Static friction between two bodies and a surface

[Isiudor]

Homework Statement

okay, I know it says be descriptive and all, but I'll present a watered down problem with some insignificant data already calculated by me.

my problem is to understand the logic behind the solution, I know the solution, I simply think my textbook is... false.

here goes.

box A with a mass of 5 kg is resting ontop of box B with a mass of 20 kg, and box B is resting on a surface

the question is how much maximum force can I apply to object B so A remains stacked exactly at the same spot on top of B without moving

the static friction between A and B is Fsmax a,b = 40
the static friction between B and the surface is Fsmax b,s = 20

now I simply fail to fathom why the answer isn't simply 60N, because if I apply 60N in a given direction, and 20 of which will be nullified by the static friction between B and surface, and 40 remains which will bring the static friction between B and A to it's max, any further application of force should move A off of B, or so it is in my mind.

Homework Equations

$$\sum$$f = ma(duh)

The Attempt at a Solution

I will list the answer to this problem, according to my textbook here.
the claim is I first calculate f = ma of body A where f is the Fsmax a,b.
this will result in a = 40 / 5 = 8, this determines the supposed "max acceleration"

so I now calculate f = ma for body B and this new a.

F - fsmax a,s = 20 * 8

F = 180N

I just cannot fathom that you could apply 180N to object B with the friction between B and surface being only 20, and still object A won't move when it's max static friction with B is only 40... where are the remaining 120N going??
I think this whole calculation is bogus, but I'm probably wrong heh... textbooks aren't wrong.

please, someone. help me understand this.

 

[Doc Al]

I will list the answer to this problem, according to my textbook here.
the claim is I first calculate f = ma of body A where f is the Fsmax a,b.
this will result in a = 40 / 5 = 8, this determines the supposed "max acceleration"

Realize that the only force available to accelerate the top block is the friction force. Using the max value of static friction thus tells you the max acceleration that the top block can have without starting to slip. And since the blocks don't slip against each other, that tells you the acceleration of both blocks.

so I now calculate f = ma for body B and this new a.

F - fsmax a,s = 20 * 8

F = 180N

Careful. Since you are looking at block B alone, you must consider all the forces on block B. There are two friction forces:
F - f(kinetic) b,s - fsmax a,b = 20*8
F - 20 - 40 = 160
So: F = 220 N

This is the force required to produce that acceleration.

I just cannot fathom that you could apply 180N to object B with the friction between B and surface being only 20, and still object A won't move when it's max static friction with B is only 40... where are the remaining 120N going??

You need to push with a force of 220 N. Friction accounts for 60 N, so the remaining 160 N is left to accelerate block B.

 

[Isiudor]

yes of course you are right, I left out the friction between a,b in the final equation by mistake.

I generally understand the point of view which the "correct" calculation is based on, but when I think of it the other way around... since I'm applying 220N to block b, and it loses 20N due to friction with the surface... shouldn't the remaining 200N who are still being applied to block B should also apply to block A which will in fact exceed the max static friction between a,b (40) which will mean that the objects will move apart. I just fail to understand what's wrong with the latter logic.

 

[Doc Al]

You seem to be thinking (at least partially) that the force applied to block B should be fully transmitted to block A. But that's not true. Since you need 160 N just to accelerate block B, only 220 - 160 = 60 N is "transmitted" to other objects (block A and the floor, in this case).

 

[Isiudor]

I'm trying to understand why it isn't true, I really am...

well I'm not thinking of it as what amount is "transmitted" but if a force of 160N is pulling on block B then wouldn't that mean that block B pulls block A with the same amount of force, due to friction? or in other words, if the theoretical max static friction between them was for example, 200... then in effect the static friction between them resulting from pulling on object b with 160N would also be equal to 160N?

had we been discussing a single object resting on the surface, with a max static friction between it and the surface of 200, and the same 160N would be applied to this one object... the the static friction between the object and the surface would be equal to the force applied, in other words 160N... I fail to see why I cannot apply the same logic when another object rests on top of the first object.

 

[songoku]

If box A has a = 8 m/s^2, then the force appplied to it will be F = ma = 40 N, not 160 N

 

[Doc Al]

well I'm not thinking of it as what amount is "transmitted" but if a force of 160N is pulling on block B then wouldn't that mean that block B pulls block A with the same amount of force, due to friction?

