Friction Force on 2 blocks on a frictionless surface....

[Arman777]

Homework Statement

The two blocks $(m=16 kg)$ and $(M=88 kg )$ , are not attached to each other.The coefficient of static friction between the blocks is $μ=0.33$, but the surface beneath the larger block is frictionless.What is the minimum magnitude of the horizontal force $\vec F$ required to keep the smaller block from slipping down the larger block ?

Homework Equations

$\vec {F_t}=m\vec a$
$F_f=μN$

The Attempt at a Solution

I made a free body diagram for m and M.Picture in attached files.
So I cannot find $F_{mM}$ or $F_{Mm}$ And ıf this was true M should move right but we don't know acceleration of $M$. Is there some other force that I don't know ?

Thanks

 

[PeroK]

1) If you apply a horizontal force to $m$, what will happen to the system?

2) What would happen if the larger block were fixed to the ground? Could you calculate the required horizontal force on $m$ in this case?

 

[TomHart]

What information do you know when the small block is right at the point of slipping - but does not slip downward?

Edit: Sorry PeroK. (Stupid autocorrect.) I should have left it alone but didn't notice you posted until after I posted.

 

[Arman777]

1) If you apply a horizontal force to $m$, what will happen to the system?

2) What would happen if the larger block were fixed to the ground? Could you calculate the required horizontal force on $m$ in this case?

1)I guess system will move ?

2)Required horizontal force for what ? I didnt understand ohh If u mean what I mean then $F=\frac {mg} {μ}$

 

[PeroK]

1)I guess system will move ?

2)Required horizontal force for what ? I didnt understand ohh If u mean what I mean then $F=\frac {mg} {μ}$

Can you put those two things together?

Hint: in the case where $M$ moves why is most of the applied force transmitted though $m$ to $M$?

 

[Arman777]

What information do you know when the small block is right at the point of slipping - but does not slip downward?

Edit: Sorry PeroK. (Stupid autocorrect.) I should have left it but didn't notice you posted until after I posted.

Well the friction force should be equal to Weight.The friction force depends the $N$ and that depends the $F$ and I guess $F_{Mm}$

 

[Arman777]

Can you put those two things together?

Hint: in the case where $M$ moves why is most of the applied force transmitted though $m$ to $M$?

Cause M has bigger mass ?

well I can I guess but I need acceleration I guess which I don't have that info $(M+m)a=\frac {mg} {μ}$

 

[PeroK]

Cause M has bigger mass ?

well I can I guess but I need acceleration I guess which I don't have that info $(M+m)a=\frac {mg} {μ}$

Even if a number in not given in the problem, you can still use that quantity. In this case, you can still consider the acceleration. It might tell you something about the forces involved. So:

$a = ?$

 

[Arman777]

Even if a number in not given in the problem, you can still use that quantity. In this case, you can still consider the acceleration. It might tell you something about the forces involved. So:

$a = ?$

I guess I see your point wait I ll solve I ll find a then from there I ll find $F_{Mm}$ then I ll find $F$ right ?

 

[PeroK]

I guess I see your point wait I ll solve I ll find a then from there I ll find $F_{Mm}$ then I ll find $F$ right ?

That's the idea.

 

[Arman777]

That's the idea.

I solved it thanks

 

[Arman777]

Or wait I have to ask something.I am so tired I thought I can figure it out but I have to ask.

$F_{mM}=Ma$ isn't this correct cause according to the question and my tryings it should be $ma$.

Why $ma$ but not $Ma$ ?

 

[PeroK]

Or wait I have to ask something.I am so tired I thought I can figure it out but I have to ask.

$F_{mM}=Ma$ isn't this correct cause according to the question and my tryings it should be $ma$.

Why $ma$ but not $Ma$ ?

There's more to it than that.

One observation is that you seem to be trying to jump straight to the answer from half way through. This is not an easy problem and it has maybe 4-5 calculations in it. There's no way to jump to the answer.

