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Q. Three blocks placed on top of one another on a table.there is a 7kg block.on top of it there is a 3kg block on which there is a 2kg block.frictional coefficient between 2kg and 3kg is 0.2,that between 3kg and 7 kg is 0.3 and the table is smooth.find their accelerations when 10N force is applied on (a)2 kg block (b)3 kg block (c) 7 kg block.[g=10m/s2]
Comments (22)
N1 = 20 N
N2 = 20+30 = 50 N ( because 2 kg and 3 kg blocksare not moving vertically..)
Hence ..max value of f1 = uN1 = 0.2x 20 = 4 N
max value of f2 = uN2 = 0.3 x 50 = 15 N
Now ..taking case (b) first ..
the force is on 3 kg block.
Lets assume all the three blocks are moving together..
considering them as a system ..frictional forces become internal.
(2+3+7)a = 10
a = 10/12 = 5/6 m/s^2
hence ..each block moves with accn 5/6 m/s^2
for 2 kg block,
2(a) shud be less than f1(max) ( because f1 is the only force that is responsible for motion.)
2a = 2x5/6 = 5/3 IS equal to f1.
also ..for 7 kg block,
7a shud be less than f2(max)
7x5/6 = 35/6 IS equal to f2.
Now ..for 3 kg block,
3(a) = F - f1 - f2
3(5/6) = 10 - 5/3 - 35/6
15/6 = 15/6
hence ..our assumption is correct...the three block really move together with accn of 5/6m/s^2..
(*love u all)
Part (c) now..
Same procedure...
Consider them to move along as a system...with accn 5/6 m/s^2
For 2 kg block,
2(5/6) = 10/6 = f1 ( f1 < f1max)
For 7 kg block,
7(5/6) = 10 - f2
f2 = 25/6 ( f2 < f2max)
Now for 3 kg block...
3(5/6) = f2 - f1
15/6 = 15/6
Hence ..again ..our assumption is correct ....accn of the system is 5/6 m/s^2
Part (a) is rather interesting..
Same procedure..
Let they move with same accn 5/6 m/s^2
For 2 kg block ..
2(5/6) = 10 - f1
f1 = 50/6 > f1(max) ...hence ..slipping occurs between 2 kg block and 3 kg block.
Solving 2 kg block independently ..
2(a) = 10 - f1(max)
2(a) = 10 - 4
a = 3 m/s^2
Now our question is reduced to this :::::
External force on 3 kg block is 4 N.
Same procedure..
consider that 3 ands 7 kg block move together..
10(a) = 4
a = 2/5 m/s^2
for 7 kg block,
7(2/5) = 14/5 = f2 ( f2 < f2 max)
and for 3 kg block ..
3(2/5) = 4 - f2
6/5 = 4 - 14/5
6/5 = 6/5
our assumption is correct..
hence ..
in this case..
2 kg block moves with accn 3 m/s^2 and the other 2 with accn 2/5 m/s^2
haan bhai ab yeh bata, ki jab force 7 kg waale block par lagi hogi, toh normal reaction on 3 and 2 kg block combined will be 50N, lekin agar woh dono saath accelerate karenge toh frictional force kul mila kar 15 N hogi, joki 10 N se jyaada hai, toh yeh teeno 1 saath accelerate kaise kar paayenge..............?
15 N is the max frictional force ..(f2 max) ..
Not to forget .friction is a self adjusting force ..
As done above, the clculations say the value of friction needed in every case...
For force on 7 kg block..
for a nominal acceleration of 5/6 m/s^2 of the system ..the values came out to be :
f1 = 10/6 and f2 = 25/6 ..both less than their max.
Conclusion : 15 N is not the frictional force..( its the max )
bhai mere hisaab se to ye hona chahiye tha ki jab 7 kg pe force lagi toh sirf, 7 kg waala hi acc. karna chahiye aur slipping honi chahiye of 3 with respect to 7 and of 2 with respect to 3,backward and forward respectively.....................
bhai main thoda bewakuf type hun, tera point abhi tak nai samajh paaya.........................................
HEY anuj ..
sun ,..
jab 7kg par 10 N FORCE LAGEGA....THAN ....USE ROKNE WAALA FRICTIONAL FORCE HOGA JO KI
DUE TO ROUGH SURFACE BETWEEN 7Kg and 3Kg ..hoga ..
so is 15N frictional force ke karan 7kg akele nhi sli-p ho sakta kyuki 15N (frictional force ) > 10N (APPLIED FORCE)
but now the floor is slippery so ..ab teeno block ek system ban kar hi ..move karege ...
to total sytem ek block ban gaya of 12kg ..(7+3+2)kg
to ..now u see ..
accn = 10/12 m.sec^2
samjha..
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