One-dimensional dynamics question

  • Thread starter Thread starter Ascendant78
  • Start date Start date
  • Tags Tags
    Dynamics
AI Thread Summary
The discussion revolves around a physics problem where a 60kg student lowers herself down 40m at a constant speed, and the force of the rope needs to be calculated. The initial calculation yielded a force of 465.5N, but the textbook answer is 490N, leading to confusion about the discrepancy. The participant questions whether the force of the rope should be considered for each side, suggesting that the total force would be double if accounting for both ropes. Responses indicate that the participant's approach is correct, and the discrepancy may stem from a potential error in the textbook or changes in parameters. The conversation highlights the challenges of understanding physics concepts and the reliability of educational materials.
Ascendant78
Messages
327
Reaction score
0

Homework Statement



A 60kg student lowers herself down 40m at a constant speed of 1.0 m/s. The chair has a mass of 35kg. Find the force of the rope. (See image)

aZxccET.jpg


Homework Equations



I was using force equations in attempts to solve for the problem.

The Attempt at a Solution



Fchair+Fstudent-Frope-Frope=m(a)
35(9.8)+60(9.8)-2Frope=95(0)
343+588-2Frope=0
2Frope=931
Frope=465.5N

Although the answer I got above was 465.5N, the answer listed in the back of the book was Frope=490N. I am at a complete loss as to what I am doing wrong here? I did a similar problem where the person was traveling upwards like this at a constant velocity and got it right. Not sure what changed here that I am not taking into consideration?

I am also curious after having done a similar problem, since the force is split into 2 parts, the force quantity for the rope would be the value for each rope individually, correct (one attached to chair and the one held by the student)? I just want to make sure I am understanding this correctly. Also, if that is the case, then if I were asked what the total force being applied to the rope was, it would be double the value to account for each side being pulled, correct?
 
Last edited:
Physics news on Phys.org
Your solution looks fine to me. Could be the student in question lost 5kg in the time between editions of the book :smile:
 
gneill said:
Your solution looks fine to me. Could be the student in question lost 5kg in the time between editions of the book :smile:

Lol, in this book, I would not be surprised. As you can tell from the image, this book is not exactly top of the line (Spiral Physics).

Anyway, thanks for letting me know because it was driving me absolutely crazy.
 
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