Tough Friction Problem.

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Homework Statement


Figure shows a small block of mass m kept on the left end of a larger block of mass M and length l. The system can slide on horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is μ and that between blocks is μ/2. Find the time elapsed before the smaller block separates from the bigger block.

The Attempt at a Solution



For the block M,
Ma2 = μ(M+m)g - μmg/2
a2 = Acceleration of M w.r.t. ground is μg(1+m/2M) towards right.
For the block m,
ma1 = μmg/2
a1 = acceleration of m w.r.t M is μg/2 towards right.

l = ut + 0.5a1 t2
u=0 so l = 0.5 μg/2 t2
t = (4l/μg)1/2
The answer is different.
Please help.
 

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Answers and Replies

  • #2
The force acting on the block M is F1=uN1 towards the left.

If we draw the free body diagram, we get N1=(M+m)g

This acts as a pseudo force on the block m.

Thus, on the block m we have F1 acting towards the right and the frictional force F2 acting towards the right.

F2=u/2 (N2)

N2=mg.

Thus, Fnet acting on the smaller block m is given by:

[tex]ma=\mu g(M+\frac{m}{2})[/tex]

From here you get the acceleration on the smaller block m.

Using the equation of motion s=ut + 1/2 at^2,

u=0, s=l.

Substituting values,

[tex]t=\sqrt\frac{4lm}{\mu g(2M+m)}[/tex]
 
  • #3
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Your answer is wrong. Try again :tongue:
 
  • #4
What's the solution? I can't figure out what I did wrong. By any chance, are these questions out of HC Verma? Or some other IIT prep material?
 
  • #5
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Hey how do you know about that book?
Yeah it is from H.C. Verma Friction last question.

Answer is [tex]t=\sqrt\frac{4lM}{\mu g(M+m)}[/tex]
 
  • #6
Because I gave the same enterances about 4 years ago. I'm in final year of Electrical Engineering right now. I'll look at the question to see if I havent misunderstood it. It has been a while since I did this stuff.
 
  • #7
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Hi

Here's my attempt :cry:. First set up free body diagrams for both the blocks. The equation
of motion for the bottom block is

[tex]\mu (m+M)g(-\hat{i})+\frac{\mu}{2}mg(\hat{i})=M\vec{a}[/tex]

where 'a' is the acceleration of the bottom block with respect to the ground. So

[tex]\vec{a}=\mu g\left(1+\frac{m}{2M}\right)(-\hat{i})[/tex]

Now the equation of motion for the top block is

[tex]\vec{a_1}=\left(\frac{\mu}{2}\right)g(-\hat{i}) [/tex]

where a1 is the acceleration of the top block with respect to the ground.
So acceleration of the top block with respect to the bottom block is

[tex]\vec{a_{rel}}=\vec{a_1}-\vec{a}[/tex]

which can be simplified using the above equations.

[tex]\vec{a_{rel}}=\frac{\mu g}{2}\left(1+\frac{m}{M}\right)(\hat{i})[/tex]

once we get this relative acceleration, we have a standard kinematics problem. Let distance
traveled by the top block be l. its initial velocity is 0. So

[tex]l=(0)t+\frac{1}{2}a_{rel}t^2[/tex]

[tex]l=\frac{1}{2}\frac{\mu g}{2}\left(1+\frac{m}{M}\right)t^2[/tex]


solving we get

[tex]t=\left(\frac{4l}{\mu g}\right)^{1/2}\sqrt{\frac{M}{m+M}}[/tex]
 
  • #8
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Now the equation of motion for the top block is
[tex]\vec{a_1}=\left(\frac{\mu}{2}\right)g(-\hat{i})[/tex]

When you observe the smaller block from the frame of reference of ground, it accelerates forward i.e. along +[tex]\hat{i}[/tex]. What made you write -[tex]\hat{i}[/tex]?

Its the forward acceleration of small block which causes it to traverse a length l and ultimately fall.
 
Last edited:
  • #9
912
19
Both of blocks will experience net force in the left direction. So both of them are decelerating with respect to the ground frame. Thats why the negative sign
 
  • #10
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Both of blocks will experience net force in the left direction. So both of them are decelerating with respect to the ground frame. Thats why the negative sign

Thats what I don't understand.
If the upper block experiences a net force in the left, why does it cross the bigger block and fall?
 
  • #11
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Hello ^.^
If the upper block experiences a net force in the left, why does it cross the bigger block and fall?
Here is mine F.B.D : http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=untitled.jpg" [Broken]
So,you will get Ns= g(M+m) And Nb = mg
Thus, the kinetic friction between surfaces when they slide :-
>> B/w A and floor , Fs = uNs = ug(M+m)
>> B/w A and B , Fb = (u/2)Nb = (u/2)mg
We are sure that when block A slides towards right(due to its intial velocity),friction on it by floor will be in (-x) direction.
--Thus, due to Fs , aA = ug(M+m)/M (-x) [(-x) shows direction]
Now,the magnitude of acceleration of B , aB = (u/2)g
Now [for thinking in which direction aB,I took both cases] whether aB is in (+x) or (-x), A will start to lag behind the B block .
Velocity of A dec. quickly . Thus,friction by B on A will surely act on A in forward direction (+x)to oppose the lagging of A i.e. relative motion of A in backward direction. And by action-rxn pair,force of friction on B by A will be in backward direction (-x).

So, actual aA = ug(M+m)/M (-x) + (u/2)mg/M = ug(2M+m)/2M (-x)
and aB = (u/2)g (-x)
>>> aBA = ug(M+m)/2M (+x)
>>> uBA = v-v = 0
>>> Relative Displacement to be achieved 'L'
thus, L = 0 + ug(M+m)t2/4M
or t = [ (4ML)/{ug(m+m)} ] 1/2

Hope this helps :)
H.C. verma really got Nice questions for Building concepts ^.^
 
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  • #12
1,384
0
Hello ^.^

Here is mine F.B.D : http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=untitled.jpg" [Broken]
So,you will get Ns= g(M+m) And Nb = mg
Thus, the kinetic friction between surfaces when they slide :-
>> B/w A and floor , Fs = uNs = ug(M+m)
>> B/w A and B , Fb = (u/2)Nb = (u/2)mg
We are sure that when block A slides towards right(due to its intial velocity),friction on it by floor will be in (-x) direction.
--Thus, due to Fs , aA = ug(M+m)/M (-x) [(-x) shows direction]
Now,the magnitude of acceleration of B , aB = (u/2)g
Now [for thinking in which direction aB,I took both cases] whether aB is in (+x) or (-x), A will start to lag behind the B block .
Velocity of A dec. quickly . Thus,friction by B on A will surely act on A in forward direction (+x)to oppose the lagging of A i.e. relative motion of A in backward direction. And by action-rxn pair,force of friction on B by A will be in backward direction (-x).

So, actual aA = ug(M+m)/M (-x) + (u/2)mg/M = ug(2M+m)/2M (-x)
and aB = (u/2)g (-x)
>>> aBA = ug(M+m)/2M (+x)
>>> uBA = v-v = 0
>>> Relative Displacement to be achieved 'L'
thus, L = 0 + ug(M+m)t2/4M
or t = [ (4ML)/{ug(m+m)} ] 1/2

Hope this helps :)
H.C. verma really got Nice questions for Building concepts ^.^

Thanks. It helped :)
 
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