10.9 Friction between horizontal surfaces (application)  (Page 2/3)

$$a=\frac{50}{\left(2+3\right)}=10m/s^2$$

Let us use subscript “1” and “2” to denote block and plank respectively. Since friction is the only force on the block (1), the friction is

$$F_1=m_{1}a=2X10=20N$$

The limiting friction, however, is :

$$F_S=\mu mg=0.5X2X10=10N$$

It means that our assumption that they move together is wrong. They move with relative velocity between them. The friction, therefore, is equal to kinetic friction and is nearly equal to limiting friction 10 N.

The free body diagram of two entities are shown in the figure.

$$a_1=\frac{F_K}{m_1}=\frac{10}{1}=10m/s^2$$

$$a_2=\frac{50-F_K}{m_2}=\frac{50-10}{2}=20m/s^2$$

The relative acceleration of 1 (block) with respect 2 (plank) is :

$$a_{12}=a_1-a_2$$

$$a_{12}=10-20=-10m∕s^2$$

Thus, block has relative acceleration in the opposite direction to the direction of individual accelerations i.e. it is directed towards left. Now, initial velocity is zero and total displacement is 5m. Applying equation of motion for constant acceleration, we have :

$$x=ut+\frac{1}{2}a{t}^2$$

$$\Rightarrow 5=0+\frac{1}{2}X10t^2$$

$$\Rightarrow t=\sqrt{\frac{10}{10}}=1s$$

Problem 3 : In the arrangement shown, a force of 4mg, acts on the plank of mass “4m”. The coefficient of friction between block “A” and plank “B” is 0.25, whereas friction between plank and the surface underneath is negligible. Find the acceleration of “A” with respect to ground and its acceleration with respect to block “B”.

Solution : External force acts on the plank “B”. As such, it has tendency to move to right with respect to block “A”. The friction on the plank is, therefore, directed in the opposite direction i.e. towards left.

According to Newton’s third law, the friction acts on block in equal measure but in opposite direction i.e. towards right. We, however, need to know the nature of friction to proceed with the force analysis.

Here, limiting friction is :

$$F_S=\mu mg=0.25X2mg=\frac{mg}{2}$$

We see here that friction is the only external force on block “A”. We assume here that friction is less than limiting friction (if assumption is wrong then we would get wrong result). In this condition, plank and block will together with a common acceleration given by :

$$\Rightarrow a=\frac{F}{6m}=\frac{4mg}{6m}=\frac{2g}{3}$$

The external force on the block would, then, be :

$$\Rightarrow F_A=2ma=\frac{2mX2g}{3}=\frac{4mg}{3}$$

Clearly, external force on block “A” is greater than limiting friction, which is not possible as friction is the only external force on it. Friction, we know, can not exceed limiting friction. It means our assumption that block and plank move together has been wrong. It further means that two entities move with relative motion. As such, friction between them is kinetic friction (about equal to limiting friction).

Now, we know the magnitude and direction of friction force operating between the surface and hence, we can draw the free body diagrams of each of the entities as shown in the figure.

The acceleration of the block with respect to ground, “ $a_A$ ”, is given by :

$$\Rightarrow a_A=\frac{F_S}{2m}=\frac{\frac{mg}{2}}{\mathrm{2m}}=\frac{g}{4}$$

The acceleration of plank with respect to ground, “ $a_B$ ”, is given by :

$$a_B=\frac{4mg-F_S}{4m}=\frac{4mg-\frac{mg}{2}}{4m}=\frac{7g}{8}$$