10.8 Friction between horizontal surfaces  (Page 4/4)

Here, we adopt the strategy of assuming nature of friction before hand. First, we calculate the limiting friction as before,

$$F_S=\mu N=\mu mg$$

Then, we consider a friction, “ $f_S$ ”, which is static friction, but whose magnitude is not known. If we are wrong in our assumption as brought out by the analysis, then we will correct our assumption; otherwise not. The free body diagrams of the block and the plank are shown in the figure.

Since friction is static unknown friction, two bodies move together without any relative motion between them. It means that :

$$a_A=a_B$$

$$\frac{F_B-f_s}{M}=\frac{f_s-F_A}{m}$$

$$\Rightarrow m{F}_B-m{f}_s=M{f}_s-M{F}_A$$

$$\Rightarrow f_s\left(m+M\right)=m{F}_B+M{F}_A$$

$$\Rightarrow f_s=\frac{m{F}_B+M{F}_A}{m+M}$$

If the value of “ $f_s$ ” as calculated, using expression derived above, is less than limiting friction, then we conclude that two bodies are indeed moving together. Otherwise, our assumption were wrong and the bodies are moving with different accelerations. In that case, we calculate accelerations separately as :

$$\Rightarrow a_A=\frac{f_s-F_A}{m}$$

$$\Rightarrow a_B=\frac{F_B-f_s}{M}$$

Velocity .vs. force

We have so far analyzed motion, considering external force on the bodies. Under certain condition, if data is provided in the form of velocity of the individual body, then the analysis is simplified significantly. Consider the set up as shown in the figure. At a certain instant, the block is imparted a velocity (alternatively, we have simply allowed a block with certain velocity to slide over the plank underneath).

An initial velocity to block, here, ensures that there is relative motion at the interface. This, in turn, ensures that the friction between the surfaces is kinetic friction. This means that nature of friction is known and need not be investigated as in the case when external force is applied.

Example

Problem : A block with initial velocity "v" is placed over a rough horizontal plank of the same mass. The plank resides over a smooth horizontal plane. Plot the velocities of the block and the plank with respect to time.

Solution : Let the mass of the block and plank each be “m” and “μ” be the coefficient of kinetic friction between them. Since block is given a velocity with respect to ground, the friction between block and plank is kinetic friction, given by :

$$F_K=\mu N=\mu mg$$

The direction of friction on the block is opposite to the direction of motion. The friction here retards the motion. Let us denote block and plank by subscripts “1” and “2” respectively.The deceleration of the block is given by :

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

It means that velocity of the block decreases at constant rate with time. The velocity – time plot of block, therefore, is a straight line with negative slope. On the other hand, friction is the only force on the plank in the forward direction. The magnitude of acceleration is same as that of block, because it has the same mass and it is worked by force of same magnitude. Since plank starts from zero velocity, the velocity – time plot of the plank is a straight line of constant slope, starting from the origin of the plot.

The block looses motion due to deceleration, whereas plank acquires motion due to acceleration. A situation comes when speeds of the two entities are equal. In that situation,

$$v_1=v_2$$

$$v-at=0+at$$

$$t=\frac{v}{2a}=\frac{v}{2\mu g}$$

The common velocity is given by :

$$v_c=v_2=at=\mu gX\frac{v}{2\mu g}=\frac{v}{2}$$

This means that velocity of the block and plank becomes equal to half of the initial velocity imparted to the block. There is no relative motion between two entities. They simply move together with same velocity as combined mass with the common velocity, “v/2”. The required plot is shown in the figure :

It must also be understood that the above approximates an ideal condition; the plank and block will eventually stop as there is some friction between plank and the underneath horizontal surface.