No. If block B pulled on block A with the same force that was pulling block B, then block B wouldn't accelerate--the net force on it would be zero.

or in other words, if the theoretical max static friction between them was for example, 200... then in effect the static friction between them resulting from pulling on object b with 160N would also be equal to 160N?

No, static friction is only what it needs to be to prevent slipping between the surfaces. Once you know the acceleration, then you know how much static friction will be produced. If the maximum static friction between A and B were 200 N, and you pulled on B with a force of 220 N (as before), then the friction force between A and B would still be only 40 N.

had we been discussing a single object resting on the surface, with a max static friction between it and the surface of 200, and the same 160N would be applied to this one object... the the static friction between the object and the surface would be equal to the force applied, in other words 160N... I fail to see why I cannot apply the same logic when another object rests on top of the first object.

When dealing with the static friction of a single object resting on the surface there is no acceleration to worry about--big difference. Again you ask yourself how much friction is required to prevent slipping? In this case it's easy. You pull B with a force of 160 N so the static friction must equal 160 N to prevent it from sliding.

 

[Isiudor]

I forgot to reply, I finally made the connection in my head and realized the only thing causing the friction between bodies A and B is the displacement of B or in other words it's acceleration.

I just asked myself what will be the friction between B and A be if a force applied on B would be less than the static friction between B and the surface, and the answer is zero, then everything fell into place.

thanks for your patience with the help

cheers,

p.s

You are applying the force to B? Then, since the force is greater than the static friction between B and the floor, B will start to move. Once it has started to move, the problem reverts to "kinetic friction". The net force on B will be the total force minus the kinetic friction force. If that is larger than the static friction between A and B, then A will slip backwards.

https://www.physicsforums.com/showthread.php?t=255527"

I just noticed at the bottom a similar question to mine and went into read the responses, and read the above reply... but in fact this fella is wrong in his answers isn't he? even if the net force on B is greater than the static friction A won't necessarily slip... for reasons written above.

unless I'm missing something, maybe the mentor title is a bit too liberally given around here.

 

[Doc Al]

I just noticed at the bottom a similar question to mine and went into read the responses, and read the above reply... but in fact this fella is wrong in his answers isn't he?

Yes, that answer is only partly correct.

 

[Isiudor]

Hello again, I'm afraid I've ran into the same "understanding block" regarding the same question, only now the forces is applied to the top object which rests ontop of the second object.

the textbook to this problem is nearly the same as when the force is being applied to the "middle" object. I will assume for example that B rests ontop of A which rests on the surface

you calculate the acceleration of A , then calculate the force applied on B with said acceleration and minus static friction max a,b.

Now I understand that when the force is applied to A object, the only thing that causes any friction between a,b is the displacement of A.

however when the force is applied to B(the top object) it should immiedietly effect their static friction, or otherwise B will slip

so when the force is applied to B, If asked what's the maximum force you can apply to B so the two bodies still won't slip doesn't it HAVE to be equel to the maximum static friction between a,b? according to my textbook it isn't, and when calculted via the f=ma routine, it turns out GREATER than the static friction between them.

I could give exact numbers if the question is coeherent enough, thanks.

 

[Doc Al]

so when the force is applied to B, If asked what's the maximum force you can apply to B so the two bodies still won't slip doesn't it HAVE to be equel to the maximum static friction between a,b? according to my textbook it isn't, and when calculted via the f=ma routine, it turns out GREATER than the static friction between them.

As before, if the force applied to B equaled the friction between the blocks, then the net force on B would be zero and it would not accelerate. The force on B must be greater than the friction in order to accelerate block B.

 

[Isiudor]

yes you are of course right about that, the thing that bogs me is that if I apply force slightly equal to the static friction max, the static friction at work between a,b will be equal to the force applied(the static friction max), correct? otherwise B will slip.

so how can I continue to increase the force applied to B, with the static friction force staying the same?

 

[Doc Al]

yes you are of course right about that, the thing that bogs me is that if I apply force slightly equal to the static friction max, the static friction at work between a,b will be equal to the force applied(the static friction max), correct? otherwise B will slip.

No, not correct. If you only apply a force to B equal to the static friction max, then the blocks will not have the maximum acceleration and the actual static friction will be less than the maximum.

so how can I continue to increase the force applied to B, with the static friction force staying the same?

The static friction does not stay the same. It starts out small (just enough to prevent slipping) and increases to its max value as needed.

 

[Isiudor]

No, not correct. If you only apply a force to B equal to the static friction max, then the blocks will not have the maximum acceleration and the actual static friction will be less than the maximum.


