How Does Elevator Physics Affect the Work Done by a Cable?

Click For Summary
SUMMARY

The discussion centers on the physics of work done by a cable on an elevator system, specifically analyzing a 0.250 kg block of cheese within a 900 kg elevator. The participant critiques the book's solution regarding the normal force (##F_N##) and its treatment as an internal force in the system. The correct approach involves calculating the tension in the cable and recognizing that ##F_N## cancels out in the overall force analysis. The participant concludes that the book's method is flawed due to its incorrect handling of internal forces.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the work-energy principle
  • Knowledge of force analysis in systems
  • Ability to apply kinematic equations
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn about tension forces in pulley systems
  • Explore advanced applications of Newton's third law
  • Investigate the implications of internal versus external forces in physics problems
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of elevator systems and force interactions.

titansarus
Messages
62
Reaction score
0

Homework Statement


Question from fundamental of physics, Halliday Resnick Walker
In Figure below, a ##m=0.250## kg block of cheese lies on the floor of a ##M=900 kg## elevator cab that is being pulled upward by a cable through distance ##d1 =2.40 m## and then through distance ##d2 = 10.5 m##. (a) Through d1, if the normal force on the block from the floor has constant magnitude ##F_N## = 3.00 N, how much work is done on the cab by the force from the cable? (b) Through d2,if the work done on the cab by the (constant) force from the cable is 92.61 kJ, what is the magnitude of ##F_N##?
halliday phy.png

Homework Equations


##W = F d## (in 1 dimension)
##\Sigma F = m a##
##\vec F_{12} = - \vec F_{21}##

3. The Attempt at a Solution

I don't have problem at the solution itself, My problem is with the solution of the book.
Answer of the book:
answer halliday.png
My solution: ##F_N - mg = m a## (##*## equation) and we get ##a = 2.2 m/s^2## now for the whole system, I write ##F - (m+M)g = (m+M)a## (##**## equation) and by substituting, I get ##F = 10800 N##. The book also get the same approximate answer but It wrote ##F + F_N - (m+M) g = (m+M) a##. I think these ##F_N## part is wrong because when we are speaking about the whole system ##F_N## and its reaction (Newton 3rd Law) cancel each other. (i.e. they are Internal Forces for the system). Also for the second part, I think we must find F with ##W = F d## and then find ##a## using (##**## equation) and then find ##F_N## with (##*## equation). Is my solution correct?
 

Attachments

  • halliday phy.png
    halliday phy.png
    7.1 KB · Views: 632
  • answer halliday.png
    answer halliday.png
    25 KB · Views: 662
Physics news on Phys.org
titansarus said:
I think these ##F_N## part is wrong because when we are speaking about the whole system ##F_N## and its reaction (Newton 3rd Law) cancel each other. (i.e. they are Internal Forces for the system). Also for the second part, I think we must find F with ##W = F d## and then find ##a## using (##**## equation) and then find ##F_N## with (##*## equation). Is my solution correct?
You are correct.
 
  • Like
Likes   Reactions: titansarus
The error in the text is canceled by the fact that it then computes the work done on cab+cheese, not just work on the cab.
The whole would be fine if F were defined as tension in the cable -FN.
 

Similar threads

Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
14
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
9K
  • · Replies 5 ·
Replies
5
Views
3K