #5
Thank you so very much!

I was able to figure it out after a bit of tweaking.

This place is the best =D. I've tried to go through answering peoples questions- but I find that I either cannot answer them well enough, or someone has already beat me to it. I'll try to give back as much as I can
 
  • #6
WPCareyDevil said:
Thank you so very much!

I was able to figure it out after a bit of tweaking.

This place is the best =D. I've tried to go through answering peoples questions- but I find that I either cannot answer them well enough, or someone has already beat me to it. I'll try to give back as much as I can

You're welcome. Glad that it helped. All contributions welcome.

Cheers.
 
  • #7
WPCareyDevil said:
So, 740.625=55.792a
a=13.27475.
Maybe I'm wrong, but I think it should be 546.762=75.574a so a=7.234.
 

Related to Apparent weight on an elevator. Magnitude of its acceleration?

1. What is apparent weight on an elevator?

Apparent weight on an elevator is the perceived weight experienced by a person inside an elevator. It is the combination of the actual weight of the person and the forces acting on the person due to the elevator's movement.

2. How is apparent weight related to the magnitude of elevator acceleration?

The apparent weight on an elevator is directly proportional to the magnitude of the elevator's acceleration. This means that as the elevator accelerates, the apparent weight of the person inside will also change accordingly.

3. How does the direction of elevator acceleration affect apparent weight?

The direction of elevator acceleration plays a crucial role in determining the apparent weight. If the elevator is accelerating upwards, the apparent weight will be greater than the actual weight. On the other hand, if the elevator is accelerating downwards, the apparent weight will be less than the actual weight.

4. What factors can affect the apparent weight on an elevator?

The apparent weight on an elevator can be affected by various factors such as the mass of the person, the direction and magnitude of elevator acceleration, and the presence of any external forces acting on the person (such as wind or turbulence).

5. How is the apparent weight calculated on an elevator?

The apparent weight on an elevator can be calculated using the formula: Apparent weight = Actual weight + (Mass x Acceleration). This takes into account the actual weight of the person and the forces acting on them due to the elevator's acceleration.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
12
Views
3K
  • Introductory Physics Homework Help
Replies
24
Views
7K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Classical Physics
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Back
Top