# PHYSICS Very confusing forces question help please, who’s right ?

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#1

Here’s the thing. My answer is B.
Idk which is right because he has a point.
What he did is he calculated the frictional force using the 2g*0.5

What I did is 24.5-5g(0.3)=5a
Then I got a=1.96
And I said frictional force=2a so B

Who’s right and why is the other wrong ?
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#2
RDKGames Physics Enemy @anybody who sees this help please
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1 year ago
#3
(Original post by Leah.J)
...
Find out if sliding occurs between m1, m2; determines moving as 1 lump or seperately. Fs = 0.5*2g < 24.5, so m2 slides under m1. So m1 experiences Fk = 0.3*2g = 5.9 N (1 d.p), to the right.

EDIT: See discussion below, I now agree with b)
Last edited by Physics Enemy; 1 year ago
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#4
(Original post by Physics Enemy)
Find out if sliding occurs between m1, m2; determines moving as 1 lump or seperately. Fs = 0.5*2g < 24.5, so m2 slides under m1. So m1 experiences Fk = 0.3*2g = 5.9 N (1 d.p), to the right.
I don’t get it
If the frictional force is less than 24.5, the boxes don’t move together ? m2 slides and m1 falls to the ground ? So the force it experiences is 2g*0.3 ?

Isn’t it correct to say that 24.5-5g(0.3)=5a ?
Is it not correct because of whatever calculation you did ?
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1 year ago
#5
(Original post by Leah.J)

Here’s the thing. My answer is B.
Idk which is right because he has a point.
What he did is he calculated the frictional force using the 2g*0.5

What I did is 24.5-5g(0.3)=5a
Then I got a=1.96
And I said frictional force=2a so B

Who’s right and why is the other wrong ?

The kinetic friction on m2 is 5g*0.3=1.5g=~15N

That leaves 9.5N to accelerate the two masses. If m1 doesn't slip, then common acceleration requires that the net force divides in proportion to their mass. That would require ~3.8N for m1. Its static friction limits it to 2g*0.5=g=~10N, so it doesn't slide.
Last edited by RogerOxon; 1 year ago
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1 year ago
#6
(Original post by Physics Enemy)
24.5 N is more than the static friction so m2 slides underneath m1. For m1 this is motion relative to the surface, so it has kinetic friction (only) of 5.9 N to the right. You modelled a 5 kg lump but they're 2 separate blocks, sliding relative to each other.
No. 24.5N is applied to m2 - it is not the force that m2 exerts on m1 (and vice versa).

You model it as one block, then test to see if the static friction between the two is sufficient for them to accelerate at the same rate. In this case, it is.
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1 year ago
#7
(Original post by RogerOxon)
No. 24.5N is applied to m2 - it is not the force that m2 exerts on m1 (and vice versa).

You model it as one block, then test to see if the static friction between the two is sufficient for them to accelerate at the same rate. In this case, it is.
I didn't say that at all - the diagram makes it clear it's on m2. As it's more than the static friction, m2 slides underneath m1. So I don't see why you'd model them as 1 lump. Unsure about the test you refer to. Interested to know the answer you reach.
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1 year ago
#8
(Original post by Physics Enemy)
I didn't say that at all - the diagram makes it clear it's on m2. As it's more than the static friction, m2 slides underneath m1. So I don't see why you'd model them as 1 lump. Unsure about the test you refer to. Interested to know the answer you reach.
You said that m2 slides under m1 - how did you conclude that? It appears that you compared it with static friction somewhere, which doesn't make sense - the applied force is countered by the force between the blocks and friction from the horizontal surface. No single interface sees all of it.

