IIT JEE , AIEEE Forum - Physics , Mechanics

anusha92

A block of mass M placed a horizontal surface is connected to a nylon rope of mass M. A force of 1 N is applied to the free end of the rope in the horizontal direction. The tension at the midpoint of the rope is

A) 0.25N B) 0.50N

C) 0.75N D) 1N

destinationIIT09

B) 0.50 newton

nishant13940

0.75 newton

pottermania1990

madman

let the tension in the middle of the rope be T
considering the half which is not connected to the block
F-T=Ma/2
considering the half which has the block
T=3Ma/2
from the two equations
F=2Ma
a=1/2M
from the first equation
1-T=1/4
T=3/4=0.75N

sunip_the_mini_mastermind

is it B)

anusha92

no, it's not B it is C. but I am not satisfied with MADMAN's answer. for me it seems the explianation is not so apt

ganesha1991

to the right the mass of rope = mass on left = M/2
there fore
T = [M +M/2]a=3/2Ma
or a =[ 2/3M ]T
for the right side
F- T=Ma
1-T = [2/3M]TM/2
1-T = T/3
[4/3]T=1
T=3/4 =0.75

ganesha1991

Re:newtons laws of motion............

problemkiller0.1

F=2ma and F-N=ma(n is force acting and N=ma

on rope at other end) so F=2N

at mid point f rope

F-N'=ma\2 and N'--N=ma\2

so

F--N'=N--N' SO F+N=2N'

N'=F+N\2 so N'=F+F\2

-------

2

SO N'=0.75 IS AWNSER

prateekdhariwal

T = [M +M/2]a=3/2Ma
or a =[ 2/3M ]T
for the right side
F- T=Ma
1-T = [2/3M]TM/2
1-T = T/3
[4/3]T=1
T=3/4 =0.75

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