How to solve questions in pseudo force frames?

[burian]

Homework Statement

Block on ground with another block on top. All surfaces of contact have friction

Relevant Equations

$$ F_{net} = ma$$\

$$f_{s} = u mg$$

$$ \frac{62 - .4mg + (.2) (m_{net} g)}{m} = a(t)$$

And, then I put ths into

$$S=\frac{1}{2} at^2$$

But, my answer is not quite coming with what's in the options?

So, my main confusions are,
SHould the bottom most surface exert a force on the top most block, also where am I making mistakes in pseudo force applications?

 

[haruspex]

You need to explain your first equation. I assume m in the .4mg refers to the top block, and that you are trying to calculate the acceleration of the lower block, but why are you adding (.2)m_net g?
Having got the acceleration of the lower block, how does that relate to the time it takes the top block to fall? Remember the top block is accelerating too.

 

[burian]

Ok, so when the motion starts, the big block is accelerating some Newtons to the left and hence the block on top accelerates leftwards by same amount relative to it. Now this amount precisely is (62 - frictional force acting at bottom of big block) N( note: Itook leftward as positive)

As, top block moves, the surface puts friction against it's motion so that's the -.4mg term.

So, the idea is once I Got the relative acceleration in pseudo force frame, the motion is like a particle moving in line under constant acceleration and we have to find time where it reaches some point. So that's what I did, but doesn't seem to being go so well for...reasons

 

[haruspex]

the big block is accelerating some Newtons to the left

It accelerates to the right, and N is not a unit if acceleration.

block on top accelerates leftwards by same amount relative to it.

If it accelerates the other way by relatively the same amount it is stationary in the lab frame. Or did you mean some amount?

this amount precisely is (62 - frictional force acting at bottom of big block) N

(62N - frictional force acting at bottom of big block) is the net force on the two block system, not an acceleration.

once I Got the relative acceleration in pseudo force frame

If you intend to use a non inertial frame you need to state what it is, or readers cannot tell what you are doing.

 

[burian]

burian said:
the big block is accelerating some Newtons to the left

It accelerates to the right, and N is not a unit if acceleration.

burian said:
block on top accelerates leftwards by same amount relative to it.

If it accelerates the other way by relatively the same amount it is stationary in the lab frame. Or did you mean some amount? [\quote]

oops I mispoke I meant the big block is accelerating to right
The top block accelerates leftwards.

burian said:
once I Got the relative acceleration in pseudo force frame

If you intend to use a non inertial frame you need to state what it is, or readers cannot tell what you are doing.
[\quote]

I was trying to use frame of bottom block

 

[haruspex]

I meant the big block is accelerating to right
The top block accelerates leftwards.

It accelerates leftward relative to the lower block, but rightward in the lab frame.
If you want to use the frame of the lower block then the lower block is stationary and the upper block accelerates leftward. Try not to mix frames in the description without explanation.

So what is the a(t) you wrote the expression for? Is it the acceleration of the lower block in the lab frame, of the upper block in the lab frame, or the upper block in the lower block's frame? The expression you wrote is not correct for any of those.

 

[burian]

It accelerates leftward relative to the lower block, but rightward in the lab frame.
If you want to use the frame of the lower block then the lower block is stationary and the upper block accelerates leftward. Try not to mix frames in the description without explanation.

So what is the a(t) you wrote the expression for? Is it the acceleration of the lower block in the lab frame, of the upper block in the lab frame, or the upper block in the lower block's frame? The expression you wrote is not correct for any of those.

Mhm yes I meant, in the frame of top block. Everything is in frame of top block ( made a mistake initially) , so that's why the pseudo force came.

I tried writing the expression for acceleration of top block in the accelerating frame of top block, but since it's in an accelerated from it has pseudo force, Hence I included the force between big block and ground and the force which the small block on top is being ulled in it

 

[haruspex]

acceleration of top block in the accelerating frame of top block

The acceleration of an object in its own frame is necessarily zero.
If you want to work in the frame of an object that has acceleration a then for each object in the system you invent a pseudo force; if an object has mass m then the pseudo force on that object is -ma.

It looks to me that you were trying to find the acceleration of the lower block in the lab frame so that you could calculate the pseudo force on the top block.
In the expression, you have the force F and the friction from the ground, but you add them, which is surprising to say the least. Both of these act on the lower block. You also appear to have a frictional force from the top block, but you just write "m", so this is unclear, but so far it looks like this is all about forces on the lower block. Then you divide by "m", but surely this is a different m now?

 

[burian]

Ok, so I realized I was making some fundamental mistakes, I tried doing it all again, and this is now what I got. Does my application of pseudo force now seem correct? I tried to work in a frame relative to the block on bottom i.e block B

 

[kuruman]

Still not correct. You show a "net" force of 62 N-18 N = 44 N. If that is supposed to be the net force on the bottom block, it is missing the force of friction exerted by the top block. Make the correction, then find the acceleration of the bottom block relative to the ground and use that to find the pseudo force acting on the top block. Remember that in a frame where the bottom block is stationary, an object of mass $m$ (e.g. the top block) experiences a pseudo force $F_{pseudo}=-ma_{bot.}$ where $a_{bot.}$ is the acceleration of the bottom block.

