Forces/Dynamics: A block sliding on a plate

[jisbon]

Homework Statement

Block of mass m = 0.5kg sits on top of large plate with mass M = 1.5kg. Contact surface of plate and ground is friction less and coefficient of static and kinetic friction is 0.9 and 0.5 respectively. System is originally at rest and external force F(t) = 2t N (where t is time in seconds) is used to pull the block. Until block starts to slide, after which F is maintained at value in part (a). Assume plate is long enough such block doesn't fall off plate.

(a) Calculate external force F at moment before block starts to slide on the plate.
(b) Calculate how far plate will travel relative to ground 1s after block slips.
(c) Calculate how far block travel relative to the ground 1s after it slips.

Relevant Equations

s=ut+0.5at^2
F=ma

Question as follows:

Block of mass m = 0.5kg sits on top of large plate with mass M = 1.5kg. Contact surface of plate and ground is friction less and coefficient of static and kinetic friction is 0.9 and 0.5 respectively. System is originally at rest and external force F(t) = 2t N (where t is time in seconds) is used to pull the block. Until block starts to slide, after which F is maintained at value in part (a). Assume plate is long enough such block doesn't fall off plate.
(a) Calculate external force F at moment before block starts to slide on the plate.

My answer/explanation:
Force needed to make block slide = static friction between block and plate.
So friction = normal of block * static friction = (0.5∗9.8)∗0.9=4.41N(0.5∗9.8)∗0.9=4.41N

(b) Calculate how far plate will travel relative to ground 1s after block slips.

My thoughts/wrong answer probably?: after block slips, plate experience
kinetic friction = (1.5∗9.8)∗0.5=7.35N(1.5∗9.8)∗0.5=7.35N
To find distance , use
s=ut+0.5at2 where u=0, so s=0.2at2s=0.2at2
To find a, F=ma
Since plate only experience kinetic friction(?),
a=7.35/2a=7.35/2 (including block on plate) = 3.6753.675
s=0.5(3.675)=1.8375s=0.5(3.675)=1.8375m to the left?
Do help explain if it's wrong. Thanks

(c) Calculate how far block travel relative to the ground 1s after it slips.

Using similar concepts as above, After 1s,
F= 2t N= 2(1)(4.41)=8.822(1)(4.41)=8.82
Friction = 0.5∗0.5∗9.8=2.450.5∗0.5∗9.8=2.45
a=8.82−2.45/0.5=12.74a=8.82−2.45/0.5=12.74
s=0.5(12.74)(1)2=6.37ms=0.5(12.74)(1)2=6.37m to the right?
Help will really be appreciated. Thanks

 

[haruspex]

Force needed to make block slide = static friction between block and plate.

That would be true if the plate were held fixed. But plate and block can accelerate.
Consider the net horizontal force on the block. What equation can you write?

 

[jisbon]

That would be true if the plate were held fixed. But plate and block can accelerate.
Consider the net horizontal force on the block. What equation can you write?

Net force on block = external force - static friction between block and plate ?
I can't seem to figure out how if plate moves in the opposite direction, it will affect the force on the block :(

 

[haruspex]

Net force on block = external force - static friction between block and plate ?

Right. Now consider acceleration (a) of the block and (b) of the plate.

 

[jisbon]

Right. Now consider acceleration (a) of the block and (b) of the plate.

Acceleration of block = (F-static friction)/mass of block
Acceleration of plate= static friction/mass of plate?

 

[haruspex]

Acceleration of block = (F-static friction)/mass of block
Acceleration of plate= static friction/mass of plate?

View attachment 247473

Right. And as long as the block is not sliding on the plate, how can you connect those equations?

 

[jisbon]

Right. And as long as the block is not sliding on the plate, how can you connect those equations?

If block is not sliding, does that mean both block and plate has the same acceleration?

 

[haruspex]

If block is not sliding, does that mean both block and plate has the same acceleration?

Yes.

 

[jisbon]

Yes.

