friction nd newton laws

A plank of mass 10 kg rests on a smooth horizontal surface . Two blocks A and B of masses m = 2kg and m = 1 kg resp. lie at a distance of 3 m on the plank as shown in the figure . The friction coeff. b/w the blocks and plank r 0.3 and 0.1 . Now a force F = 15 N is applied to the plank in horizontal direcn . Find the time ( in sec ) after which block A collides with block B .

8 Answers

30
Ashish Kothari ·

oh yeah.. calc mistake

Considering the plank and the blocks to be a system,

F=m.a

15N = 13.a

a=15/13 ms-2 [ a is the acceleration of the system]

Now, considering the blocks...

Block A

f ≤ μN
≤ 0.3 * 20 ≤ 6N

Force acting on the block due to motion of plank,

F= m.aA = 2.15/13 = 2.30 N

Hence, force of friction balances this force and hence there is no relative motion between the block A and the plank.

Block B

f ≤ μN
≤ 0.1*10 ≤ 1 N

Force acting on the block = m.aB = 1*15/13= 1.153N

Net force,

m.aB= 1.15 - 1
= 0.15
aB= 0.15/1

Time taken ≈ 6s

30
Ashish Kothari ·

Considering the plank and the blocks to be a system,

F=m.a

15N = 10.a

a=3/2 ms-2 [ a is the acceleration of the system]

Now, considering the blocks...

Block A

f ≤ μN
≤ 0.3 * 20 ≤ 6N

Force acting on the block due to motion of plank,

F= m.aA = 2.3/2 = 3N

Hence, force of friction balances this force and hence there is no relative motion between the block A and the plank.

Block B

f ≤ μN
≤ 0.1*10 ≤ 1 N

Force acting on the block = m.aB = 1*3/2 = 1.5N

Net force,

m.aB= 1.5 - 1
= 0.5N
aB= 0.5/1

Time taken = 2√3 sec.

6
AKHIL ·

no wrong!!

the ans is 6s.

21
Shubhodip ·

..how r u doin F = (sum of mass)*a here?? its wrong...
it should be 15 - fr1 - fr2 = 10a (a acc of plank)..

fr1=M1(a-B)
fr2=M2(a-b)

|B-b| gives relative accleration
actually the magnitude of friction depends upon acc. of the plank...where it itself depends upon the friction....can sm1 say if i m ri8??

i think i cant take fr1 or fr2 as Fmax....

6
AKHIL ·

see if fricn was absent
then commom acc. wud hav been 15/13 m/s2

now if its there B will slip.......
for A and M , common acc. is 14/12 ( net force = 15 - 1 )

for B acc. is 1.
relative acc. = 14/12 - 1 = 1/6.

rel. displacement = 3.
so s = 1/2 a t2
and t = 6s...

so the ans is not almost 6
it is exactly 6..

30
Ashish Kothari ·

Ok! I got my mistake.. Thanks akhil!

21
Shubhodip ·

@AKHIL...this is nt true
".see if fricn was absent then commom acc. wud hav been 15/13 m/s2"

if there wer no friction mass M wil be goin wid acc 1.5 m/s2..nd block A nd B would be in rest w r t the ground.....or will be goin wid an acc 1.5 m/s2 w r t block M..

here friction between block B and plank is 1 N,wich is max friction,(as 15N>11N)..and friction between block A and plank is adjusting friction..so acc. of block A is 7/6...so relative acc 1/6.....and T = 6

6
AKHIL ·

arey main bhi to yehi keh raha hun
aasan bhaasha mein

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