Force on the blocks and its acceleration

[theunloved]

A slab of mass m1 = 40 kg rests on a frictionless floor. a block of mass m2 = 10 kg rests on top of the slab ? Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. The block is pulled by a horizontal force with a magnitude of 100 N. (coefficient u between the slab and floor = 0)
(a) What is the resulting acceleration of the block
(b) What is the resulting acceleration of the slab

The 10 kg block is sitting on the slab. The slab is frictionless on the floor. The block and slab will be moved by the 100 N force.

This is how I did for part a
Taking upwards as positive, x+ going from right to left

From FBD
m2:
Fcontact - m2g = 0
Fc = m2g

F - fk2 = m2a2
F - uk2 * m2g = m2a2

a2 = (F - ukm2g) / m2 = 6.08m/s^2

Part B, I have no idea how, since according to my FBD, there not a single force acting on m1 horizontally.

y dir
m1:
N1 - m1g - Fc = 0

x dir
0 = 0 (since there is not a single force acting on it ?)

So how can I find acceleration of the slab ? Did I miss anything ? Is there a force acting on m1 that I don't know ? Thanks

 

[collinsmark]

Part B, I have no idea how, since according to my FBD, there not a single force acting on m1 horizontally.

You've already figured out above that m₂ feels a force of friction (separate from the external 100 N horizontal force acting on it). Where does the frictional force emanate? In other words, which object is exerting this frictional force on m₂?

What is Newton's third law of motion?

 

[theunloved]

You've already figured out above that m₂ feels a force of friction (separate from the external 100 N horizontal force acting on it). Where does the frictional force emanate? In other words, which object is exerting this frictional force on m₂?

What is Newton's third law of motion?

So you're saying that the frictional force is the only force that makes m¹ move ?

 

[collinsmark]

So you're saying that the frictional force is the only force that makes m¹ move ?

That's right. There you go. (just be careful about the direction)

 

[theunloved]

I think it will be opposite side with the horizontal force F ?
so fk2 = -m1a
a = -fk2 / m1 = -uk(m2g) / m1 ?
I still don't get it why frictional force fk2 will be the pulling force on m1, weird...

 

[collinsmark]

I think it will be opposite side with the horizontal force F ?
so fk2 = -m1a
a = -fk2 / m1 = -uk(m2g) / m1 ?

Be careful about the direction (and thus the minus sign). You might wish to look at it again.

I still don't get it why frictional force fk2 will be the pulling force on m1, weird...

Let's step back a bit to the forces acting on M₂ (the block, not the slab), and forget about M₁ for just a moment (we'll come back to it below). Which direction is the frictional force on M₂? Is it in the same direction as the external, 100 N horizontal force or is it in the opposite direction? Whichever the answer, realize that this frictional force is coming from the slab M₁. (The block is atop of the slab.)

So we've established that the M₁ slab exerts a force on the M₂ block. And according to Newton's third law of motion, the block must also exert a force on the slab. Which direction is this block to slab force (M₂ to M₁)? And what is its magnitude?

Newton's third law of motion states "For every action, there is an equal and opposite reaction."

 

[theunloved]

A slab of mass m1 = 40 kg rests on a frictionless floor. a block of mass m2 = 10 kg rests on top of the slab ? Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. The block is pulled by a horizontal force with a magnitude of 100 N. (coefficient u between the slab and floor = 0)
(a) What is the resulting acceleration of the block
(b) What is the resulting acceleration of the slab

The 10 kg block is sitting on the slab. The slab is frictionless on the floor. The block and slab will be moved by the 100 N force.

This is how I did for part a
Taking upwards as positive, x+ going from right to left

From FBD
m2:
Fcontact - m2g = 0
Fc = m2g

F - fk2 = m2a2
F - uk2 * m2g = m2a2

a2 = (F - ukm2g) / m2 = 6.08m/s^2

Part B, I have no idea how, since according to my FBD, there not a single force acting on m1 horizontally.

y dir
m1:
N1 - m1g - Fc = 0

x dir
0 = 0 (since there is not a single force acting on it ?)