The static friction does not stay the same. It starts out small (just enough to prevent slipping) and increases to its max value as needed.

well the latter is a given once you consider your first statement, but how can it be that the force applied to B isn't equal to the static friction between a,b, what difference does it make if B rests on A and I apply a force to B, or B rests on a surface and I apply the force to B? In both cases the static friction between B and w/e it rests on should be equal to force applied to B, assuming it's smaller than static friction max, if not... why not?

 

[Doc Al]

well the latter is a given once you consider your first statement, but how can it be that the force applied to B isn't equal to the static friction between a,b, what difference does it make if B rests on A and I apply a force to B, or B rests on a surface and I apply the force to B? In both cases the static friction between B and w/e it rests on should be equal to force applied to B, assuming it's smaller than static friction max, if not... why not?

As before, things change when the blocks accelerate. If block B didn't move when you pushed it, then your reasoning would be correct. But it does move.

 

[Isiudor]

okay, you just confused me a whole lot more hehe. you're saying I'm part wrong?
I'll just state a reminder that the objects have switched locations and in all current references. object B is on TOP of block A

first of all, who says block B moves? and then even if it does, relative to what? to A or the surface?

my point is this. whether or not B accelerates relative to the surface, depends on the mass of A and it's static friction with the surface. keeping in mind our goal is to not allow B to start slipping off of A.

It would be impossible to accelerate either block, without B slipping, if the static friction max between A and the surface was greater than the static friction between A,B... when the force is applied to B. correct?

so In effect, when suppose A has a mass of 1,000,000,000 kg, and hence the static friction between it and the surface is, say, 50,000,000 if a force equal to static friction max was applied to B, then that would be the same force acted on A(and B) due to static friction, in other words the force applied WILL be equal to the active static friction between them.

are you telling me that changes, when the mass of A changes? it sounds ridicules to me.

are you saying, that if now A was small enough in mass, to have a static friction with the surface of say half the static friction max of A, B, and the kinetic friction of A and the surface would for this purpose be the same(as the static friction of A, surface)... then if the same amount of force is applied to B (a force equal to the static friction max between A, B) the active static friction between A and B will be less than the force applied? just because the objects are accelerating relative to the surface?

 

[Doc Al]

okay, you just confused me a whole lot more hehe. you're saying I'm part wrong?

Yes, you're partly wrong.

I'll just state a reminder that the objects have switched locations and in all current references. object B is on TOP of block A

OK.

first of all, who says block B moves? and then even if it does, relative to what? to A or the surface?

Block B will accelerate--with respect to the surface--if there's a net force on it.

my point is this. whether or not B accelerates relative to the surface, depends on the mass of A and it's static friction with the surface. keeping in mind our goal is to not allow B to start slipping off of A.

Whether or not B accelerates depends on the net force on it.

It would be impossible to accelerate either block, without B slipping, if the static friction max between A and the surface was greater than the static friction between A,B... when the force is applied to B. correct?

Correct.

so In effect, when suppose A has a mass of 1,000,000,000 kg, and hence the static friction between it and the surface is, say, 50,000,000 if a force equal to static friction max was applied to B, then that would be the same force acted on A(and B) due to static friction, in other words the force applied WILL be equal to the active static friction between them.

This is unclear. What's the max static friction between A and the surface and between A and B? What force are you exerting on B?

are you telling me that changes, when the mass of A changes? it sounds ridicules to me.

I'm not understanding your point.

are you saying, that if now A was small enough in mass, to have a static friction with the surface of say half the static friction max of A, B, and the kinetic friction of A and the surface would for this purpose be the same(as the static friction of A, surface)... then if the same amount of force is applied to B (a force equal to the static friction max between A, B) the active static friction between A and B will be less than the force applied? just because the objects are accelerating relative to the surface?

Absolutely.

Let's look at a very simple example. Let's say that each block has a mass of 10 kg. Let's say the static friction max between A and the surface is 10 N and between A and B is 50 N. What happens if you:
- Push B with a force of 5 N?
- Push B with a force of 10 N?
- Push B with a force of 20 N?
- Push B with a force of 50 N?
For each case, figure out the friction forces acting (both of them) and decide if anything accelerates.

