Elevator acceleration with mass

AI Thread Summary
The discussion revolves around calculating the scale reading for a woman standing in an accelerating elevator. The combined mass of the woman, elevator, and scale is 875 kg, and the force exerted by the cable is 9410 N. To find the scale reading during acceleration, it's essential to apply Newton's second law and consider the normal force acting on the woman. The acceleration of the elevator is determined to be approximately 10.754 m/s², which affects the scale reading. Analyzing both the elevator and the woman separately, while using free body diagrams, is crucial for accurate calculations.
tennisman03110
Messages
5
Reaction score
0
Woman stand on a scale in a moving elevator. Her mas is 60 kg, and combined mass of elevator and scale is additional 815 kg. From rest, elevator accelerates upwards. During Acceleration, the cable applies a force of 9410N. What does the scale read during acceleration. Formula or solution would be appreciated. Thanks.
 
Physics news on Phys.org
Show how you started, please.
 
Well I know F=mass*acceleration. So 9410= 875*A. A=10.754. I don't really know what the weight of the scale would read though at that time.
 
tennisman03110 said:
Well I know F=mass*acceleration. So 9410= 875*A. A=10.754.

You need to avoid the urge to plug numbers into equations without thinking about what it is that you are doing. Newton's second law says that the net force on an object is equal to the product of its mass and acceleration. The force exerted by the cable is not the net force on the elevator. You were given 3 weights, and you need to include them in your analysis.

You should start every single one of these problems by drawing a free body diagram, and then by applying Newton's 2nd law.

I don't really know what the weight of the scale would read though at that time.

Hint: If you're standing on a scale, it doesn't read your weight, but rather the normal force applied to you.
 
I don't really understand what the normal force would be. IS it 9410. If so then the weight would equal 9410 with no acceleration. I can't really figure out what to do to find this though, becasue their is acceleration and a lady on it.
 
You need to analyze the problem in two parts. First, analyze the elevator and then analyze the lady on the scale.

In the first part you will get the acceleration of the elevator. You'll use that in the second part because the acceleration of the lady is the same as the acceleration of the elevator (they move together).

And again, for each part, draw a free body diagram and apply Newton's 2nd law.
 
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...

Similar threads

Back
Top