Solve Elevator Problem with Weight & Acceleration on Moon

  • Thread starter Thread starter seizureboi
  • Start date Start date
  • Tags Tags
    Elevator
AI Thread Summary
The discussion focuses on calculating the apparent weight of a 65kg astronaut in a moon elevator under different conditions. For part (a), the astronaut's apparent weight when the elevator accelerates upward at 5 m/s² is calculated to be 219N. In part (b), to achieve an apparent weight of 500N, the required acceleration of the elevator is determined to be 9.32 m/s². A correction is noted regarding the direction of forces, emphasizing that the apparent weight and true weight must act in opposite directions. The calculations are confirmed to be correct with this understanding of force dynamics.
seizureboi
Messages
8
Reaction score
0

Homework Statement



The effect of gravity is approximately 1/6 of what it is on Earth. If one were to build an elevator on the moon, (a) what would be the apparent weight of a 65kg astronaut when the elevator accelerates upward at 5 m/s2? (b) What is the acceleration of the elevator if the apparent weight of the astronaut is 500N?

Homework Equations



F = ma
m = mass
m = 65 kg
w = weight
w = mg
a = acceleration
a = 5 m/s2
gM = 1.63 m/s2

The Attempt at a Solution



The equation I set up for (a) was: ma - mg = w
OR (65)(5) - (65)(1.63) = w
OR 325 - 106 = w
Finally, w = 219N

For (b) I had the same exact forumla: ma - mg = w
OR (65)a - (65)(1.63) = 500
OR 65a - 105.95 = 500
OR 65a = 605.95
Finally, a = 9.32 m/s2

Just wondering, again, if this is right or not. =)
 
Physics news on Phys.org
seizureboi said:

Homework Statement



The effect of gravity is approximately 1/6 of what it is on Earth. If one were to build an elevator on the moon, (a) what would be the apparent weight of a 65kg astronaut when the elevator accelerates upward at 5 m/s2? (b) What is the acceleration of the elevator if the apparent weight of the astronaut is 500N?

Homework Equations



F = ma
m = mass
m = 65 kg
w = weight
w = mg
a = acceleration
a = 5 m/s2
gM = 1.63 m/s2

The Attempt at a Solution



The equation I set up for (a) was: ma - mg = w
OR (65)(5) - (65)(1.63) = w
OR 325 - 106 = w
Finally, w = 219N

For (b) I had the same exact forumla: ma - mg = w
OR (65)a - (65)(1.63) = 500
OR 65a - 105.95 = 500
OR 65a = 605.95
Finally, a = 9.32 m/s2

Just wondering, again, if this is right or not. =)
Ouch, you let the plus and minus sign bite you. The acceleration is always in the direction of the net force. The apparent weight (the normal force) and the true weight must act in opposite directions.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top