 

[Isiudor]

Yes, you're partly wrong.
Let's look at a very simple example. Let's say that each block has a mass of 10 kg. Let's say the static friction max between A and the surface is 10 N and between A and B is 50 N. What happens if you:
- Push B with a force of 5 N?
- Push B with a force of 10 N?
- Push B with a force of 20 N?
- Push B with a force of 50 N?
For each case, figure out the friction forces acting (both of them) and decide if anything accelerates.

you've neglected to mention the kinetic friction between A and the surface, so I'll just calculate it as if kinetic friction is equal to static friction for A, surface.

with 5N applied on B, the static friction between A and B is equal to 5N, which is less than max static friction of A, B which is 50, so B doesn't slip off of A.

since a force of 5N is now applied to A due to said friction, the static friction of A with the surface is also equal to 5N, which is less than the max static between A and surface, so nothing moves(fnet on A, and B both equal zero... if you prefer).

same answer for 10N, obviously.

I'll just skip the 20N and go straight for the 50N since 20N is basically the same.


with 50N applied on B, the static friction between A and B is equal to 50N, which is equal than max static friction of A, B which is 50, so B still doesn't slip off of A.

since a force of 50N is now applied to A due to said friction, the static friction of A with the surface will exceed it's static friction max of 10N, and object A will begin to accelerate, with kinetic friction being also 10N the equation for object A should be

50 - 10 = 10 * a

a = 4;

which now that i think of it is somewhat odd, because now the fnet for B must also be, 40

since fnet = 10 * a... but it isn't, technically the fnet on B is zero.

what am I missing?

 

[Doc Al]

with 50N applied on B, the static friction between A and B is equal to 50N, which is equal than max static friction of A, B which is 50, so B still doesn't slip off of A.

This is not correct. Again, you assume that if you push B with some force X, that B in turn will push against A with that same force X. This is not true if B accelerates.

since a force of 50N is now applied to A due to said friction, the static friction of A with the surface will exceed it's static friction max of 10N, and object A will begin to accelerate, with kinetic friction being also 10N the equation for object A should be

50 - 10 = 10 * a

a = 4;

which now that i think of it is somewhat odd, because now the fnet for B must also be, 40

since fnet = 10 * a... but it isn't, technically the fnet on B is zero.

what am I missing?

(1) B does not exert a force of 50 N on A.
(2) The net force on B is not zero. If it were zero, then it wouldn't accelerate!

Here's how I analyze this problem. First I ask, If there were no slipping between A and B what would the acceleration be? Since they stick together, I can view A and B as a single object of mass 20 kg. The forces on it are 50 - 10 = 40 N. Thus the acceleration is 40/20 = 2 m/s^2. Then I ask, Is there enough static friction between A and B to prevent slipping? What must be the static friction force between A and B to produce this acceleration of B? 50 - X = 10*2, thus X = 30 N. The friction force between A and B only needs to be 30 N to prevent slipping (not 50 N), and since that's well within the max value of 50 N, I conclude that indeed there is no slipping between A & B. (In order for B not to slip with respect to A, the required static friction must be less than the max.)

 

[Isiudor]

well I can definitely understand your solution, I DO understand it... and I could probably now solve any similar question... but what I'm still a bit confused about is the reason why rules seems to change, when A & B are accelerating, and when they are not.

in the example of a force of 5N applied to B, then the static friction between A and B is equal to B(in your specific example with the given max static frictions). It's equal to force applied.

but when B & A start accelerating, this logic no longer applies and I must calculate the friction back from the acceleration.

I don't understand why when the two objects are accelerating, you can apply a greater force to B than when the objects weren't accelerating(because the static friction of A with the surface is larger than the force applied) without B slipping.

is it because some of the "force" applied to B is "exhausted" to accelerate B? is it because the dissplacment of A and B "negates" some of the friction since they are moving as one?

 

[Doc Al]

Reread what I wrote in post #4. You are thinking that if you push B with some force, that force should be transmitted to whatever B is pushing on. But if B accelerates, some of that force is needed to produce the acceleration, thus reducing the amount of force transmitted by ma. (This is just Newton's 2nd law.) Since B pushes against A with less force, the static friction required to prevent slipping is less.

