How Can I Solve for Acceleration in This Friction Problem?

[sankalpmittal]

Homework Statement

See the image: http://postimage.org/image/4ukoszng1/
Look at question 28 there.
If you feel uncomfortable, you can click on the image to increase its size.
Note, T:Tension in string
a: acceleration in big block
R₁: reaction force by small block on big block..

Homework Equations

f≤f_s=µ_sN
f_k=µ_kN

The Attempt at a Solution

This problem is different from what I have done. Now, I think that the acceleration of small block down must be twice the acceleration of big block, because small block is connected by one segment while the big block by two.

Considering FBD of big block:

T-µ₂Mg-R₁=Ma

Considering FBD of small block:

mg-T-µ₁R₁=2ma

But, there are two equations and three unknowns. How can I solve for "a" ?

Please help !

Thanks in advance...

 

[MostlyHarmless]

The acceleration of the two blocks will be the same because they are part of the same system. If one block moves with the same direction and acceleration.

EDIT: also since your problem gives no numbers, I imagine they answer your looking for isn't going be numerical, so the number of unknowns isn't relevant

 

[sankalpmittal]

The acceleration of the two blocks will be the same because they are part of the same system. If one block moves with the same direction and acceleration.

No, I think you are not correct. Small block is connected by one segment. When big block, connected by 2 segments will move say x/2 units right, each segment will shorten by x/2. But the small block is connected by one segment, so that single segment has to lengthen by x units downwards. So small block will move downward by x units. Now differentiating distance, twice we get,

a_small = d²x/dt= a
a_big=0.5d²x/dt=a/2

Q.E.D..

By distance, I meant of course displacement.

Anyone ? Any hints?

Edit:

also since your problem gives no numbers, I imagine they answer your looking for isn't going be numerical, so the number of unknowns isn't relevant

Answer given is in terms of µ₁, µ₂, m and M..

 

[SammyS]

Homework Statement

See the image: http://postimage.org/image/4ukoszng1/
Look at question 28 there.
If you feel uncomfortable, you can click on the image to increase its size.
Note, T:Tension in string
a: acceleration in big block
R₁: reaction force by small block on big block..

Homework Equations

f≤f_s=µ_sN
f_k=µ_kN

The Attempt at a Solution

This problem is different from what I have done. Now, I think that the acceleration of small block down must be twice the acceleration of big block, because small block is connected by one segment while the big block by two.

Considering FBD of big block:

T-µ₂Mg-R₁=Ma

Considering FBD of small block:

mg-T-µ₁R₁=2ma

But, there are two equations and three unknowns. How can I solve for "a" ?

Please help !

Thanks in advance...

The vertical component of the acceleration of the small block is equal to (the negative of) twice the horizontal component of the acceleration of the large block. The large block's motion being purely horizontal.

The horizontal component of the acceleration of the small block is equal the horizontal component of the acceleration of the large block.

Also:

You're missing some forces exerted on the big block.
The Normal force on the big block is not Mg. To calculate this, you should have an equation involving the vertical components of the forces on the big block.

Added in Edit:

I inserted the word "twice" above.

 

[MostlyHarmless]

I'm not sure what you are doing to justify, the accelerations of the blocks being different, but if they are connected, the acceleration is going to the same for both masses. Take Newtons second law. We have a system of masses Fnet=(M+m)a. Also, just think about holding a stick with one hand on either end of the stick. If you move one hand the the other will move in exactly the same waY.

 

[sankalpmittal]

The vertical component of the acceleration of the small block is equal to (the negative of) the horizontal component of the acceleration of the large block. The large block's motion being purely horizontal.

Thanks for replying SammyS!

Huh ? I do not understand. The directions of forces perpendicular to each other, do not have effect on each other. So how can downward motion of small block be negative of...
Oh ! Sorry ! I understand. So you are taking right as positive and downward as negative.

But hold on ! I do not agree.The vertical component of the acceleration of the small block should be twice the (the negative of) the horizontal component of the acceleration of the large block. This is because small block is connected by one segment but large block by two. See the reasoning in my previous post.

The horizontal component of the acceleration of the small block is equal the horizontal component of the acceleration of the large block.

Uhhh, Yes.

Also:

You're missing some forces exerted on the big block.
The Normal force on the big block is not Mg. To calculate this, you should have an equation involving the vertical components of the forces on the big block.

