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.
Homework 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.
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

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haruspex

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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

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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? Last edited:

haruspex

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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

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

jisbon

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

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

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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

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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 travelled by the plate, from rest, at a point in time 1 second after slipping starts, or does it want the distance travelled 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 travelled by the plate, from rest, at a point in time 1 second after slipping starts, or does it want the distance travelled 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

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= 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

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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

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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

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

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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 $\frac {2tN}{mass of plate} = acceleration$?

"Forces/Dynamics: A block sliding on a plate"

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