Okay, so going back to the question:
(a) Calculate external force F at moment before block starts to slide on the plate.
So I assume to slide, Acc of block must be more than acc of plate.
So in this case,
$$\frac {F-friction} {mass of block} = \frac {friction} {mass of plate+block?}$$
And solve for F?

I have also been thinking about part b: (b) Calculate how far plate will travel relative to ground 1s after block slips.

So after it slips, the question states that F is maintained at value in part (a), yet there is formula given $$F=2t N$$ In this case, which should I actually follow?

Thank you

 

[haruspex]

$$\frac {F-friction} {mass of block} = \frac {friction} {mass of plate+block?}$$

Where did the +block come from?
That was not in your post #5.

 

[jisbon]

Where did the +block come from?
That was not in your post #5.

Oh because I figured to make the mass includes the block since the plate is technically carrying the block, or is it not?

 

[haruspex]

Oh because I figured to make the mass includes the block since the plate is technically carrying the block, or is it not?

In post #5 you wrote, correctly

Acceleration of plate= static friction/mass of plate

That comes from consideration of the forces on the plate.
If you want to treat the plate and block as an item the frictional force between them becomes an internal force and will not appear in the ΣF=ma equation.

 

[jisbon]

In post #5 you wrote, correctly

That comes from consideration of the forces on the plate.
If you want to treat the plate and block as an item the frictional force between them becomes an internal force and will not appear in the ΣF=ma equation.

oh yep that actually makes sense now. i got part a to be 4.41N. Now regarding part b:
Calculate how far plate will travel relative to ground 1s after block slips.

So after it slips, the question states that F is maintained at value in part (a), yet there is formula given $$F=2tNF=2tN$$
In this case, which should I actually follow?

 

[haruspex]

i got part a to be 4.41N

No, that's what you had originally. You should be getting a different answer using your equations in post #5.

the question states that F is maintained at value in part (a), yet there is formula given

The formula ceases to apply once the block starts to slip.

I'm not sure what part b is asking. Does it want the whole distance traveled by the plate, from rest, at a point in time 1 second after slipping starts, or does it want the distance traveled during the 1 second after slipping starts? It sounds like the latter.
Either way, you will need to find the velocity when slipping starts.

 

[jisbon]

No, that's what you had originally. You should be getting a different answer using your equations in post #5.

The formula ceases to apply once the block starts to slip.

I'm not sure what part b is asking. Does it want the whole distance traveled by the plate, from rest, at a point in time 1 second after slipping starts, or does it want the distance traveled during the 1 second after slipping starts? It sounds like the latter.
Either way, you will need to find the velocity when slipping starts.

Hm that's weird let me see again:
So part (a) is:
$$\frac {F-friction} {mass of block} = \frac {friction} {mass of plate}$$
Friction = $$(0.5*9.8)*0.9 = 4.41N$$
So$$F = (\frac {friction} {mass of plate} * mass of block) + friction$$
= $$(\frac {4.41} {1.5} * 0.5) + 4.41$$
= 5.88N
Will this be the final answer?
Regarding part b, I think the questions wants the distance traveled during the 1 second after slipping starts. So how should I start? By finding the initial velocity when it starts to slip? Or..? Thanks!

 

[haruspex]

= 5.88N
Will this be the final answer?

Yes.

finding the initial velocity when it starts to slip?

Yes, you will need to do that, and find the subsequent acceleration.

 

[jisbon]

Yes.

Yes, you will need to do that, and find the subsequent acceleration.

Thanks for part a.

Regarding b, I can;t figure out how plate has a initial velocity, because what I thought is it initially at rest and only starts accelerating due to the kinetic friction, hence my workings as follows:

Kinetic friction =$$(1.5∗9.8)∗0.5=7.35N$$
To find distance , use
$$s=ut+0.5at^2$$ where u=0, so $$s=0.5at^2$$
To find a, $$F=ma$$
Since plate only experience kinetic friction(?),
$$a=7.35/2a=7.35/2$$ (including block on plate) = 3.6753.675
$$s=0.5(3.675)=1.8375m$$

Could I have a hint or sort on where do I get my initial velocity? Thank you

 

[haruspex]

Regarding b, I can;t figure out how plate has a initial velocity, because what I thought is it initially at rest and only starts accelerating due to the kinetic friction,

There is no friction between plate and ground, but as soon as F > 0 there is friction between plate and block, so the plate accelerates.

