Understanding Acceleration and Friction in a Block Over Block System

[vissh]

Hello :D
The block over block questions today got on my mind And i got some doubts in which i need some guidance :)
Let the two blocks as shown http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=LOM.jpg" . I want to find the accelerations of blocks when :-
<1>F = 20 N <2>F = 40 N <3>F = 60 N <4>F = 100 N <5>F = 150 N {g=10m/s²}

Homework Equations

a=F_net/m
Limiting static friction b/w 2 surfaces, f_l = uN

The Attempt at a Solution

The Limiting friction b/w blocks ,f_l = 40 N
<1>I do these problems in following manner :- F < f_l .Thus,the 2 blocks have no relative motion and both have same acceleration , a.
a = F/(4+6) = 20/(10) = 2m/s²

<2> F = f_l . Still, both have no relative motion and have same acceleration,a.
a = 40/10 = 4m/s²

<3> F > f_l . The f_l will act on 4 kg block on right side while F - f_l will act on 6 kg block . So, now they will have different accelerations .
a_6kg = (60 - 40)/6 = 20/6 m/s² [right side]
a_4kg = (40)/6 m/s² [right side]

Will do <4>,<5> in same manner . Am i right or wrong ?oO

The other thing which is bugging me is this :-
>When we apply , let F = 20 N ,the 6kg block tend to move right . But the 4 kg block will[as F < f_l] apply 20 N in opposite direction to stop it.
>The 6kg block is stopped . But by action-rxn, The 6kg block apply a force 20 N in right side and thus now 4 kg is tending to move now.
>To stop it, 6 kg will apply 20 N in opposite direction and thus cycle continues [for me].Lol infinite cycles ,then , How will motion occur :O

 

[zorro]

These are my views-
You can't find the accelerations of the blocks when F >40N as you don't know the coefficient of kinetic friction which will be less than 1.

Say it is 0.8
If F = 60N,
the 6kg block accelerates with F-f/6 = 4.66m/s² where f = 0.8 x 6 x 10
Now for the 4kg block, there is kinetic frictional force acting towards right.
so a = 0.8 x 4 x 10/4 = 8 m/s²
But wait! acceleration of 'm' w.r.t to ground will be 8 - 4.66 = 4.33m/s² (relative)

The other thing which is bugging me is this :-
>When we apply , let F = 20 N ,the 6kg block tend to move right . But the 4 kg block will[as F < fl] apply 20 N in opposite direction to stop it.
>The 6kg block is stopped . But by action-rxn, The 6kg block apply a force 20 N in right side and thus now 4 kg is tending to move now.
>To stop it, 6 kg will apply 20 N in opposite direction and thus cycle continues [for me].Lol infinite cycles ,then , How will motion occur :O

When we apply F=20N on the 6kg block, there is a frictional force acting on it.
But that f is not equal to 20N! (friction is not maximum).
It is just that value of 'f' which causes no relative motion between the two blocks

Hope it clears

 

[vissh]

Thanks for reply :)

These are my views-
You can't find the accelerations of the blocks when F >40N as you don't know the coefficient of kinetic friction which will be less than 1.

Hmm I forgot to mention,coef. of kinetic friction =1 [ by the way is it necessary for it to be less than 1 :O]
But ok let's take it 0.8

If F = 60N,
the 6kg block accelerates with F-f/6 = 4.66m/s² where f = 0.8 x 6 x 10

now on drawing The FBD of 4 kg block,you can easilt see that the normal rxn,N b/w the 2 blocks is (4)*(10) and the knectic friction b/w 2 surfaces is uN [where N is normal rxn b/w the 2 surfaces] . I think you wrote wrong "f = 0.8 x 6 x 10: - it sud be f = 0.8 x 4 x 10

But wait! acceleration of 'm' w.r.t to ground will be 8 - 4.66 = 4.33m/s² (relative)

Don't got why and how you wrote this . Can you pls explain :)

When we apply F=20N on the 6kg block, there is a frictional force acting on it.
But that f is not equal to 20N! (friction is not maximum).

The limiting friction b/w the surfaces is 0.8 x 4 x 10 = 32 N
What you mean by Maximum !

 

[zorro]

Thanks for reply :)

Hmm I forgot to mention,coef. of kinetic friction =1 [ by the way is it necessary for it to be less than 1 :O]
But ok let's take it 0.8

Coefficient of kinetic friction is always less than coefficient of static friction.

now on drawing The FBD of 4 kg block,you can easilt see that the normal rxn,N b/w the 2 blocks is (4)*(10) and the knectic friction b/w 2 surfaces is uN [where N is normal rxn b/w the 2 surfaces] . I think you wrote wrong "f = 0.8 x 6 x 10: - it sud be f = 0.8 x 4 x 10

Printing mistake. It is 4 only.

Don't got why and how you wrote this . Can you pls explain :)

By using the formula for relative acceleration.
acceleration (m w.r.t M) = a (m w.r.t. ground - M w.r.t. ground)

The limiting friction b/w the surfaces is 0.8 x 4 x 10 = 32 N
What you mean by Maximum !

Limiting friction = Maximum friction.
Here for limiting static friction you have to use the value of coefficient of static friction (not coeff. of kinetic friction).
When we apply 20N force, the frictional force between the two blocks would not be 20N. It will be less than that.

 

[vissh]

Thanks Abdul :D
But i wanted to know [mostly] if i am right in solving this ?[If coeff of kinetic friction is very very nearly equal to 1] :-

<3> F > f_l . The f_l will act on 4 kg block on right side while F - f_l will act on 6 kg block . So, now they will have different accelerations .
a_6kg = (60 - 40)/6 = 20/6 m/s² [right side]
a_4kg = (40)/6 m/s² [right side]

Will do <4>,<5> in same manner . Am i right or wrong ?oO

Thanks in advance :D

 

[vissh]

Can Anyone tell if i am right or wrong ??
Have done the whole solution ,Just want to know if the solution is right or Wrong ??
Will be Thankful to you :)