# friction type question

• Jul 6th 2010, 09:04 AM
jiboom
friction type question
A block of mass m sits atop a mass M which rests on a frictionless table.
The coeffecient of friction between the two masses is $\mu$

The mass M is connected to a spring of force constant k attached to the wall.
(i) How far can mass M be pulled so that upon release, the upper mass m does not slip off?
(ii) Repeat if $\mu'$s the coeffecient of friction between M and the table.

im getting bit confused about the signs.

for big+small block we have upon release

$F=(M+m)a=-kx$
giving $a= \frac{-kx}{(M+m)}$

for small block
F= $\mu R$= $mg\mu$

now as big block goes back upon release then does small block move opposite way so friction acts in same direction as acc?

so
ma= F= $mg\mu$?

then

$\frac{-kx}{(M+m)}=$ $mg\mu$?

so as we want max x we need

$\frac{-kx}{(M+m)}<$ $mg\mu$?

but answers negate the Lhs. Why is this?

for part (ii) for the big block with friction the answers have

$(M+m)a=-xk+\mu'(M+m)g$

but once again should the friction not be in opposite direction to acc of the block?
• Jul 7th 2010, 11:11 AM
Ackbeet
I'm not sure I agree with your equation of motion for the big block. Where's the frictional force? If the big block acts on the small block via friction, then surely the small block acts on the big block via friction by Newton's Third Law.

[EDIT]: I would not write an equation of motion for big + small, incidentally.

The key to avoiding sign errors in problems like this is to draw a picture, assign a coordinate system, and then assign all variables according to the coordinate system.
• Jul 8th 2010, 10:05 AM
jiboom
what i give in first post is as in the answer. to quote the whole answer:
For the case where there is no friction between the block and the table, the force on the two block combination
is F=(M+m)a=-kx
Assuming the block of mass m does not slip, both blocks accelerate at the same rate. The force that is
accelerating the smaller block is the force due to friction. Intuitively, if there were no friction between
the blocks, the smaller mass would stay at the stretched position and not be pulled back by the spring
when the mass M was released. (It would then, of course, fall to the ground.) Therefore, in this case,
friction must act in the direction of the motion in order to keep the smaller block on top of the larger
block. ma = Ff < μmg. Since we want to know the maximum distance the block can be pulled, we want
the maximum value for the friction.using the value for a found above:

ma=[m(kx)]/(M+m)< μmg

now im not even sure the sign of the kx is correct. I assume the a is acc when blocks are released, so does not the kx act in same direction? It resists the motion as i pull the blocks so when i release the acc is toward the spring and so is the Kx?? as for the sign of the friction i still cant see what way the block will move. in the solution it says friction acts in same direction of motion??

let me say i pull blocks to the right,then when i let go acc is to the left. is the small block moving right so friction acts to left? But then the spring force is also acting to the left so has same sign has acc??
• Jul 9th 2010, 03:10 AM
Ackbeet
The frictional force is going to act as follows:

The big block is going to pull the smaller block along with it. Thus, the big block exerts a force on the smaller block in the direction of motion. On the other hand, the small block (by Newton's third law) is going to exert a restraining force (via friction) on the big block. So that force is going to be opposite the direction of motion. however, since you're not concerning yourself with writing an equation of motion for just the big block, that force is not going to show up. And it looks as though it does not need to.

Let's say the wall is to the left of both blocks. We will define x = 0 as the equilibrium position of the spring, x is positive to the right. We are going to pull both blocks to the right a maximum distance such that the small block stays on top of the big block. Clear so far? Now then, the spring is going to act on both blocks (treated as a single block with mass M + m, since they are not going to move relative to one another, by assumption) in the negative x direction. That is,

$F_{\text{spring}}=-kx.$

This must equal, as you've already demonstrated, the total mass times the acceleration, by Newton's Second Law:

$-kx=(M+m)a.$

This acceleration will be in the negative x-direction, correct? That is, we could just as easily write the above equation thus:

$-kx=-(M+m)|a|,$ or

$kx=(M+m)|a|.$

Now the only force in the direction of motion for the smaller block is the frictional force, which we've already said is going to act in the negative x direction. Newton's Second Law for the smaller block reads as follows:

$-\mu m g=-m|a|,$ where we've used the same trick as before to write the magnitude of a.

