Question Of the Day - 18-06-2011 Mechanics

30

Ashish Kothari ·2011-06-18 09:07:01

Now, first things first, let us assume the acceleration of the big block is A(wrt to ground) and the small blocks is a(wrt to the big block).

Then,

how do we decide that in which direction will the big block move wrt to the ground. We notice that the contact force N₂ on the small block is unbalanced. So, there must be some balancing force so as the keep the block in horizontal equilibrium. Where does this force come from? From the given situation, obviously, only the pseudo force due to the motion of the big block will act and will balance N₂. So for the pseudo force to act to the left, the big block must move to the right!

So we get the following set of equations:

T - N₂=4mA
T+N₁+4mg=N_g

N₁=mg
T+mA=ma

N₂=mA
mg-T=ma

Solving we get the following relations,

T=5mA
6mA=ma => A=a/6

Now,
mg-T=ma
mg - 5mA=6mA
mg=11mA
A=g/11

a= 6g/11

Edits: Accelerations are now according to the frames in which they were calculated.

UP 0 DOWN 0
Post on FacebookPost anonymously?

49

Subhomoy Bakshi ·2011-06-19 03:36:42

We did not need that method anyway...

But I did by that method just to make you familiar to "How strong tools we have acquired, but never used, really are!" :D

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

Debosmit Majumder ·2011-06-19 00:25:02

i solved this by ashish`s method,but i think subhomoy da`s method is great.havn`t thought of applying cnsrvtn of momentum ever in these sums.

UP 0 DOWN 0
Post on FacebookPost anonymously?

49

Subhomoy Bakshi ·2011-06-18 10:39:30

My solution:

Concepts used:
1) law of conservation of linear momentum.
2) net work done by tension=0
2) F_ex=m.a

P.S. all accn are with respect to inertial frame of reference.

Let the velocity of the 3 masses m₁, m₂ and M at the given instant respectively be v₁, v₂ and V.
Next, the masses m₂ and M move in horizontal direction together...
thus, V=v₂

And also, net force in horizontal direction on the system comprising of all the three masses and the string = 0

thus, momentum can be conserved.
so, 4mV+mV-mv₁=0 (initially everything was at rest)

thus, 5V = v₁
differentiating once w.r.t time we get,
b=5a.

Next,
using constraint equation on string,
work done by tension on m₁=T.x_b
work done by tension on pulley in horizontal direction=T.x_a
work done by tension on pulley in vertical direction=0
work done by tension on mass m₂=-T.x_c

thus, T.x_a+T.x_b-T.x_c=0
gives, x_a+x_b=x_c

differentiating twice w.r.t. time,

a+b=c
or, 6a = c

Thus,

T=m(5a)
mg-T=m(6a)

adding,
mg=11ma
a=g11

b=5g11

c=6g11

net accn of block m₂= √37g11

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

kunl ·2011-06-18 10:34:23

u said u r more interested in fbd and acceleration relations...i gave both[3]...+bonus equations(6 equations in 6 unknowns==>perfectly solvable)...now i m wondering wats left in this problem???LOL!

i miss TWO pinks[3][3]

UP 0 DOWN 0
Post on FacebookPost anonymously?

30

Ashish Kothari ·2011-06-18 10:11:20

Thats a mistake. a is the acceleration of the block with respect to the bigger block!

UP 0 DOWN 0
Post on FacebookPost anonymously?

49

Subhomoy Bakshi ·2011-06-18 10:07:42

but u already mentioned that a is w.r.t ground?? [12]

UP 0 DOWN 0
Post on FacebookPost anonymously?

30

Ashish Kothari ·2011-06-18 10:04:15

Actually in the ground frame my equations should've have been in this form:

N₂=mA

T=m(a-A)

UP 0 DOWN 0
Post on FacebookPost anonymously?

49

Subhomoy Bakshi ·2011-06-18 10:03:49

yes.. that is right! :D

UP 0 DOWN 0
Post on FacebookPost anonymously?

30

Ashish Kothari ·2011-06-18 10:01:51

Yeah right. I made a mistake. In the ground frame, after the system is released from rest, the block moving horizontally will have the acceleration (a-A) and the block moving down will have two components of acceleration.