Forces/Dynamics: A block sliding on a plate

[jisbon]

No.
In post #5 you had two equations which I will abbreviate as $a=\frac{F-f_s}{m_b}$ and $a=\frac{f_s}{m_p}$.
Previously, you combined them so as to eliminate a and found the relationship between F and f_s. Now I am asking you to combine them so as to eliminate f_s instead, and hence find the relationship between F and a.

Ok.. If I understand your statement correctly, do you mean by:
$a=\frac{F-f_s}{m_b}$ and $a=\frac{f_s}{m_p}$
$m_b*a_b= F-f_s$
$f_s = F - (m_b*a_b)$
$a_p = \frac {(F - (m_b*a_b))}{m_p}$
?
Thank you so much for your patience and guidance

 

[haruspex]

Ok.. If I understand your statement correctly, do you mean by:
$a=\frac{F-f_s}{m_b}$ and $a=\frac{f_s}{m_p}$
$m_b*a_b= F-f_s$
$f_s = F - (m_b*a_b)$
$a_p = \frac {(F - (m_b*a_b))}{m_p}$
?
Thank you so much for your patience and guidance

Yes, but remember we are dealing with the time up to when slipping starts, so the two accelerations are the same.
Get it into the form a=..., and substitute the given formula for F as a function of t.

 

[jisbon]

Yes, but remember we are dealing with the time up to when slipping starts, so the two accelerations are the same.
Get it into the form a=..., and substitute the given formula for F as a function of t.

$m_b*a_b= F-f_s$
$f_s = F - (m_b*a_b)$
$a_p = \frac {(F - (m_b*a_b))}{m_p}$
Referring to your highlighted part, if I wish to substitute $F = 2tN$
Should't I just take $f_s = F - (m_b*a_b)$ = $F= f_s + (m_b*a_b)$ = $2tN$?
I feel that my though processes are just different from yours. So I would like to check the process again:

1) I need to find distance traveled by plate for 1s after it slips.
2) So I need to find out the velocity of the plate at the moment the block slips, and the acceleration of the plate at that point (which will be constant since only friction is involved for plate)
3)I can find the acceleration of the plate by using, F=ma, thus acceleration of plate is #5.88N/1.5## . Is this correct? (I'm thinking that acceleration is ALWAYS constant for plate since it only experience friction all the time)
4) Then to find out velocity of plate, I need to find out time it takes for block to slip. In this case, is finding t from:
$F= f_s + (m_b*a_b)$ = $2tN$ correct?
If I can find time taken for block to slip, won't the velocity of plate when block slip = $v=u+at$ where a is found in step 3?

Thank you.

 

[haruspex]

$m_b*a_b= F-f_s$
$f_s = F - (m_b*a_b)$
$a_p = \frac {(F - (m_b*a_b))}{m_p}$

You seem to have missed the part where I pointed out that until the block slips the two accelerations are the same. There is not an a_p and an a_b, just a. So get it into the form a=(a function of masses and F) then use F=2t.

 

[jisbon]

You seem to have missed the part where I pointed out that until the block slips the two accelerations are the same. There is not an a_p and an a_b, just a. So get it into the form a=(a function of masses and F) then use F=2t.

So something like:
$a = \frac {(F - (m_b*a_b))}{m_p}$
$a = \frac {(2TN - (m_b*a_b))}{m_p}$
?

 

[haruspex]

So something like:
$a = \frac {(F - (m_b*a_b))}{m_p}$
$a = \frac {(2TN - (m_b*a_b))}{m_p}$
?

Why do you still have a_b terms on the right? There is only the one acceleration during this phase of the problem. Turn all a_p and a_b terms into plain a and rearrange to get the equation into the form a=some function of F, m_b and m_p. There should be no a terms on the right.
When you have that, substitute 2t for F and see if you can get an equation for the velocity.

 

[jisbon]

Why do you still have a_b terms on the right? There is only the one acceleration during this phase of the problem. Turn all a_p and a_b terms into plain a and rearrange to get the equation into the form a=some function of F, m_b and m_p. There should be no a terms on the right.
When you have that, substitute 2t for F and see if you can get an equation for the velocity.

Oh sorry. Corrected it as followed:
$a = \frac {(F - (m_b*a))}{m_p}$
$a*m_p = (F - (m_b*a))$
$(a*m_p) + (m_b*a) = F$
$a(m_p + m_b) = F$
$a = \frac {F}{(m_p + m_b)}$
$a = \frac {2tN}{(m_p + m_b)}$
To be clear, I will be using N for the plate right?
This a I'm finding will be the acceleration before it slips. So from there, I am able to find the velocity of plate when it starts to slip with this acceleration value? Thanks!

 

[haruspex]

from there, I am able to find the velocity of plate when it starts to slip with this acceleration value?

I would call it an acceleration function, not a value; it is not constant.
But yes, you now have the acceleration as a function of time. How do you find the velocity as a function of time from that?

 

[jisbon]

I would call it an acceleration function, not a value; it is not constant.
But yes, you now have the acceleration as a function of time. How do you find the velocity as a function of time from that?

I will integrate it to become velocity?
So from my previous equation:

$a=\frac {2tN}{m_p+m_b}$

$a = \frac {2t*9.8*1.5}{1.5+0.5}$
$a = 14.7t$
$v = (\frac {14.7}{2})t^2$?

 

[haruspex]

I will integrate it to become velocity?
So from my previous equation:

$a = \frac {2t*9.8*1.5}{1.5+0.5}$
$a = 14.7t$
$v = (\frac {14.7}{2})t^2$?

Yes, your integration is fine, but where did the factor g come from?

 

[jisbon]

Yes, your integration is fine, but where did the factor g come from?

Oh wait isn't $N= g* mass$?

 

[haruspex]

Oh wait isn't $N= g* mass$?

No. N stands for Newtons. A Newton is the force required to accelerate a mass of 1kg at 1m/s². Nothing to do with gravity.

 

[jisbon]

No. N stands for Newtons. A Newton is the force required to accelerate a mass of 1kg at 1m/s². Nothing to do with gravity.

Oh no. I always thought that the formula was F=2tN where N is the normal force, turns out N is supposed to be the units :/
If that's the case will $a= \frac {2t}{1.5+0.5} = t$
Hence $v = \frac {1}{2}t^2$
?

 

[haruspex]

Oh no. I always thought that the formula was F=2tN where N is the normal force, turns out N is supposed to be the units :/
If that's the case will $a= \frac {2t}{1.5+0.5} = t$
Hence $v = \frac {1}{2}t^2$
?

Yes,

 

[jisbon]

So with a and v, I can find out initial velocity when plate before it slips?

 

[haruspex]

So with a and v, I can find out initial velocity when plate before it slips?

Yes - do you see how?