# S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
 It is currently Wed, 1 Jun 2016 02:41:48 UTC

 All times are UTC [ DST ]

 Page 1 of 1 [ 4 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: Elevators and ForcePosted: Sat, 18 Sep 2004 15:40:50 UTC
 S.O.S. Oldtimer

Joined: Tue, 20 May 2003 20:19:56 UTC
Posts: 226
An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is .0600 g. What are the maximum and minimum forces the motor should exert on the supporting cable?

For the maximum (assuming elevator goes up):

Is the tension force the same as the motor's exerting force (which has opposite direction)?

F_t - mg = T - 4850 kg*9.8 m/s^2 = T- 47530 N
T - 47530 N = 4850 kg*0.0588 m/s^2
T - 47530 N = 2851.8 N
T = 50,381.8 N = F_motor ???

How do I get the minimum force?

_________________
Gracias for your time, patience, and input.

"Which is a kind of integrity, if you look on every exit being an entrance somewhere else."
- Main Player from Stoppard's Rosencrantz and Guildenstern Are Dead

Top

 Post subject: Posted: Sun, 19 Sep 2004 05:31:42 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 30 Nov 2003 04:59:29 UTC
Posts: 720
F = MA
M = 4850

Max A = 1.06g (gravity + acceleration up) = 10.388
Max F = 50381.8

Min A = .94g (gravity - acceleration down) = 9.212
Min F = 44678.2

Top

 Post subject: Posted: Sun, 19 Sep 2004 15:11:17 UTC
 S.O.S. Oldtimer

Joined: Tue, 20 May 2003 20:19:56 UTC
Posts: 226
Could you explain where the .94g and 1.06g come from?

_________________
Gracias for your time, patience, and input.

"Which is a kind of integrity, if you look on every exit being an entrance somewhere else."
- Main Player from Stoppard's Rosencrantz and Guildenstern Are Dead

Top

 Post subject: Posted: Mon, 20 Sep 2004 04:47:56 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 30 Nov 2003 04:59:29 UTC
Posts: 720
The statement of the problem said that the maximum acceleration of the elevator is .06 g. The tension (=force) in the cable holding the elevator must support the weight of the elevator (mg) as modified my any additional acceleration: +.06mg going up, -.06mg going down.

There is more than one way to attack this problem. (Usual for most problems.)

Instead of writing mg+.06mg {or mg-.06mg} and factoring out mg to get (1+.06)mg {or (1-.06)mg}, I considered acceleration by itself: the cable must provide enough force to counter the acceleration due to gravity plus extra going up, or minus extra going down. Having established the accelerations that must be provided by the cable, I then calculated the required forces.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 4 posts ]

 All times are UTC [ DST ]

#### Who is online

Users browsing this forum: No registered users

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forum

Search for:
 Jump to:  Select a forum ------------------ High School and College Mathematics    Algebra    Geometry and Trigonometry    Calculus    Matrix and Linear Algebra    Differential Equations    Probability and Statistics    Proposed Problems Applications    Physics, Chemistry, Engineering, etc.    Computer Science    Math for Business and Economics Advanced Mathematics    Foundations    Algebra and Number Theory    Analysis and Topology    Applied Mathematics    Other Topics in Advanced Mathematics Other Topics    Administrator Announcements    Comments and Suggestions for S.O.S. Math    Posting Math Formulas with LaTeX    Miscellaneous