This implies

$\mu g=|a|.$

Now the first part of the problem requires the spring force to be less than or equal to the frictional force (in magnitude!), otherwise the smaller block would slip off. Thus, from the first equation, we have

$x=\frac{(M+m)|a|}{k}\le\frac{(M+m)\mu g}{k}.$

• Jul 9th 2010, 09:30 AM
jiboom
Quote:

Originally Posted by Ackbeet
The frictional force is going to act as follows:

The big block is going to pull the smaller block along with it. Thus, the big block exerts a force on the smaller block in the direction of motion. On the other hand, the small block (by Newton's third law) is going to exert a restraining force (via friction) on the big block. So that force is going to be opposite the direction of motion. however, since you're not concerning yourself with writing an equation of motion for just the big block, that force is not going to show up. And it looks as though it does not need to.

what way is the motion of the small block? When we release the spring,if we say it starts in the middle of the big block, does is its motion towards the left or the right end of the big block?

Quote:

Originally Posted by Ackbeetl

$F_{\text{spring}}=-kx.$

This must equal, as you've already demonstrated, the total mass times the acceleration, by Newton's Second Law:

$-kx=(M+m)a.$

This acceleration will be in the negative x-direction, correct? That is, we could just as easily write the above equation thus:

$-kx=-(M+m)|a|,$ or

$kx=(M+m)|a|.$

Not sure what is going on here.

For F=ma you state acc is in neg x-direction, so why do we have opposite signs in your -kx=(M+m)a if the spring force and acc are going in the same direction?

Eg if i throw ball up with acc a and mass m then the F=ma is ma=-mg as acc is up and mg is down. Im clearly misunderstanding the F=ma bit as i would just write (M+m)a=kx for the block.

Sorry if im being bit dim but i have never studied mechanics and my friend has started doing some and wanted my help. seems im just as lost as him :)
• Jul 10th 2010, 03:52 PM
Ackbeet
Question: Which way is the motion of the small block?
Answer: Well, we are trying to set up the problem and our solution so that its position relative to the big block is unchanged throughout. I can tell you that its motion, if it were to move, would want to move to the end of the big block opposite to the spring.

Question: For F=ma, you state that the acceleration is in the negative x-direction; so why do we have opposite signs in your -kx=(M+m)a if the spring force and acc are going in the same direction?
Answer: Signs that appear explicitly and actual signs are not always the same thing. x is positive, by my definition of the coordinate system. Therefore, -x is negative. a is negative already. Therefore, both -kx and (M+m)a are negative.

Hopefully, this will explain your example as well. I would agree that if you throw a ball straight up, the force is -mg (where x is defined positive upward). That just means that when we apply Newton's Second Law: ma = -mg, we see that the acceleration a is negative. Remember that displacements, velocities, accelerations, and forces are all vector quantities: they have magnitude and direction. In 1-dimensional motion such as the problem at hand, all that means is that the vector quantities I've just mentioned have a magnitude and a sign. That sign could well be hidden "inside" the variable name. Does that help?
• Jul 11th 2010, 08:35 AM
jiboom
thank you for your time with this matter. I am understanding what you are saying but still confusion exists.

For the ball with acceleration up and mg down we have ma=-mg so you now say acceleration is negative so acts down ? how can this be if i give the ball acceleration in positive (up) direction? I can see deccelartion will occur but how do i see from a=-mg there is an upward acceleration? Or is your remark acceleration is negative wrong and i use the first paragraph to say as mg acts down it is negative to -(mg) is positive?

If i wanted to accelerate the spring in the positive direction (to the right) how would i write F+ma for this case?

As -kx is spring force would i not end up with ma=-kx >? (although i not sure this is good eg of what im trying to ask as i expect i need extra to -kx as the extension would change with time )

Further to my post i have the followin questions that confuse me fiurther as one uses signs for acc while the other does not.

