Direction of the work done by force of friction

[dreamz25]

Topic : Direction of the work done by force of friction

Consider a case when a block is placed on a horizontal platform at rest. It then starts accelerating with a constant acceleration such that,
Friction present between the block and the horizontal platform > mass times its acceleration
so clearly sliding won't take place..
but can someone tell me how with reference to the fixed frame is the work done by force of friction +ve?

actually one more doubt of mine is ...

we know that friction opposes motion of a body.
It always acts in a direction opposite to the motion of the body.
and
as stated "the motion of a body with constant acceleration, the firction being greather than the mass times its acceleration"... how can this be possible?
is it all about limiting friction..?

i m muddled ... kindly help...
and i m a beginner...!

 

[tiny-tim]

hi dreamz25!

… how with reference to the fixed frame is the work done by force of friction +ve?

to find the direction of friction, always ask "which way would the body go if the surface suddenly turned to ice?"

as stated "the motion of a body with constant acceleration, the firction being greather than the mass times its acceleration"... how can this be possible?
is it all about limiting friction..?

kinetic friction will be greater than ma if µ_k > a/g

 

[sophiecentaur]

Hang on. 'Friction' doesn't do any work. It's the force overcoming the friction that does the work. And, because Work is Energy and a scalar quantity, it doesn't actually have a direction - although the Force does (because it's a vector).

"the motion of a body with constant acceleration, the firction being greather than the mass times its acceleration"... how can this be possible?

I'm not sure what this means. If the MassXacceleration (the force?) is less than the friction, then the object will slow down from its initial speed or never get moving in the first place. I think the original statement should have had a 'less than' rather than a 'greater than' in it.

 

[rcgldr]

Consider a case when a block is placed on a horizontal platform at rest. It then starts accelerating with a constant acceleration

Note that it's the horizontal platform that starts accelerating with a constant acceleration (assumed to be horizontal acceleration). The block accelerates at the same speed because the static friction is greater than (or it could be equal to) mass of block times acceleration, let u = static coefficient of friction, then as already posted, (except it's static friction):

u m g > m a
u > a / g

work done by force of friction is +ve?

Normally static friction isn't considered to be able to do work. It's the platform that's doing the work, and static friction is just keeping the block from sliding on the platform.

Kinetic friction (sliding) does negative work by converting kinetic energy into heat.

 

[sophiecentaur]

? How can friction 'do work'? It's a LOSS MECHANISM. That is the equivalent of saying that a Resistor can generate Electrical Power. If anyone is 'doing work' it's Gravity - because Gravitational Potential Energy is decreasing if something moves downhill.
Also, a force cannot be less than the friction force if you want to get something to start moving. The block will only move when the weight component acting down the slope is greater than the static friction force.
imo, this thread should start again at the beginning with a question that doesn't involve Friction doing Work.

 

[tiny-tim]

'Friction' doesn't do any work.

yes it does …

if i move a block horizontally by pulling it with a rope, the tension does work on it

if i move a block horizontally by pressing my hand on the top, and using static friction to move it, the static friction does the same work on it

(and the work done will always be positive, as opposed to negative)

Normally static friction isn't considered to be able to do work.

but that's because work = force "dot" displacement, and with static friction there's normally no displacement

here, there is displacement ​

 

[sophiecentaur]

If your hand weren't there, nothing would happen. ergo, it's your hand that's doing the work. What is Friction? It's a reaction force that is only there 'in response' to another force.
If you can't tell me that a resistor can supply electrical power then you can't tellme that 'friction' (a rough surface) can do work. If it could then you'd have a new inexhaustible source of free energy.

There never is a displacement in the same direction as the friction force.
I think there must be a confusion of terms here because I know you don't believe in perpetual motion.

 

[rcgldr]

The block will only move when the weight component acting down the slope is greater than the static friction force.

Note that the orginal post describes a situation where the block rests on a "horitzontal" platform, and the "horizontal" platform has constant "horizontal" acceleration. There is no "slope". The block is not sliding on the horizontal platform because friction force is greater than mass of block times rate of acceleration of block and platform.

 

[tiny-tim]

If your hand weren't there, nothing would happen. ergo, it's your hand that's doing the work. What is Friction? It's a reaction force that is only there 'in response' to another force.

suppose there are two blocks on ice, side by side, i push one and it pushes the other …

that's a reaction force on the second block, and it does do work!​

now suppose there are two blocks on ice, one on top of the other, i push the top one and that makes the other move through friction …

why do you say the friction force doesn't do work? ​

the engine of a car turns the axle, and the car goes uphill …

that's an increase in mechanical energy of the car …

the only external force uphill is the friction (from the road on the tyres) …

are you saying that the only external force doesn't do work? ​

 

[A.T.]

we know that friction opposes motion of a body.

