8.12 Unbalanced force system (application)  (Page 2/2)

$\text{For “A”}$

$$\begin{array}{l}{a}_1=\frac{Mg-F}{M}\\ \Rightarrow a_1=g-\frac{F}{M}\end{array}$$

Similarly,

$$\begin{array}{l}\Rightarrow a_2=g-\frac{F}{m}\end{array}$$

As M>m,

$$\begin{array}{l}\Rightarrow a_1>a_2\end{array}$$

Motion in horizontal direction

Problem 4 : Three blocks “A”, “B” and “C” of identical mass “m” are in contact with each other. The blocks are pushed with a horizontal force "F". If $N_1$ and $N_2$ be the normal forces in horizontal direction at the contact surface between the pairs of blocks “A and B" and "B and C" respectively, then find (i) the net force on each of the blocks and (ii) normal forces, $N_1$ and $N_2$ .

Solution : Net force means resultant force on individual block. Here, the three blocks in contact under the given situation have same acceleration as block on left pushes the block ahead. As such, we can treat the motion of three blocks as that of a single composite block of mass “3m”. Let the common acceleration of the composite mass be “a”. Then,

$$\begin{array}{l}a=\frac{F}{3m}\end{array}$$

Note that we consider only horizontal forces as net force in vertical direction is zero. Since acceleration of each block of identical mass,"m", is same, the net force on each block must also be same. This is given by :

$$\begin{array}{l}{F}_{\mathrm{net}}=ma=\frac{F}{3}\end{array}$$

In order to analyze forces for normal forces at the interface in horizontal direction, we need to carry out force analysis of the blocks appropriately. We start force analysis with the third block (C), as there is only one external force on the block in the horizontal direction.

$\text{Free body diagram of block “C”}$

$$\begin{array}{l}\sum F_x=N_2=ma=mx\frac{F}{3m}=\frac{F}{3}\end{array}$$

We can now consider block “B” and “C” as the composite block of mass “2m”. Considering Free body diagram of block “B” and “C” together, we have :

$$\begin{array}{l}\sum F_x=N_1=2ma\\ \Rightarrow N_1=2ma=2mx\frac{F}{3m}=\frac{2F}{3}\end{array}$$

Problem 5 : Three blocks of mass 5 kg, 3 kg and 2 kg are placed side by side on a smooth horizontal surface as shown in the figure. An external force of 30 N acts on the block of 5 kg. Find the net external force on the block of mass 3 kg.

Solution : Here, we know that there is no motion in vertical direction. It means that forces in vertical directions are balanced force system and consideration in vertical direction can be conveniently excluded from the analysis. We can find net external force on the middle block by determining normal forces due to blocks on each of its sides. Evidently, the net external force will be the difference of normal forces in horizontal direction. This approach, however, will require us to analyze force on each of the blocks on the sides.

Alternatively, we find common acceleration, “a”, of three blocks as :

$$a=\frac{F}{M}=\frac{30}{5+3+2}=\frac{30}{10}=3m/s^2$$

According to second law of motion, the net force on the block is equal to the product of its mass and acceleration. Let the net force on the block be “F”. Then,

$$\Rightarrow F=ma=3X3=9N$$

Constrained motion

Problem 6 : At a given instant, two beads, “A” and “B” of equal mass “m” are connected by a string. They are held on a circular grove of radius “r” as shown in the figure. Find the accelerations of beads and the tension in the string, just after they are released from their positions.

Solution : Since motion is initiated, velocity of the beads are zero. No centripetal acceleration is associated at this instant. At this instant, the beads “A” and “B” have tangential accelerations in horizontal and vertical direction respectively. As the beads are connected with a string, magnitude of accelerations of the beads are equal in magnitude. As such, magnitude of tangential accelerations are equal when beads are released.

We can find acceleration and tension by considering free body diagram of each of the beads. The initial positions of the beads are at the end of a quarter of circle. It means that triangle OAB is an isosceles triangle and the string makes an angle 45° with the radius at these positions. Let $N_1$ and $N_2$ be the normal force at these positions on the beads. Let also consider that the magnitude of acceleration of either bead is “a”.

The free body diagrams of two beads are shown in the figure.

Considering FBD of “A” and forces in horizontal direction, we have :

$$T\cos {45}^0=ma$$

$$\Rightarrow T=\sqrt{2}ma$$

Considering FBD of “B” and considering forces in vertical, we have :

$$mg-T\sin {45}^0=ma$$

Substituting for “T”,

$$\Rightarrow mg-\sqrt{2}maX\frac{1}{\sqrt{2}}=ma$$

$$\Rightarrow mg-ma=ma$$

$$\Rightarrow a=\frac{g}{2}$$

We find tension in the string by putting value of acceleration in its expression as :

$$T=\frac{\sqrt{2}mg}{2}=\frac{mg}{\sqrt{2}}$$