So the friction µ₁R₁ (kinetic) acts on the big block in upward direction also ? Other forces also act? I cannot think of one?

Edit:

@Jesse H.

I'm not sure what you are doing to justify, the accelerations of the blocks being different, but if they are connected, the acceleration is going to the same for both masses. Take Newtons second law. We have a system of masses Fnet=(M+m)a. Also, just think about holding a stick with one hand on either end of the stick. If you move one hand the the other will move in exactly the same waY.

That is not what I say. That is what my textbook say.

Well,

Suppose one block is connected by a single segment from pulley. Other block of same mass is connected by the two segments, tension in each segment being T. Will the acceleration(downward) of two blocks be same ? Think it yourself.

 

[SammyS]

Thanks for replying SammyS!

Huh ? I do not understand. The directions of forces perpendicular to each other, do not have effect on each other. So how can downward motion of small block be negative of...
Oh ! Sorry ! I understand. So you are taking right as positive and downward as negative.

But hold on ! I do not agree.The vertical component of the acceleration of the small block should be twice the (the negative of) the horizontal component of the acceleration of the large block.

Right. It should be TWICE. After putting in all those other adjectives, I left the word "twice" out.

I edited that post.

 

[Maiq]

I'm not 100% sure but i would think that the tension for the M block would be 2T. Also since the reaction force between the two blocks is the only force acting on the m block in the horizontal direction, you could say R1=ma.

 

[MostlyHarmless]

Sorry, taking a second look at the diagram. I see where I was mistaken. Apologies.

 

[sankalpmittal]

I'm not 100% sure but i would think that the tension for the M block would be 2T. Also since the reaction force between the two blocks is the only force acting on the m block in the horizontal direction, you could say R1=ma.

Why you say,R₁=ma ?@Jesse H.
You are welcome.

Right. It should be TWICE. After putting in all those other adjectives, I left the word "twice" out.

I edit that post.

Ok thanks.

Now let me write my solution comprehensively so that you can freely look and analyze it.

FBD for small block:

Forces acting are:

1.Downward force mg
2.Upward force tension T
3.Friction upward: µ₁R₁
4.Pseudo force of Ma on it towards left.
5.Reaction force R₁ by big block on it towards right for its horizontal equilibrium.

So, my equation becomes:

R₁=Ma ..(i)
mg-T- µ₁R₁=2ma
On using (i),
mg-T- µ₁Ma=2ma ... (A)

FBD of big block:

Forces acting on it are:

1.Downward force Mg
2.Tension T due right
3.Friction by ground of µ₂R due left
4.Reaction force or third law pair of friction, exerted by small block downward:µ₁R₁
5.Reaction force R by floor on big block

So accounting for vertical equilibrium of big block,

R=Mg+µ₁R₁=Mg+µ₁Ma ...(ii)

T-µ₂R=Ma

On using (ii)

T-µ₂Mg-µ₂µ₁Ma=Ma ...(B)

On adding A and B, Tension gets canceled out, and I get,

a=(mg-µ₂Mg)/(µ₂µ₁M+µ₁M+2m+M)

This is not the correct answer as per the book. Where did I go wrong ?

 

[Maiq]

Why do you say R1=Ma? The net force acting an object is equal to its mass times its acceleration. R1 is the only force acting on block m so R1 would be equal to block m times the acceleration a.

 

[SammyS]

How did you get that Pseudo force?

Looking at the components of the acceleration of the small block, the block travels in a straight line, the slope of which is -2.

It looks like you have let a = the acceleration of the big block. I would do that too.

I would say that R₁ = ma .

For each FBD , you should have one equation for horizontal components and one for vertical components.

 

[sankalpmittal]

Why do you say R1=Ma? The net force acting an object is equal to its mass times its acceleration. R1 is the only force acting on block m so R1 would be equal to block m times the acceleration a. Also R1 affects block M as well, because of Newton's third law.

Yes. Thank you. I get it now. Now please see my solution and tell where I went wrong. Addresses to Maiq's below post.

Dang it ! Again thanks. I forgot to include R₁ in horizontal motion of big block.

How did you get that Pseudo force?

Looking at the components of the acceleration of the small block, the block travels in a straight line, the slope of which is -2.

It looks like you have let a = the acceleration of the big block. I would do that too.

I would say that R1 = ma .

For each FBD , you should have one equation for horizontal components and one for vertical components.

Ok, sorry.

Again with this correction, and maiq's correction, I will continue it tomorrow. For now, I am off to sleep.