That is too vague. Friction opposes the relative motion of the two interacting bodies. Depending on the situation and chosen reference frame, one body can be doing positive or negative work on the other body, via the force of friction (static or sliding).

 

[A.T.]

There never is a displacement in the same direction as the friction force.

The displacement depends on the reference frame. You can always find a frame where there is displacement in the same (or opposite) direction as the friction force.

If I accelerate a rolling table with a book on it, and the book starts moving in my frame of reference, then in that frame the table is doing positive work on the book via the force of friction.

 

[sophiecentaur]

Ok. I think we can resolve this. Static friction (when there is no slippage) is just the same as the intermolecular bonds in the rest of the rigid structure. They all TRANSMIT a force but I don' think you could say that they do work.
If there is slippage and energy loss at the interface then work is done 'against' the friction force because the relative motion is against the friction. But I don't see how one could say that work is done 'by' that friction force. It is just a normal lossy reaction force.

 

[sophiecentaur]

Ok. I think we can resolve this. Static friction (when there is no slippage) is just the same as the intermolecular bonds in the rest of the rigid structure. They all TRANSMIT a force but I don' think you could say that they do work.
If there is slippage and energy loss at the interface then work is done 'against' the friction force because the relative motion is against the friction. But I don't see how one could say that work is done 'by' that friction force. It is just a simple lossy reaction force.

 

[A.T.]

Ok. I think we can resolve this. Static friction (when there is no slippage) is just the same as the intermolecular bonds in the rest of the rigid structure. They all TRANSMIT a force but I don' think you could say that they do work.

Of course you can say that. If the two bodies in static contact are moving along the direction of the forces between them, then they are doing work on each other. In this simple sticky case:

work_done_by_A_on_B = - work_done_by_B_on_A

(the displacements are equal, but the forces opposite)

because the relative motion i against the friction.

Talking about relative motion implicitly means choosing the rest frame of one of the objects. In that frame the resting object is always doing negative work on the other sliding object. But that is just one possible frame.

But I don't see how one could say that work is done 'by' that friction force. It is just a normal lossy reaction force.

A force is a force. The definition of work doesn’t distinguish between different types of forces.

 

[Dale]

? How can friction 'do work'?

If friction doesn't do work than what force increases the KE and momentum of an automobile?

Better yet, consider a box on the back of a truck. There are three forces acting on the box: the weight, the normal force, and friction. As the truck moves the box accelerates, gaining KE. Now, if we make the back of the truck frictionless what happens? As the truck moves the box does not accelerate, not gaining KE. So KE is gained in the presence of friction and KE is not gained in the absence of friction. Therefore friction can be used to increase KE, so friction can do work.

 

[dreamz25]

lemme post the original question out here...!

A 5kg block is kept on a horizontal platform at rest. At time t = 0, the platform starts moving with a constant acceleration of 1m/s². The coefficient of friction μ between the block and the platform is 0.2. The work done by the force of friction on the block in the fixed reference frame in 10 s is
(A) +250 J
(B) -250 J
(C) +500 J
(D) -500 J

soln.

Assuming that the block does not slide on the platform
F_f = ma = 5 N, N = mg = 50 N and μN = 10 Newton => F_f < μN
The block will remain at rest relative to the platform

Displacement D relative tot he ground = 1/2 (1)(10)^2 = 50 m
Thus, Work done by force of friction = F_rD cos 0 = + 250 J

again, a more question below it says...

In the previous problem, if μ = 0.02, the work done by the force of friction on the block in the fixed reference frame in 10 seconds is
(A) +10J (B) -10J (C) +250J (D) -250J

soln,
Limiting friction = 0.02(50) = 1N
thus, the block will slideo n the platfor.
F_f = ma = 1; a = 1/5 m/s²
Thus, Displacement D = 1/2(1/5)(10)^2 = 10m
Work done by F_f = 1N (10)m = +10J

-----------------------------------------------------

Just tell me,

in the first case i quite accept it that the F_f will be +ve but
how in the 2nd one...!

 

[dreamz25]

See if the box remains at rest relative to the platform then why is it in motion
with reference to the ground frame... I mean how is the body moving..!
is it something like.. a book is placed on a trolley and the trolley is moving.. ?

 

[dreamz25]

See if the box remains at rest relative to the platform then why is it in motion
with reference to the ground frame... I mean how is the body moving..!
is it something like.. a book is placed on a trolley and the trolley is moving.. ?