 

[burian]

I tried again and I got this, btw is

$$ F_{relative} = F_{actual} - F_{pseudo} =5* a_{rel} ?$$

 

[kuruman]

You most recent work is incomplete. We cannot help you pinpoint your mistakes if you don't show all your work, not just the bottom line.
1. Draw a free body diagram (FBD) showing all the forces acting on the bottom block.
2. Write Newton's second law for the bottom mass based on your FBD that you drew.
3. Find the acceleration of the bottom mass in the lab frame.
4. Find the fictitious force acting in the accelerated frame.
5. Draw a FBD of the top mass in the accelerated frame.
6. Find the net force on the top mass in the accelerated frame.
7. Find the acceleration of the top mass in the accelerated frame.

When you draw your FBDs, pay particular attention to the direction of the force(s) of friction acting on each block. Being systematic and organized should help a lot, you and anyone else reading this thread who wishes to offer help.

 

[haruspex]

View attachment 267795
I tried again and I got this, btw is

$$ F_{relative} = F_{actual} - F_{pseudo} =5* a_{rel} ?$$

In post #1 you had a term -.4mg, where I think the m referred to the 5kg. Now this has become +20N. Why the sign change?

The next two lines I cannot read. Please take the trouble to type in your algebra. It makes it a lot easier for anyone trying to help you.
But I think you are then subtracting the 64N you (wrongly) calculated from other forces, as though this 64N is the pseudo force to be applied to the top block. It is not!
When you have correctly found the net force on the lower block you can use that to find the lower block's acceleration. Then you can multiply that acceleration by the top block's mass to find the pseudo force on the top block.
I explained all this in an earlier post.

 

[burian]

NOTE: signs to right are positive, signs to left are negative

In the bottom block, net force is,

$$ F_{B} = 62 N+.4 (N_A ) - .2 (N_a + N_b) = 62 + .4 * 5 * 10 - .2 ( 5*10 + 4*10) =62 + 4 *5 - 2(9) =64$$

$$F_{A} = 62 N - .4N_{A} = 62N -.4 * 10 * 5 =42N$$

Now in the non inertial frame of block-B, the acceleration on block-A,

$$ 5*a = 42-64 $$

$$ a = \frac{-22}{5} =- 4.5$$

Now does my acceleration seem right?

 

[haruspex]

In the bottom block, net force is,

$$ F_{B} = 62 N+.4 (N_A ) - .2 (N_a + N_b) $$

You still do not have the signs right here.
Which way is the top block accelerating? What force makes it accelerate? So which way is that force acting on it?
What exerts that force on it? So what force does the top block exert in return?

 

[burian]

You still do not have the signs right here.
Which way is the top block accelerating? What force makes it accelerate? So which way is that force acting on it?
What exerts that force on it? So what force does the top block exert in return?

In lab frame, top is moving to right but in the frame relative to bottom block, it's moving to left.

The force is due to friction and pseudo force.

I'm not sure where my signs are wrong hmmm

 

[haruspex]

but in the frame relative to bottom block, it's moving to left.

Right, so which way does the frictional force between the blocks act on the lower block?

 

[burian]

Right, so which way does the frictional force between the blocks act on the lower block?

Depends on which friction you are talking about, if it's one with bottom surface then it's leftward and if it's between top surface than it is rightward

 

[haruspex]

Depends on which friction you are talking about, if it's one with bottom surface then it's leftward and if it's between top surface than it is rightward

I said "between the blocks", so that's the top of the lower block.
In post #16 you wrote correctly of the top block that

relative to bottom block, it's moving to left.

If the top block is sliding left on the lower block, which way is the frictional force it exerts on the lower block??

 

[burian]

I said "between the blocks", so that's the top of the lower block.
In post #16 you wrote correctly of the top block that

If the top block is sliding left on the lower block, which way is the frictional force it exerts on the lower block??

To the right!

$$F_{B} = 100N$$

$$ F_{A}=42N$$

The relative acceleration is:

$$ F_{B} - F_{A} = ma_{rel} = 58N$$

$$5a_{rel} = 58N$$

$$a_{rel} = \frac{58}{5} \approx 12$$

did I do it right now?

 

[kuruman]

Perhaps this will help. Imagine standing on the bottom block, moving with it as one. You look around and you see the top block slide towards the end of the bottom block 16 m away. If you can find the net force on the top block according to you, then you can easily find the time it takes to reach the end if you can find the acceleration according to you.

To find the acceleration according to you, you need to find the net force according to you. How many forces act on the block according to you? Answer: Two. The fictitious force and friction. What are the magnitudes and directions of these according to you?

 

[haruspex]

To the right!

In post #17 I asked which way the friction from the top block acts on the lower block, and the answer to that is to the left. But you now correctly have 42N for the net force on the lower block, so maybe you just didn't read my question aright.

$$F_{B} = 100N$$
$$ F_{A}=42N$$

The relative acceleration is:
$$ F_{B} - F_{A} = ma_{rel} = 58N$$

In the diagram, the lower block is B, so I am not sure how you are defining $F_A$ and $F_B$. The net force on B is 42N to the right.
What force is 100N?

But you still seem confused about forces and accelerations. You get a relative acceleration by subtracting one acceleration from another, not by subtracting one force from another.
I have told you this twice already, in the first part of post #8 and the last part of post #13.