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

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Here’s the thing. My answer is B.

My friends answer is C.

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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#3

(Original post by

...

**Leah.J**)...

_{s}= 0.5*2g < 24.5, so m2 slides under m1. So m1 experiences F

_{k}= 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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(Original post by

Find out if sliding occurs between m1, m2; determines moving as 1 lump or seperately. F

**Physics Enemy**)Find out if sliding occurs between m1, m2; determines moving as 1 lump or seperately. F

_{s}= 0.5*2g < 24.5, so m2 slides under m1. So m1 experiences F_{k}= 0.3*2g = 5.9 N (1 d.p), to the right.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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#5

(Original post by

Here’s the thing. My answer is B.

My friends answer is C.

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 ?

**Leah.J**)Here’s the thing. My answer is B.

My friends answer is C.

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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#6

(Original post by

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.

**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.

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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#7

(Original post by

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.

**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.

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#8

(Original post by

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.

**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.

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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#9

(Original post by

You said that m2 slides under m1 - how did you conclude that?

**RogerOxon**)You said that m2 slides under m1 - how did you conclude that?

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#10

(Original post by

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.

**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.

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#11

(Original post by

A force of 5.9N on m2 would see it accelerate faster than m1 ..

**RogerOxon**)A force of 5.9N on m2 would see it accelerate faster than m1 ..

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#12

(Original post by

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.

**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.

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#13

(Original post by

You have omitted the force that M1 exerts on m2.

**RogerOxon**)You have omitted the force that M1 exerts on m2.

Last edited by Physics Enemy; 1 year ago

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#14

**Leah.J**)

Here’s the thing. My answer is B.

My friends answer is C.

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 ?

_{1}and m

_{2}should be stated as an inequality:

*μ*

_{s}≤ 0.5

stating that the

**maximum**coefficient of static friction between m

_{1}and m

_{2}is 0.5, instead of stating that the coefficient of static friction between m

_{1}and m

_{2}is 0.5.

The inequality of the coefficient of static friction between m

_{1}and m

_{2}implies that the static friction acting on m

_{1}is

Static friction ≤ 2(0.5)

*mg*Without additional info such as the m

_{1}is on the verge of sliding, we cannot conclude the static friction is at the maximum.

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#15

(Original post by

The answer is B.

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.

**RogerOxon**)The answer is B.

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 = m

_{2}aand 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:*μ*

_{s}m

_{1}g= m

_{1}a

Anyway, this is just my view.

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#16

(Original post by

Anyway, this is just my view.

**Eimmanuel**)Anyway, this is just my view.

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 - F

_{k}= 5a and then F

_{s}= 2a acting on m1.

But modelling masses individually, noting F

_{s}is unknown at the outset (whereas F

_{k}is at max):

For m2: 24.5 - F

_{k}- F

_{s}= 3a, For m1: F

_{s}= 2a

Eliminating F

_{s}gives 24.5 - F

_{k}= 5a as above

Where F

_{s}⩽ g so a ⩽ g/2, else sliding: a2 > a1

Last edited by Physics Enemy; 1 year ago

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