Post by: TyErd on June 02, 2010, 04:41:28 pm
Post by: Stroodle on June 02, 2010, 05:01:36 pm
Post by: lachymm on June 03, 2010, 08:00:40 pm
Post by: kyzoo on June 03, 2010, 10:15:29 pm
Post by: kenhung123 on June 03, 2010, 10:44:40 pm
Post by: kenhung123 on June 03, 2010, 11:37:55 pm
Post by: m@tty on June 04, 2010, 12:08:12 am
Post by: kenhung123 on June 04, 2010, 12:17:28 am
Post by: m@tty on June 04, 2010, 12:21:46 am
Post by: kenhung123 on June 04, 2010, 12:28:36 am
How is that showing n=0? You are assuming thatAlright. Thanks matty. I think it is better to simply say that apparent weightlessness occurs when there is no reaction force and this happens when a body's acceleration is the same as its surroundings.
Post by: Chavi on June 04, 2010, 12:29:55 am
For these 4 mark questions I'd say for full marks you need to:
1) Define Weight, Apparent Weight, Apparent Weightlessness and True Weightlessness
2) Mention that the object is in free fall, so the Normal reaction force = 0
3) Prove the N=0 using calculations
4) Add that a gravitational force, g, is still acting on the object. Thus, there is only a sensation of weightlessness (as Apparent weight is measured by the Normal force), which differs from true weightlessness where g=0 (in deep space)
Post by: kenhung123 on June 04, 2010, 12:36:37 am
Post by: Chavi on June 04, 2010, 12:45:16 am
So how do I should N=0?
Net force = W - N
If Net Force = Weight, then the Normal force = 0
To understand this concept, consider yourself sitting in front of your PC/mac at time of reading, 1am. Your vertical accel = 0 (I'm assuming you're not floating up, or falling into a sinkhole :) ), so Net force = 0 [Newton's 2nd]. That implies that 0 = W - N, so N = W in this case, and you feel as heavy as the value for the magnitude of your normal force.
Now imagine yourself skydiving. You're falling at an accel of 9.8 m/s2, so your net force = 9.8m.
So 9.8m = W - N (and W = mg)
so N = 9.8m - 9.8m = 0.
Your normal reaction force is zero. This means that you are not in contact with any surfaces, and you feel weightless.
Hope that helps
Post by: kenhung123 on June 04, 2010, 07:21:36 am
Post by: kyzoo on June 04, 2010, 08:13:04 am
implies that the object is stationary on a flat surface.
For these 4 mark questions I'd say for full marks you need to:
1) Define Weight, Apparent Weight, Apparent Weightlessness and True Weightlessness
2) Mention that the object is in free fall, so the Normal reaction force = 0
3) Prove the N=0 using calculations
4) Add that a gravitational force, g, is still acting on the object. Thus, there is only a sensation of weightlessness (as Apparent weight is measured by the Normal force), which differs from true weightlessness where g=0 (in deep space)
Why would you have to define true weightlessness when it asks for apparent weightlessness?
Post by: Chavi on June 04, 2010, 02:08:14 pm
For these 4 mark questions I'd say for full marks you need to:
1) Define Weight, Apparent Weight, Apparent Weightlessness and True Weightlessness
2) Mention that the object is in free fall, so the Normal reaction force = 0
3) Prove the N=0 using calculations
4) Add that a gravitational force, g, is still acting on the object. Thus, there is only a sensation of weightlessness (as Apparent weight is measured by the Normal force), which differs from true weightlessness where g=0 (in deep space)
Why would you have to define true weightlessness when it asks for apparent weightlessness?
So that you show the VCAA examiner that you know the difference between these two easily confused concepts.
Post by: lachymm on June 04, 2010, 04:10:29 pm
Post by: m@tty on June 04, 2010, 05:51:58 pm
~ Object is in free-fall, there is no reaction force.
~ Weight is still present, but is not realised as reaction is zero.
~ Hence he/she is experiencing apparent weightlessness.
Are these really four mark questions? I've forgotten how many they were in the papers I've done..
@Chavi If there is a non-zero net force acts on a body then it is impossible for the body to be stationary. I was talking about net force, not reaction force.
And I still don't see how the 'calculations' actually help further your argument.