 

[dreamz25]

Does that mean in all the cases the work done by kinetic friction is +ve?
i mean can never it be -v... Right..!

 

[rcgldr]

In the previous problem, if μ = 0.02, the work done by the force of friction on the block in the fixed reference frame

In the first case i quite accept it that the F_f will be +ve but how in the 2nd one?

Because it's the work done by force of friction ... in the fixed reference frame.

 

[tiny-tim]

hi dreamz25!

(ooh, new page! )

Assuming that the block does not slide on the platform
F_f = ma = 5 N, N = mg = 50 N and μN = 10 Newton => F_f < μN
The block will remain at rest relative to the platform

Displacement D relative tot he ground = 1/2 (1)(10)^2 = 50 m
Thus, Work done by force of friction = F_rD cos 0 = + 250 J
…
thus, the block will slideo n the platfor.
F_f = ma = 1; a = 1/5 m/s²
Thus, Displacement D = 1/2(1/5)(10)^2 = 10m
Work done by F_f = 1N (10)m = +10J
…
in the first case i quite accept it that the F_f will be +ve but
how in the 2nd one...!

in the first one, the friction force is forward, and so is the displacement

in the second one, the friction force is still forward, even though it's smaller, and so the displacement is also forward, and smaller, and so …

… in all the cases the work done by kinetic friction is +ve?
i mean can never it be -v... Right..!

right!

 

[dreamz25]

@Tiny tim -
That means ur this statement
"which way would the body go if the surface suddenly turned to ice?"
is noteworthy.

just one thing..

what u mean by ur statement is

if the surface suddenly turned to ice then the body would go in the direction where
the net force acts and the direction of this net force is the direction of friction.

 

[tiny-tim]

if the surface suddenly turned to ice then the body would go in the direction where
the net force acts and the direction of this net force is the direction of friction.

no, i think i mean the exact opposite …

if you're standing on a hill, and it suddenly turns to ice,

then you'll slide down,

so the friction must have been keeping you up ​

(friction always opposes the motion, or the "intended" motion)

 

[sophiecentaur]

I am having difficulty with the concept that Work Done would be different in different reference frames. To achieve a particular motion requires a fixed amount of energy in whatever (non relativistic) frame you choose to calculate it, surely?

Also, if an object is accelerated and moved by a force then the work done on it is defined by its motion and the force relative to it (i.e. + or - acceleration / increasing KE or decreasing KE).

Diagrams showing the forces and movement during 'traction' problems are often confusing and keeping an eye on the Energy Situation is usually the best way to get it right. An inappropriate choice of which two of the four possible four arrows can give the 'wrong answer'.

As with Electronic control, if a switch has zero or infinite resistance then no energy is lost (equivalent to non-friction slipping and rigid connection) In no circumstances is energy delivered by a simple switch and neither can work be 'done' by a frictional slipping contact - Energy is just lost.

The "work done by the friction force" is just the work done on the target object. Would you ever dream of talking about the work done by a series or parallel resistor in a motor circuit? That's what I am having a problem with. You might, of course, talk in terms of the PD across a resistor and use that to calculate the work or power from a motor.

If a question is posed to a student in those terms then it is tailor made to confuse; it is a totally non-practical way of viewing friction problems, in which the relevant forces are the 'real ones' and not the 'reaction ones'.

Does that mean in all the cases the work done by kinetic friction is +ve?
i mean can never it be -v... Right..!

Can this be right? If a car skids to a halt then the work done on the car "by friction" is Negative (energy taken out but not transferred to a massive Earth) but if a slippery block is on a moving travelator, then the work done is Positive (KE increases but the Earth's KE doesn't change). So can there be a general rule?

 

[A.T.]

I am having difficulty with the concept that Work Done would be different in different reference frames.

Since displacement is frame dependent, it's obvious that work is frame dependent as well.

To achieve a particular motion requires a fixed amount of energy in whatever (non relativistic) frame you choose to calculate it, surely?

No. Kinetic energy is frame dependent as well.

Also, if an object is accelerated and moved by a force then the work done on it is defined by its motion and the force relative to it (i.e. + or - acceleration / increasing KE or decreasing KE).

And the motion is frame dependent, so is work and KE.

neither can work be 'done' by a frictional slipping contact

Sure it can. A force is a force. The definition of work doesn’t distinguish between different types of forces.

The "work done by the friction force" is just the work done on the target object. Would you ever dream of talking about the work done by a series or parallel resistor in a motor circuit? That's what I am having a problem with.

