# Direction of the work done by force of friction

1. Mar 21, 2012

### 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.....!!

2. Mar 21, 2012

### tiny-tim

hi dreamz25!
to find the direction of friction, always ask "which way would the body go if the surface suddenly turned to ice?"
kinetic friction will be greater than ma if µk > a/g

3. Mar 21, 2012

### 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).
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.

4. Mar 21, 2012

### rcgldr

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

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.

Last edited: Mar 21, 2012
5. Mar 21, 2012

### 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.

6. Mar 21, 2012

### tiny-tim

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

here, there is displacement

7. Mar 21, 2012

### 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.

8. Mar 21, 2012

### rcgldr

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.

9. Mar 22, 2012

### tiny-tim

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?

10. Mar 22, 2012

### A.T.

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).

11. Mar 22, 2012

### A.T.

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.

12. Mar 22, 2012

### 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.

13. Mar 22, 2012

### 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.

14. Mar 22, 2012

### A.T.

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)

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.

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

15. Mar 22, 2012

### Staff: Mentor

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.

Last edited: Mar 22, 2012
16. Mar 22, 2012

### 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
Ff = ma = 5 N, N = mg = 50 N and μN = 10 Newton => Ff < 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 = FrD 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.
Ff = ma = 1; a = 1/5 m/s2
Thus, Displacement D = 1/2(1/5)(10)^2 = 10m
Work done by Ff = 1N (10)m = +10J

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

Just tell me,

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

17. Mar 22, 2012

### 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.. ???

18. Mar 22, 2012

### 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.. ???

19. Mar 22, 2012

### dreamz25

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

20. Mar 22, 2012

### rcgldr

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