So how can I find acceleration of the slab ? Did I miss anything ? Is there a force acting on m1 that I don't know ? Thanks

Thanks collinsmark, I finally got it and understand how third Newton's law works. On the other hand, I'm still a bit confused about the sign of the frictional force acting on m1.
I uploaded the Picture and my FBD, and as you can see, if m2 is moved from right to left, m1 will be sliding from left to right, therefore, its frictional force will be opposite to its direction which is positive, so isn't it that fk2 = -m1a ? acceleration should be negative because its opposite to x+ ?

 

[collinsmark]

I uploaded the Picture and my FBD, and if m2 is moved from right to left, m1 will be sliding from left to right,

Whoa, stop right there!

In your second attachment, notice that m₂ has two horizontal forces acting on it. Forces are vectors and they need to be added together as vectors. To find the net force, take the head of one horizontal vector and put it up to the tail of the other horizontal vector. The net force on m₂ is from the remaining tail to the remaining head.

But now look at horizontal forces acting on m₁. There is only one force, and it's pointing to the left. (You don't need to add any horizontal vectors together for this part, since there is only 1 horizontal vector.) It points to the left, meaning that m₁ accelerates to the left. (Meaning that both objects accelerate to the left.)

therefore, its frictional force will be opposite to its direction which is positive, so isn't it that fk2 = -m1a ? acceleration should be negative because its opposite to x+ ?

The FBD has the horizontal forces correct.

The frictional force applied on an object will be opposite the object's direction if the frictional force is due to that object sliding on something stationary (or at least sliding on something that's moving slower than itself). That is the case for m₂, but not m₁. Instead, in the m₁ situation the frictional force is due to a different object sliding on top of m₁. Sure, m₂ feels a frictional force in the opposite direction of its motion, but m₁ feels a frictional force in the same direction as its motion (in this problem) due to Newton's third law.

[STRIKE]By the way, not that it matters for this problem, but there is one force vector missing for a vertical force in the FBD (second attachment, second figure). Among other forces, the figure shows N₂ pushing down on m₁, which is correct (nothing wrong with that). But due to Newton's third law, m₁ is also pushing back up on m₂ by the same amount (N₂), and you're missing that force vector.[/STRIKE]

And one last thing, I agree with your original answer for part (a) . It's only part (b) that I was commenting about.

 

[collinsmark]

Actually, I take that last part back about the vertical forces. Nevermind about that. The figure is correct if you define N₁'s magnitude to be the sum of N₂ and m₁g.

 

[theunloved]

Whoa, stop right there!

In your second attachment, notice that m₂ has two horizontal forces acting on it. Forces are vectors and they need to be added together as vectors. To find the net force, take the head of one horizontal vector and put it up to the tail of the other horizontal vector. The net force on m₂ is from the remaining tail to the remaining head.

But now look at horizontal forces acting on m₁. There is only one force, and it's pointing to the left. (You don't need to add any horizontal vectors together for this part, since there is only 1 horizontal vector.) It points to the left, meaning that m₁ accelerates to the left. (Meaning that both objects accelerate to the left.)

The FBD has the horizontal forces correct.

The frictional force applied on an object will be opposite the object's direction if the frictional force is due to that object sliding on something stationary (or at least sliding on something that's moving slower than itself). That is the case for m₂, but not m₁. Instead, in the m₁ situation the frictional force is due to a different object sliding on top of m₁. Sure, m₂ feels a frictional force in the opposite direction of its motion, but m₁ feels a frictional force in the same direction as its motion (in this problem) due to Newton's third law.

[STRIKE]By the way, not that it matters for this problem, but there is one force vector missing for a vertical force in the FBD (second attachment, second figure). Among other forces, the figure shows N₂ pushing down on m₁, which is correct (nothing wrong with that). But due to Newton's third law, m₁ is also pushing back up on m₂ by the same amount (N₂), and you're missing that force vector.[/STRIKE]

And one last thing, I agree with your original answer for part (a) . It's only part (b) that I was commenting about.

Alright, so for m1, it should be fk = m1a ;)

Very clear explanation, wish that you were our professor college, physics would be fun then ;) Thank you very much for your help.