1)a string passes through 2 smooth pulleys A and B and carrying a mass of 5m at C (end of string that goes through B).The pulley A is not fixed and has mass 3m while B is fixed.

THe solution says pulley A has acceleration a down and the mass at C has acceleration 2a up then gives

NL2 for A : 3mg-2T=3ma
NL2 for C: T-5mg=10ma

why is one of the RHS -a as the act in opposite direction ? It seems a is just a nop matter what direction it acts in. BUT

in question 2)

from a point A in a plane sloping 20 degrees to the horizontal,a ball is thrown up the plane at 40 degrees with a speed of 15m/s...

here they resolve parallel to plane and use s=ut+at^2/2 to get

s=15tcos(40) -(t^2/2) gsin 20!!!!!!

so here they have put a - sign in with the acceleration as up the plane is +ve down the plane is -ve.

What is going on? im at a loss to see in these 3 cases (fspring force and acceleration acting in same direction, accelerations acting in opposite directions on pulleys and throwing ball up plane) why the sign of the acceleration is not indicative of the direction except for the last question??? .
• Jul 12th 2010, 02:38 AM
Ackbeet
Question: For the ball with acceleration up and mg down we have ma=-mg, so you now say acceleration is negative, and so it acts down?
Answer: This is not really correct language. When you look at Newton's Second Law, F = ma, what that is really saying is that the sum of the forces on an object produces or implies the acceleration. Forces act, and are causes. Accelerations react to forces, and are effects.

Question: How can this be if i give the ball acceleration in positive (up) direction?
Answer: All of this depends on when you "start the clock", so to speak. If you were to give the ball an initial known upward velocity and position, and then tell me to predict what happens next, I would start the clock at the instant the ball left your hand. That's t = 0, and at that point exactly, your hand is no longer a force on the ball. The only force on the ball is gravity, with the usual -mg formula. So you've accelerated the ball initially with the force of throwing, but once the ball leaves your hand, gravity takes over and is the only force on the ball (if you neglect air resistance). Does that make sense?

Question: I can see deccelartion will occur but how do i see from a=-mg there is an upward acceleration?

Question: Or is your remark acceleration is negative wrong and i use the first paragraph to say as mg acts down it is negative to -(mg) is positive?
Answer: No, m and g are always positive, by convention. You assign a sign to the quantity based on your choice of coordinate system. Gravity always acts down (to a good approximation) while on Earth, so if y is positive upward, the gravitational force is approximated by -mg. If, on the other hand, you define y positive downward, which is a good choice in some problems, then the gravitational force is +mg. Make sense?

Question: If I wanted to accelerate the spring in the positive direction (to the right) how would I write F+ma for this case?
Answer: I'm assuming you meant to write F = ma in that question. In this situation, if I wanted to accelerate the spring in the positive direction, to the right, and if the spring was already to the right of its equilibrium position, then you would have to exert an outside force on the system to make that happen. If you were then to let the system go (in other words, remove your outside force), and you were interested in the motion, then I would start the clock at the instant you removed the outside force, use the new initial position and velocity, and go from there. It would be simpler, I think.

Question: As -kx is spring force would i not end up with ma=-kx >?
Answer: Could you please re-phrase this question? If you mean "will the spring force not end up being in the positive x-direction at some point", I would say, "Yes, any time the masses on the spring go farther left than the equilibrium position of the spring, the spring force is going to change sign (essentially, x will then be negative, thus canceling out the negative sign in front of -kx)." Is that what you meant? The spring force is always opposed, in direction, to the position of the spring relative to its equilibrium point, at least if you're in the region governed by Hooke's Law. You can exceed that if you wish, probably damaging the spring in the process.
• Jul 12th 2010, 02:39 AM
Ackbeet
• Jul 12th 2010, 09:43 AM
jiboom
Question: If I wanted to accelerate the spring in the positive direction (to the right) how would I write F=ma for this case?

here i mean the spring starts at x=0 and is given an accleration. i was thinking that f=ma would still be ma=-kx but your 2 replies answer that. another force is needed and the spring force would change.

