10.1 Friction (application)  (Page 2/2)

Solution : The initial velocity of box is zero. When we put the box on the belt, the friction on the belt, acts opposite to the direction of motion of the conveyor belt. The friction on the box, opposite to the friction on the belt. acts in the direction of motion of the belt.

The friction force accelerates the box till there is no relative motion between box and belt. In other words, the box accelerates till the box achieves the velocity of the belt i.e. “v”. Since there is relative motion between two bodies, the friction is equal to kinetic friction till their velocities are equal. Thus, acceleration of the box with respect to ground is :

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

Here, initial velocity is zero, final velocity is “v” and acceleration is “μg”. Thus, applying equation of motion for constant acceleration, we have :

$$v=u+at$$

$$v=0+\mu gt$$

$$\Rightarrow t=\frac{v}{\mu g}$$

Blocks

Problem 4 : A variable increasing force is applied on a rod in the arrangement shown in the figure. The blocks are identical and are placed over a rough horizontal surface. Determine which of the block will move first?

Solution : The blocks undergo translation in opposite directions. Since blocks are identical, the friction between block and horizontal surface is same for both blocks. It is evident that the block, having greater external force parallel to surface, will overcome friction first.

It is, therefore, evident that we need to analyze forces in the horizontal direction. The forces on the rod are shown in the figure below. The components of forces in horizontal direction form balanced force system before the motion is initiated. From the force analysis in horizontal direction, we have :

$$\Rightarrow N_2\sin \theta =N_1+F$$

The contact normal force are equal and opposite between rod and blocks. The horizontal component of normal force ( $N_2\sin \theta$ ) as applied by the rod on the block “B” is greater than the horizontal normal force, ( $N_1$ ) as applied by the rod on the block “A”. It means the external horizontal force on “B” will exceed the limiting friction first. As such, block “B” will move first.

Stack of blocks

Problem 5 : The masses of three blocks stacked over one another are $m_1$ = 30 kg, $m_2$ = 10 kg and $m_3$ = 40 kg. The coefficients of friction between block “1” and ground is ${\mu }_0$ = 0.1, between “1” and “2” is ${\mu }_1$ = 0.2 and between “2” and “3” is ${\mu }_2$ = 0.3. An external force is applied on the block “3” as shown in the figure. What least force is required to initiate motion of any part of the stack?

Solution : In order to answer this question, we are required to determine limiting static friction at each of the interface. The limiting friction between first block and ground is :

$$F_0={\mu }_{0}N={\mu }_0\left(m_1+m_2+m_3\right)g=0.1\left(30+10+40\right)X10=80N$$

The limiting friction between first and second block is :

$$F_1={\mu }_{1}N={\mu }_1\left(m_2+m_3\right)g=0.2\left(10+40\right)X10=100N$$

The limiting friction between second and third is :

$$F_2={\mu }_{2}N={\mu }_2\left(m_3\right)g=0.3X40X10=120N$$

If the external force is less than 80 N, then none of the block will move. If external force is equal or greater than 80 N, but less than 100 N, then only the first block (the one in contact with ground) will move as external force on it is equal to the limiting friction between block “1” and the underlying horizontal surface.