A 60-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when:
a) the elevator is descending at a constant rate of 10 m/s
b) the elevator is accelerating downward a 2 m/s^2
c)the elevator is ascending and its speed is decreasing at the rate of 2 m/s^2, and
d) the elvator is accelerating upward at 2 m/s^2
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Physics problem to be solved please
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Our forum has been available for many years. In September 2014 we decided to switch it to read-only mode. Month after month we saw less posts with questions and answers from real people and more spam posts. We were spending more and more resources cleaning the spam until there were less them 1 legitimate message per 100 spam posts. Then we decided it's time to stop.
All the posts in the forum will be available and searchable. We understand there are a lot of useful information and we aren't going to remove anything. As for the new questions, you can always ask them on convert-me.com FaceBook page
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by shill » Fri Dec 23, 2005 6:16 am
If you want to do these problems on your own, read this page.
The acceleration due to gravity is 9.80665 m/s² [down].
a) Since the elevator is descending at a constant rate, the acceleration is zero. The net force will be 588.399 N, and the scale will read 60 kg.
b) (m*a)+f_gravity = (60. kg * -2.0 m/s²) + 588.399 N = 468.3993 N. Divide by acceleration due to gravity, therefore the scale will read approximately 47.8 kg.
c) This one is a little trickier. The elevator is ascending, but it is slowing down, so the acceleration is going in the opposite direction (downward). Since this acceleration (2.0 m/s² [down]) is exactly the same as in the previous question, the scale will read the same (47.8 kg).
d) Acceleration is 2.0 m/s² [up]. Use the same formula as in question b.
(m*a)+f_gravity/g = ((60. kg * 2.0 m/s²) + 588.399 N)/9.80665 m/s²) = approximately 72.2 kg.
I make no claims to the correctness of this advice. Use your own judgement to confirm this advice.
The acceleration due to gravity is 9.80665 m/s² [down].
a) Since the elevator is descending at a constant rate, the acceleration is zero. The net force will be 588.399 N, and the scale will read 60 kg.
b) (m*a)+f_gravity = (60. kg * -2.0 m/s²) + 588.399 N = 468.3993 N. Divide by acceleration due to gravity, therefore the scale will read approximately 47.8 kg.
c) This one is a little trickier. The elevator is ascending, but it is slowing down, so the acceleration is going in the opposite direction (downward). Since this acceleration (2.0 m/s² [down]) is exactly the same as in the previous question, the scale will read the same (47.8 kg).
d) Acceleration is 2.0 m/s² [up]. Use the same formula as in question b.
(m*a)+f_gravity/g = ((60. kg * 2.0 m/s²) + 588.399 N)/9.80665 m/s²) = approximately 72.2 kg.
I make no claims to the correctness of this advice. Use your own judgement to confirm this advice.
- shill
- Expert
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- Joined: Fri Feb 06, 2004 8:22 am
by Guest » Fri Dec 23, 2005 7:19 am
Mostly to be a smartass, note that we haven't established WHAT kind of a scale it is.
If it is a "dishonest weight, chock full of springs" scale, I agree with shill's method (I didn't check all the figures).
If it is a physicians' scale, or any type of "beam balance" scale, the counterweights will be equally affected as the girl, and the scale will indicate 60 kg. (the sensitivity may be slightly affected as "gravity" is higher or lower by about 20%)
If it is a "dishonest weight, chock full of springs" scale, I agree with shill's method (I didn't check all the figures).
If it is a physicians' scale, or any type of "beam balance" scale, the counterweights will be equally affected as the girl, and the scale will indicate 60 kg. (the sensitivity may be slightly affected as "gravity" is higher or lower by about 20%)
- Guest
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4 posts
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