In outline:
You push B with force F
B accelerates
B pushes A with force F - ma
A resists slipping by pushing back equally with force F - ma
Thus the static friction required is just F - ma

 

[rl.bhat]

To move the blocks A and B together,the maximum force acting on A should not exceed the force of friction between the surface a and b. Or f(a,b) = μ(a,b)*m*g where m is the mass of A block.
When the F is acted on B, let a be the acceleration of A and B. Then
F - μ(b,s)(M + m)g = (M+m)*a ...(1)
To move the blocks A and B together, a should be such that
μ(a,b)*m*g = ma Or μ(a,b)*g = a ...(2)
Substitute this value of a in eq.(1), you get
F - μ(b,s)(M + m)g = (M+m)*μ(a,b)*g
So the maximum force F = μ(b,s)(M + m)g + (M+m)*μ(a,b)*g
= [μ(b,s)+μ(a,b)](M+m)*g

 

[Doc Al]

To move the blocks A and B together,the maximum force acting on A should not exceed the force of friction between the surface a and b. Or f(a,b) = μ(a,b)*m*g where m is the mass of A block.
When the F is acted on B, let a be the acceleration of A and B. Then
F - μ(b,s)(M + m)g = (M+m)*a ...(1)

OK.

To move the blocks A and B together, a should be such that
μ(a,b)*m*g = ma Or μ(a,b)*g = a ...(2)

The net force on block A = μ(a,b)*M*g - μ(b,s)(M + m)g = ma.

 

[rl.bhat]

OK.

The net force on block A = μ(a,b)*M*g - μ(b,s)(M + m)g = ma.

As you know, the frictional force arises only due to the relative motion of the point of contact. Friction will always opposes the relative motion of the point of contact.
In our problem, the point of contact of body A with B will tend to be at rest. Due to the force F body B is moving forward. With respect to B, the point of contact of body A will move backward. Therefore friction f on A will act forward.
If f < ma, A will fall back than B.
If f = ma A will move with B. In that case a is the acceleration of (M + N)
The maximum f = μ_a,b*mg, where m is the mass of A.

 

[Doc Al]

As you know, the frictional force arises only due to the relative motion of the point of contact. Friction will always opposes the relative motion of the point of contact.
In our problem, the point of contact of body A with B will tend to be at rest. Due to the force F body B is moving forward. With respect to B, the point of contact of body A will move backward. Therefore friction f on A will act forward.
If f < ma, A will fall back than B.
If f = ma A will move with B. In that case a is the acceleration of (M + N)
The maximum f = μ_a,b*mg, where m is the mass of A.

Not sure I'm getting your point. My point was that your equation (2) in post #22 implied that there was only one friction force acting on block A. But there are two.

 

[rl.bhat]

Not sure I'm getting your point. My point was that your equation (2) in post #22 implied that there was only one friction force acting on block A. But there are two.

Frictional force on A is purely due to the relative motion of A and B. It is nothing to do with the frictional force between B and surface. Even if B rests on a frictionless surface, frictional force on A will be the same. Only the value of acceleration will change

 

[Doc Al]

Frictional force on A is purely due to the relative motion of A and B. It is nothing to do with the frictional force between B and surface. Even if B rests on a frictionless surface, frictional force on A will be the same. Only the value of acceleration will change

Note that B is the top block, not the bottom one. A is on the bottom and experiences two friction forces.

 

[rl.bhat]

Note that B is the top block, not the bottom one. A is on the bottom and experiences two friction forces.

In the problem static friction between the block B and surface, fsmax(a,b)is given as 20.

 

[Doc Al]

In the problem static friction between the block B and surface, fsmax(a,b)is given as 20.

The problem changed a while ago:

I'll just state a reminder that the objects have switched locations and in all current references. object B is on TOP of block A

Also, fsmax(a,b) would seem to stand for the max static friction between blocks A and B, not between the bottom block and the surface.

 

[rl.bhat]

The problem changed a while ago:

Also, fsmax(a,b) would seem to stand for the max static friction between blocks A and B, not between the bottom block and the surface.

Sorry. I entered the discussion from post#22. I didn't notice the change.
I was discussing according to the original problem.
But what wrong with the original problem?

 

[Isiudor]

nothing is wrong, at least not from where I'm sitting.

in fact I received back a quiz I had 2 days ago, and got 100(A+ in conventional terms, I guess).

Doc Al has been great help in helping me understand it all.

thanks again.

 

[tigeranishg]

the answer given by ur textbook is correct
u can solve the problem if u know seudo force topic

 

[tigeranishg]

if u want the block A to be stationary with respect to bolck B.
the net force acting on it must be equal to zero.
fmax a,b = 40
a = 40/5=8
now a seudo opp in direction to this frictional force should act to keep it in rest.
so the block B must move with accelaration of 8
on bolck B frictional force 20N is acting back.
F - fmaxa,b = ma
F- 20 = 160
f = 180N
any doubt contact <personal email removed>
i will explain through a diagram