A force of 5.9N on m2 would see it accelerate faster than m1 ..
Last edited by RogerOxon; 1 year ago
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1 year ago
#9
(Original post by RogerOxon)
You said that m2 slides under m1 - how did you conclude that?
As I said in the thread - 24.5 is more than the static friction between m1 and m2 of 0.5*2g, so m2 slides underneath and my ans follows. I may be wrong ofc, interested to know the book's answer and method.
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1 year ago
#10
(Original post by Physics Enemy)
As I said in the thread - 24.5 is more than the static friction between m1&m2 of 0.5*2g, so m2 slides underneath and my ans follows. I could be wrong ofc, interested to know the book's answer and method.
Sorry, but you are wrong. Even if there were no friction between m2 and the horizontal surface, m1 would not see all of the 24.5N, as it also accelerates m2.
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1 year ago
#11
(Original post by RogerOxon)
A force of 5.9N on m2 would see it accelerate faster than m1 ..
My ans says M1 (on top) has kinetic friction 5.9 N to the right. For M2, it's 24.5 - 0.3*5g = 9.8 N to the right i.e) M2 sliding under M1.
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1 year ago
#12
(Original post by Physics Enemy)
My ans says M1 (on top) has kinetic friction 5.9 N to the right. For M2, it's 24.5 - 0.3*5g = 9.8 N to the right i.e) M2 sliding under M1.
You have omitted the force that M1 exerts on m2.
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1 year ago
#13
(Original post by RogerOxon)
You have omitted the force that M1 exerts on m2.
True, 24.5 - 0.3*5g = 9.8 is what you got; not sliding. Thinking about it, your reasoning to divide it as 2:3 makes sense, so 3.9 N < g for M1 is likely correct.
Last edited by Physics Enemy; 1 year ago
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1 year ago
#14
(Original post by Leah.J)

Here’s the thing. My answer is B.
Idk which is right because he has a point.
What he did is he calculated the frictional force using the 2g*0.5

What I did is 24.5-5g(0.3)=5a
Then I got a=1.96
And I said frictional force=2a so B

Who’s right and why is the other wrong ?
Although your friend’s working is incorrect, I would say that the question is misleading in some way or it is written in this way deliberately. The coefficient of static friction between m1 and m2 should be stated as an inequality:
μs ≤ 0.5
or
stating that the maximum coefficient of static friction between m1 and m2 is 0.5, instead of stating that the coefficient of static friction between m1 and m2 is 0.5.

The inequality of the coefficient of static friction between m1 and m2 implies that the static friction acting on m1 is
Static friction ≤ 2(0.5)mg

Without additional info such as the m1 is on the verge of sliding, we cannot conclude the static friction is at the maximum.
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1 year ago
#15
(Original post by RogerOxon)

The kinetic friction on m2 is 5g*0.3=1.5g=~15N

That leaves 9.5N to accelerate the two masses. If m1 doesn't slip, then common acceleration requires that the net force divides in proportion to their mass. That would require ~3.8N for m1. Its static friction limits it to 2g*0.5=g=~10N, so it doesn't slide.

PRSOM!

By saying “The kinetic friction on m2 is 5g*0.3=1.5g=~15N” can be confusing in IMO, as you mention “That leaves 9.5N to accelerate the two masses”. At first, you are looking at m2 and using that info of m2 to “conclude” a resultant force of 9.5 N on m1 and m2. Sometimes for the simple-minded, they may write the N2L for m2 based on your conclusion as

24.5 - 5g*0.3 = m2a

and it is incorrect.

I would add something to avoid any ambiguity.
There are two frictional forces acting on m2:
Kinetic friction between m2 surface and the ground:
0.3*5g
Static friction between m2 and m1:
μsm1g= m1a

Anyway, this is just my view.
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1 year ago
#16
(Original post by Eimmanuel)
Anyway, this is just my view.
Good points. He was looking at the system as a 5 kg lump, shouldn't refer to m2 really, you'd then have to invoke static friction too.

As a 5 kg lump, it allows you to 'ignore' static friction as it forms an internal N3 pair on m1, m2 so net = 0. Hence 24.5 - Fk = 5a and then Fs = 2a acting on m1.

But modelling masses individually, noting Fs is unknown at the outset (whereas Fk is at max):

For m2: 24.5 - Fk - Fs = 3a, For m1: Fs = 2a
Eliminating Fs gives 24.5 - Fk = 5a as above
Where Fs ⩽ g so a ⩽ g/2, else sliding: a2 > a1
Last edited by Physics Enemy; 1 year ago
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