Mechanics problem

  • Thread starter kushan
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  • #1
kushan
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Hello everybody ,
I am stuck at this question :frown:,

Homework Statement



Two blocks of masses m=1kg and M =2kg are connected by a non-deformed light spring .They are lying on a rough horizontal surface . The coefficient of friction is 0.4 ,What minimun constant force has to be applied in the horizontal direction to the block m , in order to shift the order block M .

Homework Equations


I came up with work energy conservation theorem
F.x = 1/2 kx^2 +f1.x + 1/2mv^2
(where F is external force
f1is force of friction on m. and v is velocity of mass m)
F-kx-f1=ma1
and when block M is just about to move kx=f2
(here f2 is force of friction on M

, its leading me nowhere .
Any help?
 

Answers and Replies

  • #2
ehild
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Work with the forces instead of energy.

ehild
 
  • #3
kushan
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After working with forces I get
F-umg - uMg = ma

Now how do i find a here?
 
  • #4
ehild
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A has to be just a bit greater than zero, so practically zero.


ehild
 
  • #5
kushan
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But this is block 1 which will be moving , not block 2 which will be stationary till the last moment
 
  • #6
ehild
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What does "shift" mean? You need to made the second block moving. The force F has to be enough to move the first block and stretch the spring. How the first block will move and what is the maximum stretching you can have with the applied constant force F? That stretching of the spring must be enough to overcome the friction of the second block.

ehild
 
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  • #7
kushan
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I am really not getting ehild can you frame some equations ?
 
  • #8
ehild
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Newton's second law for the first mass?

ehild
 
  • #9
kushan
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let me try again .
Thanks
Ehild
 
Last edited:
  • #10
ehild
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Hello everybody ,
I am stuck at this question :frown:,

Homework Statement



Two blocks of masses m=1kg and M =2kg are connected by a non-deformed light spring .They are lying on a rough horizontal surface . The coefficient of friction is 0.4 ,What minimun constant force has to be applied in the horizontal direction to the block m , in order to shift the order block M .

Homework Equations


I came up with work energy conservation theorem
F.x = 1/2 kx^2 +f1.x + 1/2mv^2
(where F is external force
f1is force of friction on m. and v is velocity of mass m)
F-kx-f1=ma1
and when block M is just about to move kx=f2
(here f2 is force of friction on M

, its leading me nowhere .
Any help?

I reconsidered your attempt with work and energy, it might be a better approach as mine one.

Should the block of mass m have non-zero velocity when the bigger block starts moving?

But it would be useful to perform an experiment. What happens when you exert some force on mass m, bigger than f1=μmg?

ehild
 
  • #11
voko
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Does anything change in the problem if the spring is replaced with a rigid link?
 
  • #12
ehild
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Yes, it will.

ehild
 
  • #13
kushan
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yes ehild , The block with mass m will be moving and spring will be extending .
When it reaches the limit ie kx=uMg the block M will move .
I am not getting how to go about it .which equations to use? confusing!
 
  • #14
ehild
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Does the mass m need to move at the instant when kx=uMg?

ehild
 
  • #15
voko
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I'll rephrase my question: does anything change with regard to the magnitude of the force if the link is rigid?
 
  • #16
ehild
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I'll rephrase my question: does anything change with regard to the magnitude of the force if the link is rigid?

Yes. The smallest force which is enough just to move the bigger mass out of its steady state is less than the force needed to move both blocks together with the same velocity (and acceleration).

See picture. What happens when you connect the weight?

ehild
 

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  • #17
voko
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Yes. The smallest force which is enough just to move the bigger mass out of its steady state is less than the force needed to move both blocks together with the same velocity (and acceleration).

ehild

I do not see how. To move the bigger mass, the spring has to pull it with some force F. To make the spring do it, it must be extended till its reaction force is equal to F. How is that different from a rigid link?
 
  • #18
ehild
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In case of a rigid link both blocks have to move together. In case of the spring the block pulled by F will start to move if F overcomes friction, but the second block stays in rest untill the displacement x of the first block yields enough force to overcome the friction of the second block. The question is what constant force is needed to displace the first block so far.

ehild
 
  • #19
Vatsa
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I Think this is the solution....
for the body to accelerate horizontally then the forces on m are
External Force F
Frictional force -mgf
Force due to the spring -kx
Force Due Friction on M Mgf


Mgf=2mgf
F-(mgf+kx)>2mgf
F>3mgf+kx
F>12+kx If k is negligible then
F>12 Newtons
 
  • #20
kushan
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Ehild , when this force f is doing work .
Will we say F.xo ( where xo is the actual displacement )
or F.x(x is displacement had there been no friction), because of the work done in overcoming friction ?
 