I posted the other questions here as they contain the crux of my confusion namely the signs related to the "a" in the f=ma.
For the pulley one accelerates up the down but no sign difference is seen in the resulting F=ma equations but in the wedge question a minus sign is given to the acceleration in s=ut+at^2/2 equation. I am ok getting the forces and solution,just need to get to grips to the signs for the acceleration.

from your reply to my question about the ball i now think i dont need to say +a or -a in f=ma as this sign will come from the sum of the forces and tell me if a is positive or negative and so up or down, but then why does the wedge question put a minus sign in for the acceleration down the wedge? just as i thought i was getting it.....
• Jul 12th 2010, 10:20 AM
Ackbeet
Hmm. I'm trying to generalize here and hopefully clear up some confusion.

The goal of classical mechanics is the following: given the forces on an object, predict its motion for all time. The forces on an object are functions only of time, and the position and velocity of the object: $\vec{F}=\vec{F}(\vec{r},\dot{\vec{r}}\,;t)$. Newton's Second Law then gives us the differential equations of motion (other methods may be easier to apply in certain circumstances): $\sum\vec{F}(\vec{r},\dot{\vec{r}}\,;t)=m\ddot{\vec {r}}.$
This is, in general, a second-order differential equation, the solution of which is $\vec{r}(t).$
In most cases, the acceleration is initially unknown. You're trying to solve for the vector $\vec{r}$. If you can find that, then, theoretically, you know everything classical mechanics can tell you about the system.

Now, then. How do you assign signs to the acceleration? Actually, most of the time, you don't. You assign signs to the forces relative to your choice of a coordinate system. That's it, then.

When I do a physics problem, the first thing I do is copy out the problem word-for-word (I always tend to miss things if I don't do that; it's well worth the time invested, for me). The next thing I do is draw a picture of the problem. I assign a coordinate system first, then I label things. This will usually cut out most of the confusion of signs. If it doesn't, and you get to the end of solving for whatever it was you were trying to solve for, then double-check your answer and see if it makes sense.

If you're still confused, and would like help on your other two problems, please post a picture of the situation, and the statement of the problem, word-for-word.
• Jul 13th 2010, 09:30 AM
jiboom
i think your line about "we assign forces the signs" is what i am after.

So for my block question i give a negative to kx which in turn tells me,since

(M+m)a=-kx that acceleration is towards the left ??

So for my pulley question i dont give signs to the acceleration but from

3mg-2T=3ma say i will find 3mg-2T is positive and so acceleration is up

and finally for the wedge, using F=ma i have

-mgsin 20=ma and so a=-gsin20 and goes down the wedge.

If you agree then i thank you very much with your help in clearing up this matter.

As im helping a friend i should not worry and let him find this out but this is the beauty of helping people. They ask questions about things i didnt and just accepted so although i could get the answer i had no suggestion as to the sign issue.
• Jul 13th 2010, 10:27 AM
Ackbeet
"So for my block question i give a negative to kx which in turn tells me,since

(M+m)a=-kx that acceleration is towards the left ??"

No: it means that the acceleration and the position have opposite signs. Eventually, x will go negative, which will make a positive.
• Jul 13th 2010, 01:10 PM
jiboom
Quote:

Originally Posted by Ackbeet
"So for my block question i give a negative to kx which in turn tells me,since

(M+m)a=-kx that acceleration is towards the left ??"

No: it means that the acceleration and the position have opposite signs. Eventually, x will go negative, which will make a positive.

my fault, but i think you are referring to wrong question. I meant the original block question when we have acceleration upon release. Where i put my block question im hoping you mean when i suggested the acceleration was away from the equilibrium point (x=0)
• Jul 13th 2010, 01:28 PM
Ackbeet
The force on a spring is a restoring force: it always wants to go back to equilibrium, which we've defined as x=0. Initially, when you have the block to the right of equilibrium (x>0), the spring force, being -kx, is directed to the left, towards the equilibrium point. Therefore, the block is going to move left. At some point, the block will pass the equilibrium point (x=0), and be to the left of the equilibrium point, and x<0. At that point, the force, being -kx, is now going to be directed to the right, towards the equilibrium point. Remember: a spring always wants its equilibrium position! Does this make sense?