Thinking back. Why did you say tension on big block be 2T ? Again, only one segment is connected directly to big block. Though its acceleration will be halved of smaller's downward one, but force will be still T.

 

[Maiq]

Also R1 affects block M as well, because of Newton's third law.

 

[Maiq]

After thinking about it i realized the last part of my last post was incorrect. It would actually equal -R1 if all other forces equaled 0. Sorry I didn't think of the direction of R1 when I thought of that explanation. Either way I believe everything else I said is true.

 

[sankalpmittal]

Ok, so after all the correction and including other forces:

FBD for small block:

Forces acting are:

1.Downward force mg
2.Upward force tension T
3.Friction upward: µ₁R₁
4.Pseudo force of ma on it towards left.
5.Reaction force R₁ by big block on it towards right for its horizontal equilibrium.

So, my equation becomes:

R₁=ma ..(i)
mg-T- µ₁R₁=2ma
On using (i),
mg-T- µ₁ma=2ma ... (A)

FBD of big block:

Forces acting on it are:

1.Downward force Mg
2.Tension T due right
3.Friction by ground of µ₂R due left
4.Reaction force or third law pair of friction, exerted by small block downward:µ₁R₁
5.Reaction force R by floor on big block
6.Reaction force R₁ on it due left by small block as pointed out by maiq.

So accounting for vertical equilibrium of big block,

R=Mg+µ₁R₁=Mg+µ₁ma ...(ii)

T-µ₂R-R₁=Ma

On using (ii)

T-µ₂Mg-µ₂µ₁ma-ma=Ma ...(B)

On adding A and B, Tension gets canceled out, and I get,

a=(mg-µ₂Mg)/(3m+M+µ₁µ₂m+µ₁m)

Now also the answer is incorrect!

Where did I go wrong? What forces I missed in the FBDs ?

 

[SammyS]

Ok, so after all the correction and including other forces:

FBD for small block:

Forces acting are:

1.Downward force mg
2.Upward force tension T
3.Friction upward: µ₁R₁
4.Pseudo force of ma on it towards left.
5.Reaction force R₁ by big block on it towards right for its horizontal equilibrium.

So, my equation becomes:

R₁=ma ..(i)
mg-T- µ₁R₁=2ma
On using (i),
mg-T- µ₁ma=2ma ... (A)

The above looks good.

FBD of big block:

Forces acting on it are:

1.Downward force Mg
2.Tension T due right
3.Friction by ground of µ₂R due left
4.Reaction force or third law pair of friction, exerted by small block downward:µ₁R₁
5.Reaction force R by floor on big block
6.Reaction force R₁ on it due left by small block as pointed out by maiq.

The string also passes over the pulley, which is attached to the bog block.

Therefore, the string has two strands pulling to the right on the big block: That's 2T . AND one strand pulling down on the big block.

So the stuff below is incorrect.

So accounting for vertical equilibrium of big block,

R=Mg+µ₁R₁=Mg+µ₁ma ...(ii)

T-µ₂R-R₁=Ma

On using (ii)

T-µ₂Mg-µ₂µ₁ma-ma=Ma ...(B)

On adding A and B, Tension gets canceled out, and I get,

a=(mg-µ₂Mg)/(3m+M+µ₁µ₂m+µ₁m)

Now also the answer is incorrect!

Where did I go wrong? What forces I missed in the FBDs ?

 

[haruspex]

mg-T- µ₁ma=2ma ... (A)
T-µ₂Mg-µ₂µ₁ma-ma=Ma ...(B)

I agree down to there, but I don't get the result below:

On adding A and B, Tension gets canceled out, and I get,

a=(mg-µ₂Mg)/(3m+M+µ₁µ₂m+µ₁m)

I get (mg-Ma-Mgµ₂)/[m(3+µ₁µ₂+µ₁)]

 

[sankalpmittal]

The above looks good.

The string also passes over the pulley, which is attached to the bog block.

Therefore, the string has two strands pulling to the right on the big block: That's 2T . AND one strand pulling down on the big block.

So the stuff below is incorrect.

Thanks SammyS ! I got the correct answer !

And of course thanks to maiq!

And thanks Jesse H. for at least trying...

@ haruspex:

Yes, when you isolate "a", you will get the wrong answer as I was. But on following SammyS's hint, I got the correct answer. Thanks SammyS, once again. And thanks haruspex for replying.