Question

With what minimum acceleration can a fireman slide down a rope whose breaking strength is 3/4th of his weight?

Answer 1

I think we need something more. The situation or a diagram if there's one.

Answer 2

A.1/4g B.1/2g
C.3/4g D.Zero

Answer 3

no diagram

Answer 4

Okay, let me think, I think we are going to have more dirty scans in OS from me..

Answer 5

http://cnx.org/content/col10322/latest/

Answer 6

try it out

Answer 7

what should i do there?

Answer 8

change the variations

Answer 9

height and weight

Answer 10

ok
\[B= \frac{3}{4}mg\]\[a= g-\frac{B}{m}\]
\[a=g- \frac{3}{4}g= \frac{1}{4}g\]

Answer 11

Okay I have a logical conclusion of what happens, but how is acceleration equal to his height (3/4 that is)? its not even dimensionally correct.

Answer 12

or is it 3/4 of his weight???

Answer 13

YUP!!!!!!! sry

Answer 14

PLZ. WHY YOU NO TYPE CORRECTLY???? i am alien from pluto now!!! _/(o.O)\_

Answer 15

i saw similar question in google ,so just put the values and got the answer

But how is this ?
\[a= g-\frac{B}{m}\]

Answer 16

Edited the question now!

Answer 17

okay i am sacnning it now.. phew!!

Answer 18

ok lol

Answer 19

Same rules, warnings, and hazards apply again!!! read at your own risk :p

Answer 20

Tell me something, are you actually understanding what I am trying to say? Ya for bas formality..?

Answer 21

okay correction: " 'ma' amount of force from 'mg' of the gravity is DOING WORK"

Answer 22

Well ,i understood your prev solution
But this is kinda confusing

Answer 23

okay. see. i am using a bit of 'maatri-bhaasha' 'cause am a bit exasperated.
tell me first do you understand the situation?

Answer 24

Yeah ,fireman is sliding a rope

Answer 25

hmm. chalo then good.

ab dekho, what happens when the fireman is stationary holding on to the rope? what is the tension in the rope then?

Answer 26

Ok ,his entire weight is being supported by the string (from your scan)

Answer 27

so, then, T=mg. right? because his weight mg is being balanced by an equivalent force, thats why he is stationary. and that force is none other than the tension. so, T=mg. fine till here?

Answer 28

ok

Answer 29

chalo then. now very very slowly try to visualise, imagine, what I show you.

Imagine the fireman relaxing his grip on the rope by a wee bit.
so what happens? he slides.
at what rate? some acceleration 'a' we don't know.

so if he is accelerating down with 'a', the net force acting on him must me 'ma'. right?


Feel aayi?

Answer 30

Lol okayy

Answer 31

wat i have done is drawn a free-body diagram with ALL the forces acting on the man
the tension of rope iss pulling him up T the system not being at rest,is accelearting down
so mg>T
but given then max T withstandable by rope is mg*3/4
we put it in the eqn formed and hence find the maximum accleartion that can be achieved
if T was more than that then man shud fall on ground(free-fall)