Perhaps you've started to do physics now at a slightly higher level. One aspect of harder problems is that you have a longer chain of calculations to get the answer.

Anyway, in this case, the first step is to express $a$ in terms of $F, M$ and $m$.

 

[Arman777]

There's more to it than that.

One observation is that you seem to be trying to jump straight to the answer from half way through. This is not an easy problem and it has maybe 4-5 calculations in it. There's no way to jump to the answer.

Perhaps you've started to do physics now at a slightly higher level. One aspect of harder problems is that you have a longer chain of calculations to get the answer.

Anyway, in this case, the first step is to express $a$ in terms of $F, M$ and $m$.

I made the all steps,like making a referance frame drawing forces and finding a and then N ok I ll type them just wait a sec

 

[Arman777]

I guess there's something wrong in our equation.

we said $(M+m)a=\frac {mg} {μ}$ This is also equal $F$ (that we said),so we can find easily a here cause we know M and m and g and μ.
And If F=(M+m)a which its something like 476 which that's not the answer ??

I am so tired.I don't know this is true or not but I ll go to sleep

 

[Doc Al]

I would attack this problem in this order:
(1) What friction force is required between the blocks?
(2) What normal force (between the blocks) is thus needed?
(3) What acceleration is thus produced?

And so on.

 

[Arman777]

(1)$F=μN$
(2)well I guess $ma=F-N-F_{Mm}$ from there $N$ equals to zero I guess but its wrong so $a$should be zero I guess ??
or $N=F-F_{Mm}$ but this means there's no acceleration.
(3)$F=(M+m)a$

 

[Doc Al]

(1)$F=μN$

What must the friction force equal so that the block does not slip down?

 

[Arman777]

What must the friction force equal so that the block does not slip down?

weight which is $mg$

 

[Doc Al]

weight which is $mg$

Right. Now answer my question #2.

 

[Arman777]

Right. Now answer my question #2.

I already did

 

[Doc Al]

I already did

Do it again. Use your answer from question #1 to answer #2. (Your previous answer of N = zero is clearly wrong!)

 

[Arman777]

Are u talking about $N=\frac {mg} {μ}$

The total force acting on m is $ma$ which should be equal to the sum of the force on horizontally.Theres F and N.

$ma=F-N$
$N=F-ma$ ??

 

[Doc Al]

Do a force analysis of the big mass M.

 

[Arman777]

$F_{mM}=Ma$

 

[Doc Al]

$F_{mM}=Ma$

What does $F_{mM}$ equal? (Consider your answers to questions #1 and #2.)

 

[Arman777]

What does $F_{mM}$ equal? (Consider your answers to questions #1 and #2.)

Theres no connection just there's $F_{mM}$ in N

as I said If $N=F-ma-Ma$ we get zero If I can write N I ll sove the queston.Just Idk why I can't write it.I am so annoying I know.

 

[Doc Al]

Are u talking about $N=\frac {mg} {μ}$

Good! Use it.

The total force acting on m is $ma$ which should be equal to the sum of the force on horizontally.Theres F and N.

$ma=F-N$
$N=F-ma$ ??

This is OK, but unhelpful. Instead, apply Newton's 2nd law to M instead of m.

 

[Arman777]

I ll ask my prof next semester...Thanks

 

[Doc Al]

I ll ask my prof next semester...Thanks

Don't give up. You are very close to the solution!

You applied Newton's 2nd law to mass m. OK. Now apply it to mass M. (And make use of the value you already found for N.)

 

[Arman777]

Friction Force on 2 blocks on a frictionless surface...

??
ohh wait I ll need N ok wait a sec

 

[Arman777]

Is $N= F_{mM}$ ??

 

[Doc Al]

Is $N= F_{mM}$ ??

Of course! (And you found N already.)

 

[Arman777]

Of course! (And you found N already.)

Finally

 

[Arman777]

Ok I solved.Thx