# 8.15 Incline plane  (Page 3/3)

 Page 3 / 3

$\begin{array}{l}\sum {F}_{y}=mg\mathrm{sin}\theta =ma\\ ⇒a=\frac{mg\mathrm{sin}\theta }{m}=g\mathrm{sin}\theta \end{array}$

In the following sections of the module, we shall consider three specific cases of motion of a block on smooth incline :

• Motion on an incline
• Motion of two blocks on an incline
• Motion on double incline

## Motion on an incline

Problem 1 : With what speed a block be projected up an incline of length 10 m and angle 30° so that it just reaches the upper end (consider g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ ).

Solution : The motion of the block is a linear motion. The component of gravity acts in the opposite direction to the motion. Let initial velocity be “u”. Here,

u = ?, x = 10 m, a = - g sin30°, v = 0 (final velocity). Using equation of motion,

$\begin{array}{l}{v}^{2}={u}^{2}+2ax\\ ⇒0={u}^{2}-2g\mathrm{sin}{30}^{0}xx={u}^{2}-2\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}10\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}\frac{1}{2}\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}10\end{array}$

$\begin{array}{l}⇒{u}^{2}=100\\ ⇒u=10\phantom{\rule{2pt}{0ex}}m/s\end{array}$

## Motion of two blocks on an incline

Problem 1 : Two blocks of mass “m1” and “m2” are placed together on a smooth incline of angle “θ” as shown in the figure. If they are released simultaneously, then find the normal force between blocks.

The incline is smooth. Hence, there is no friction. The blocks, therefore, move down due to the component of gravitational force parallel to incline. In other words, acceleration of each of the block is “gsinθ”. As such, the blocks move with equal acceleration and hence there is no normal force between the surfaces of the two blocks. We can confirm the conclusion drawn by considering free body diagram of each of the blocks.

It is important to understand that this is a different situation than when two blocks are pushed together by an external force, “F”. The external force can produce different accelerations in the blocks individually and as such normal force appears between the surfaces when blocks move collectively with same acceleration.

In this case, however, the gravitational force produces same acceleration in each of the blocks - independent of their masses; and as such, normal force does not appear between blocks. Since blocks are together initially, they move together with same velocity in downward direction.

## Motion on double incline

Problem 3 : Two blocks “A” and “B” connected by a string passing over a pulley are placed on a fixed double incline as shown in the figure and let free to move. Neglecting friction at all surfaces and mass of pulley, determine acceleration of blocks. Take g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ .

Solution : Here, we assume a direction for acceleration. If we are correct, then we should get positive value for acceleration; otherwise a negative value will mean that direction of motion is opposite to the one assumed. We also observe here that each of the block is, now, acted by additional force of tension as applied by the string.

Let us consider that block “A” moves down and block “B” moves up the plane. The figure shows the free body diagram as superimposed on the body system.

$\text{Free body diagram of “A”}$

The forces on the blocks are (i) weight of the block “A” (ii) Normal force applied by incline, " ${N}_{1}$ ", and (iii) Tension in the string, T. Analyzing force in x - direction :

$\begin{array}{l}\sum {F}_{x}=100g\mathrm{sin}{30}^{0}-T=100a\\ ⇒100\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}g\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}\frac{1}{2}-T=100a\end{array}$

$\text{Free body diagram of “B”}$

The forces on the blocks are (i) weight of the block “B” (ii) Normal force applied by incline, " ${N}_{2}$ ", and (iii) Tension in the string, T. Analyzing force in x - direction :

$\begin{array}{l}\sum {F}_{x}=T-50g\mathrm{sin}{60}^{0}=50a\\ ⇒T-50\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}g\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}\frac{\surd 3}{2}=50a\end{array}$

Adding two equations, we have :

$\begin{array}{l}⇒50\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}g\left(1-\frac{\surd 3}{2}\right)=150a\\ ⇒a=\frac{50\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}10X\left(2-1.73\right)}{150X2}=0.45\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

Positive value indicates that the chosen direction of acceleration is correct. Thus, the block “A” moves down and the block “B” moves up as considered.

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