10.9 Friction between horizontal surfaces (application)  (Page 3/3)

The relative acceleration of block “A” with respect to plank, “ $a_{\mathrm{AB}}$ ”, is :

$$\Rightarrow a_{AB}=a_A-a_B=\frac{g}{4}-\frac{7g}{8}=-\frac{5g}{8}$$

The relative acceleration of “A” with respect to be is, thus, directed in the opposite direction of the motion.

One external force each on block and plank

Problem 4 : A block of mass 4 kg is placed on a plank of mass 8 kg. At an instant, two external forces are applied on them as shown in the figure. The coefficient of friction for surfaces between block and plank is 0.5, whereas friction between plank and underneath surface is negligible. Find the friction between block and plank.

Solution : Considering block, the maximum static friction between block and plank is :

$$F_S=\mu N=\mu mg=0.5X4X10=20N$$

The block and the plank will have relative motion with respect to each other when friction between the surfaces is kinetic friction (about equal to limiting friction). If friction between them is less than limiting static friction, then two entities will move together.

Here, we do not know the magnitude of friction. So we do not know whether two entities move together or not. Let us assume that they move together. If our assumption is wrong, then friction as determined from the force analysis under the assumed condition will equal to limiting friction, otherwise less than it.

Now, net external force on the combined body is towards left. Let the common acceleration be, “a”, and let the friction between block and plank be “ $F_F$ ”. The free body diagrams of block and plank are shown in the figure.

$$a=\frac{F_F-5}{4}=\frac{40-F_F}{8}$$

Solving for “ $F_F$ ”, we have :

$$\Rightarrow 8F_F-40=160-4F_F$$

$$\Rightarrow 12F_F=200$$

$$\Rightarrow F_F=16.66N$$

Thus, we see that friction at the interface is actually less than the limiting friction. Our assumption, therefore, was correct and hence, friction between the surface is 16.66 N.

Block with initial velocity

Problem 5 : A plank “B” of mass “2m” is placed over smooth horizontal surface. A block “A” of mass “m” and having initial velocity “ $v_0$ ” is gently placed over the plank at one of its end as shown in the figure. If the coefficient of friction between “A” and “B” is 0.5, find the acceleration of “B” with respect to “A”.

Solution : We see here that block and plank has certain initial relative velocity. It means that friction between “A” and “B” is kinetic friction. The friction acts opposite to the velocity of the block “A”. On the other hand, friction is the only external force on the plank and acts opposite to the friction on “A”. Clearly, it is the friction that accelerates plank "B".

Let the magnitudes of accelerations of the block and the plank be “ $a_A$ ” and “ $a_B$ ” respectively with respect to ground. The free body diagrams, showing the forces on the block and the plank, are shown here.

The magnitudes of accelerations are :

$$a_A=\frac{\mu mg}{m}=\mu g$$

$$a_B=\frac{\mu mg}{2m}=\frac{\mu g}{2}$$

Two accelerations are in opposite direction. The relative acceleration of “B” with respect to “A” is :

$$a_{BA}=a_B-a_A=\frac{\mu g}{2}-\left(-\mu g\right)=\frac{3\mu g}{2}$$

The relative acceleration is in the same direction as that of the acceleration of block “B” i.e. towards right.