 

[jisbon]

There is no friction between plate and ground, but as soon as F > 0 there is friction between plate and block, so the plate accelerates.

So with that I gathered, I supposed the plate is already moving even when block did not start to slide on plate, but how am I suppose to find the initial velocity of the plate? I'm kind of stuck right here :(

 

[haruspex]

So with that I gathered, I supposed the plate is already moving even when block did not start to slide on plate, but how am I suppose to find the initial velocity of the plate? I'm kind of stuck right here :(

At what time does the block start to slide?
Up until that point, they move together under the force F. What is the acceleration as a function of time, and how does that turn into speed as a function of time?

 

[jisbon]

At what time does the block start to slide?
Up until that point, they move together under the force F. What is the acceleration as a function of time, and how does that turn into speed as a function of time?

Block only starts to slide at t seconds (Using the formula $F(t) = 2t N$, so only starts to slide at $t= \frac {F}{2N} = \frac {5.88N}{2*9.8*0.5} = 0.6 seconds$
Acceleration as a function of time (not sure what you meant by this but do you mean): $a= \frac {dv}{dt}$?

I still can't figure out how finding the time taken for block to slide will give the inital velocity tho :(

 

[haruspex]

$a= \frac {dv}{dt}$?

Yes, but you know force as a function of time and you know the mass, so...

 

[jisbon]

Yes, but you know force as a function of time and you know the mass, so...

Won't the inital velocity = when t = 0, which means that$F(t) = 2t N = 0$? If F=0, won't there be any acceleration?

 

[haruspex]

Won't the inital velocity = when t = 0, which means that$F(t) = 2t N = 0$? If F=0, won't there be any acceleration?

When t=0 there is no acceleration, but a fraction of a second later there will be.
Please try to write the equation connecting the given force function with the equation I quoted in post #22.

 

[jisbon]

When t=0 there is no acceleration, but a fraction of a second later there will be.
Please try to write the equation connecting the given force function with the equation I quoted in post #22.

$F(t) = 2t N$ and
$a = \frac {dv}{dt}$
Is it correct to state that $\frac {2tN}{mass of plate} = acceleration$?

 

[haruspex]

$F(t) = 2t N$ and
$a = \frac {dv}{dt}$

Yes.

Is it correct to state that $\frac {2tN}{mass of plate} = acceleration$?

No. Use your post #5 equations again, this time eliminating frictional force.

 

[jisbon]

Yes.

No. Use your post #5 equations again, this time eliminating frictional force.

If I eliminate friction from:
$Acceleration of plate= static friction/mass of plate$
Won't there be any acceleration?

 

[haruspex]

If I eliminate friction from:
$Acceleration of plate= static friction/mass of plate$
Won't there be any acceleration?

I said to use the two equations in a way that eliminates the static friction force. You should be left with an equation relating F, acceleration and mass.

 

[jisbon]

$Acceleration of plate= static friction/mass of plate$

$F(t)=2tN$ and
$a= \frac {dv}{dt}$

Since static friction = static friction coefficient* N,
N = \frac {F}{2t}
$static friction = \frac {F}{2t} * static friction coefficient$ =
$Acceleration of plate= \frac{\frac {F}{2t} * static friction coefficient}{mass of plate}$
Do you mean this by any chance? Thanks

 

[haruspex]

$Acceleration of plate= static friction/mass of plate$

Since static friction = static friction coefficient* N,
N = \frac {F}{2t}
$static friction = \frac {F}{2t} * static friction coefficient$ =
$Acceleration of plate= \frac{\frac {F}{2t} * static friction coefficient}{mass of plate}$
Do you mean this by any chance? Thanks

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.

 

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