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Let us consider a routine mechanics problem.A block of $ 10kg $ rests on another block of $ 2kg $ on a frictionless table.Let $ \mu = 0.2 $ be the coefficient of friction(let's neglect the difference between coefficients of static and kinetic friction here)between the surfaces of the two blocks. image

A horizontal force of $ 10N $ is applied on the lower block.Now,the maximum force of friction( $F_\mu $)that can be applied on either block by the other surface is equal to $$ F_\mu\,_\max = \mu R $$ where $R$ is the normal contact force between the two blocks.As the blocks don't accelerate vertically, $$ R= 10g \approx 100N $$ Therefore, $$ F_\mu\,_\max = 0.2 \times 100 = 20N $$ The newton's 2nd law for the lower block in the horizontal direction would be $$ 10 - Friction = 2kg \times a_2 $$ and that for the upper block would be $$ Friction = 10kg \times a_1 $$ Clearly,the maximum force of friction is more than the force applied($10N$).So,frictional force will ensure that none of the blocks have any acceleration.

But,if you look at both the blocks as a single system,the only net external force is $F=10N$ and hence the acceleration of the centre of mass of the system is $ a= \frac{10}{10+2} = \frac{5}{6} m/s^2 .$

Since both of the the above results contradict each other,(unless the blocks accelerate in the opposite directions with same magnitudes of acceleration,which is not the case),I assume that its fairly wrong to say that the blocks have zero individual accelerations(with respect to ground).I seriously suspect that the acceleration of either block in my equation is of the frame of reference of the other block,rather than ground;but I fail to see how.I think my question is mostly a confusion in writing newton's 2nd law for different reference frames when friction is involved.

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The force of static friction is the minimum of the net force applied(excluding friction) and $\mu R$. Since in your situation, the friction only need be as large to keep the two blocks together, and the force required is smaller than the maximum friction ($\frac{25}{3}N<20N$), the friction will be the force required, which is $\frac{25}{3}N$.

$$ForceRequired=10kg \times \frac{5}{6}ms^{-2}=10N - 2kg \times \frac{5}{6}ms^{-2}=\frac{25}{3}N<20N=\mu R={Maximum Friction}$$

Clearly,the maximum force of friction is more than the force applied(10N).So,frictional force will ensure that none of the blocks have any acceleration.

Frictional force does NOT ensure none of the blocks have any acceleration. This is absurd as the ground is frictionless, thus effects to do with friction surely cannot use the ground as a frame of reference.

In your scenario, what friction does ensure is that the relative acceleration between the blocks are zero.

By the virtue of Newton's third law, when an external force is applied, you move. It's the same when friction force from the 2kg block acts on the 10kg block. Because of the third law, no matter what, just because 2kg block acts on the 10kg block, the 10kg block has to move. It doesn't matter the force is smaller than the maximum friction. The third law takes care of that.

EDIT: To address concerns of the OP in comments, I've made some clarifications.

(1) Because relative acceleration between the blocks are zero when the system accelerates as a whole.

(2) The friction shouldn't be 10N. Friction minimizes relative motion two bodies. A $f=10$ does not minimize relative motion between the two bodies engaged in friction within the range $f<\mu R$.

(3) Friction doesn't adjust to 10N. As udiboy's answer explains, friction opposes tendency of relative motion, and if $f=10N$, there is relative motion between the blocks, whereas there exists a value of f within the maximum possible friction force which causes minimum relative motion. So f will take on the value which causes zero relative motion. Which is $f=10kg \times \frac{5}{6}ms^{-2}=10N - 2kg \times \frac{5}{6}ms^{-2}=\frac{25}{3}N$.

(4) The reaction force is the frictional force the top block exerts on the bottom block.

(5) Yes, that is right. The friction top block exerts on bottom block and bottom block exerts on top block form an action reaction force pair.

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Why did you consider the acceleration of system[both blocks] while calculating force required?Shouldn't it be 10N? Well,I took the lower block independently.Then,the forces acting are gravity,normal force due to upper block,normal force due to ground,friction due to upper block and the external force of 10N. So,10-friction=2a. Since friction adjusts to 10N,a=0.If you say that this is relative to the upper block,what will be the forces on the lower block relative to ground? What reaction force are you taking about?The upper block exerts only friction right? –  scienceauror Sep 1 '13 at 13:20
    
@scienceauror I've made some edits. –  namehere Sep 2 '13 at 11:58
    
Oh thanks then,I got the point to note,i.e. friction would tend to reduce relative motion between the surfaces,not reduce the acceleration of the body on which its acting. –  scienceauror Sep 2 '13 at 21:35
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The most basic nature of friction is that it tries to oppose relative motion or tendency of motion. The external force creates a tendency of motion, which friction tries to oppose. Alright, so this suggests the force of friction on the lower block should be $10N$.

But then, look at the motion of the upper $10Kg$ block. There is a friction force of $10N$ acting on it forward, and no other force in the horizontal direction. This shows that the upper block will accelerate forward by $1m/s^2$. There is relative motion between the two blocks!

This defeats the initial purpose of friction, which was to oppose relative motion. If friction is $10N$, it will create relative motion between the two blocks. This is impossible.

What will happen is that friction will adjust to a value such that there is no relative motion between the two blocks, i.e. their velocity and acceleration are the same at all instants. Thus $a_1=a_2$, then you can proceed as @namehere suggests.

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