Your electrical analogies are flawed.

If a question is posed to a student in those terms then it is tailor made to confuse;

And that makes it a good question. It forces to student to think strictly and apply the definitions precisely, instead of relying on intuitive reasoning and some vague preconceived notions.

it is a totally non-practical way of viewing friction problems, in which the relevant forces are the 'real ones' and not the 'reaction ones'.

Here you don't even make sense. Reaction forces (according to Newton's 3rd) are always "real forces". Fictitious forces do not obey Newton's 3rd. And there is no difference between action- and reaction forces in Newton's 3rd: They both are forces and can do work.

Can this be right? If a car skids to a halt then the work done on the car "by friction" is Negative

Depends on the reference frame. But in the initial rest frame of the Earth, yes.

but if a slippery block is on a moving travelator, then the work done is Positive

If it accelerates in some frame, then positive work is done on it in that frame.

So can there be a general rule?

The general rule is the defintion of work:

 

[Dale]

Just tell me,

in the first case i quite accept it that the F_f will be +ve but
how in the 2nd one...!

The block starts at rest so the initial KE is 0. It can either stay at rest (no work, no change in KE), or if it moves at all then its KE must increase and so there must be positive work done on it. Positive is the only possible way to go when starting from 0 KE.

 

[Dale]

Can this be right? If a car skids to a halt then the work done on the car "by friction" is Negative (energy taken out but not transferred to a massive Earth) but if a slippery block is on a moving travelator, then the work done is Positive (KE increases but the Earth's KE doesn't change). So can there be a general rule?

For a car (flat ground, no air resistance) the only horizontal force is the friction force on the tires. So if the car is accelerating then friction is doing positive work on the car and if the car is decelerating then friction is doing negative work on the car.

 

[sophiecentaur]

For a car (flat ground, no air resistance) the only horizontal force is the friction force on the tires. So if the car is accelerating then friction is doing positive work on the car and if the car is decelerating then friction is doing negative work on the car.

This is what I meant by a 'general rule' - it makes perfect sense and is clearly correct. I was arguing against the suggestion that it is 'always positive'.

 

[A.T.]

I was arguing against the suggestion that it is 'always positive'.

And who suggested that?

 

[sophiecentaur]

Since displacement is frame dependent, it's obvious that work is frame dependent as well.

No. Kinetic energy is frame dependent as well.

And the motion is frame dependent, so is work and KE.


Sure it can. A force is a force. The definition of work doesn’t distinguish between different types of forces.


Your electrical analogies are flawed.


And that makes it a good question. It forces to student to think strictly and apply the definitions precisely, instead of relying on intuitive reasoning and some vague preconceived notions.


Here you don't even make sense. Reaction forces (according to Newton's 3rd) are always "real forces". Fictitious forces do not obey Newton's 3rd. And there is no difference between action- and reaction forces in Newton's 3rd: They both are forces and can do work.


Depends on the reference frame. But in the initial rest frame of the Earth, yes.


If it accelerates in some frame, then positive work is done on it in that frame.


The general rule is the defintion of work:

Too many points to argue individually and I agree that a lot of this is right. I was not really putting my thoughts very well. That stuff about real forces was rubbish! Clearly, the effect of the friction force of a car clutch makes the wheels turn.

I guess what I am saying about the Energy involved not depending on the frame of reference refers to the dissipated energy due to friction. It is entirely due to the relative motion of the two surfaces and the sliding friction force. So, in whatever frame you want to do the other (KE) calculation, you should get the same value for the Loss in KE due to friction - and it will always be a loss, whether there is positive or negative work done on the block. In this respect, the Electrical Resistive dissipation analogy holds and the block should and will heat up by the same amount however you care to do the sums.

I still do not quite see where the frame dependence of KE comes into this. Although KE is, in general, frame dependent, that KE is only relevant to the frame of the block or the 'fixed' Earth. Wouldn't any other calculation of KE (in a third reference frame) need to be modified back into one of those frames before it had any meaningful value? We do all our sums of Earth-bound collisions referred to the Earth's surface and do not include our rotation or orbital velocity, normally.

 

[sophiecentaur]

And who suggested that?

See post 21. - final comment

 

[tiny-tim]

See post 21. - final comment

ah … that's me! …

… in all the cases the work done by kinetic friction is +ve?
i mean can never it be -v... Right..!

right!

… by "all the cases", dreamz25 and i were referring to an initially stationary block on an accelerating platform with all possible coefficients of friction

 

[sophiecentaur]

Ok. Sorted, then.