  • #21
ehild
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F is the applied force. That force will do work if the block moves (that is, when F overcomes friction F>f).
If the displacement is x, the work of the applied force is Fx and the work of friction is fx.

ehild
 
  • #22
kushan
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Sorry vatsa , the answer isnt 12 newtons :l
 
  • #23
ehild
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I do not see how. To move the bigger mass, the spring has to pull it with some force F. To make the spring do it, it must be extended till its reaction force is equal to F. How is that different from a rigid link?

F is the applied force, exerted on the first block. On the second block, the elastic force of the spring acts which is not equal to F. It depends on the displacement of the first block.

ehild
 
  • #24
ehild
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Nobody tried an experiment????
SO: start pulling the first block. Supposed that the force is big enough, the first block starts to accelerate. ma=F-μmg-kx. (You have seen similar equation when a weight is hanged on a vertical spring. What kind of motion is it?) The block accelerates to the right, speeds up, its displacement increases, but the spring force increases, too, so the acceleration decreases, turns to negative and the speed decreases till the block stops at maximum displacement.
The force succeeds to shift the second block when the elastic force at this maximum stretching of the spring is greater than μMg.

ehild
 
  • #25
kushan
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So the equation would be
kx=uMg
and F-umg-uMg=ma
and F.x=1/2mv^2 + f.x + 1/2kx^2
Which is the nothing but anti derivative of 2nd equation .
What to do after this Ehild?
 
  • #26
ehild
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I asked already: must v differ to zero to exert force on the second block? read my previous post:"The block accelerates to the right, speeds up, its displacement increases, but the spring force increases, too, so the acceleration decreases, turns to negative and the speed decreases till the block stops at maximum displacement. "

ehild
 
  • #27
kushan
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your case seems to make block M rigid/fixed , if at a certain time during the accelration of block m
F-umg+kx=ma ,
kx become become greater than friction , the block m will still be moving and M will start to move .
It is impossible for block m to return back as we can consider spring to be internal , and
combinning both masses m and M to form m+M
will give equation
F.x= 1/2mv^2+ fx
which clearly shows v cannot fall or become negative
 
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  • #28
kushan
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Omg , I get it all now .......
 
  • #29
kushan
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Ehild is it same as a a block vertically hanging , with charge q , so two forces are acting on it and it does shm ?
 
  • #30
ehild
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Ehild is it same as a a block vertically hanging , with charge q , so two forces are acting on it and it does shm ?

something like that...
 
  • #31
voko
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It is still not obvious to me that the minimal force would be different from the rigid case. But I will have to postpone further analysis till tomorrow.
 
  • #32
kushan
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Ok this case is same as a vertical SHM in presence of gravity , like a spring tied to a tree with a mass m hanging , now if mass m is increased or in other words ( a force F is added) the SHM will still continue .
The relation to that with this question is that , For Force to be just minimum block M has to act like the stationary support and gravity as F.
Is this thing correct Ehild?
 
  • #33
ehild
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Not quite Kushan.

For this problem, Newton's second Law states for block m:

ma=F-μmg-kx.
For the mass hanging on spring:

ma=mg-kx.
You see the constant force is mg in the second case and F-μmg in the first one.

As the force of friction changes sign in every half period, it is not real SHM after the first maximum displacement.

For the time span from t=0 to the first maximum, the solution is

x=(F-μmg)/k[1-cos(wt)] (w=√(k/m)),
as initially x=0 and v=0.

How does x and v change with time? When is x maximum? What is the velocity when x is maximum?

I stop for today, and go to sleep. :zzz:

ehild
 
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  • #34
voko
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To move the bigger mass, [itex]\mu M g = k x[/itex]. On the other hand, [itex]\frac {kx^2} {2} + \frac {m v^2} {2} = (F - \mu m g)x[/itex]. Since F is minimal, [itex]\frac {m v^2} {2} = 0[/itex], so [itex]\frac {\mu Mg x} {2} = (F - \mu m g)x[/itex]. Indeed, we need a lesser force than in the rigid case.
 
  • #35
azizlwl
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I'll rephrase my question: does anything change with regard to the magnitude of the force if the link is rigid?

Is there different in force if we use different springs of different coefficient k.
What about if the value is so big where it is equivalent to rigid body.
Anyway no data given about the spring.

In calculating springs in series, the forces on each spring are equal.
 

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