Acceleration of Block on Smooth Inclined Plane

When an Inclined Plane is at Rest

$R=m g \cos \theta$ along with normal to the incline, $m g \sin \theta=m a$ along the incline $a=g \sin \theta$

When an Inclined Plane is Given Acceleration 'b'

$\begin{aligned}
& R=m g \cos \theta+m b \sin \theta \\
& m a=m g \sin \theta-m b \cos \theta \\
& a=g \sin \theta-b \cos \theta
\end{aligned}$

The condition of the body is to be at rest relative to the inclined plane.

$\begin{aligned}
& a=g \sin \theta-b \cos \theta=0 \\
& b=g \tan \theta
\end{aligned}$

For More Information On Acceleration of Block on the horizontal inclined plane, Watch The Below Video:

Solved examples based on Acceleration of Block on the horizontal inclined plane

Example 1: As shown in the diagram a small block of mass 10 kg is kept on a smooth inclined plane. What should be the value of the force F along the incline so that the block remains stationary w.r.t incline is:

1) 50

2) 20

3) 35

4) 40

Solution:

F.B.D of the block along the incline-

As the block is at rest, the net force along the incline must be zero-

$F=m g \sin 30^{\circ} F=10 \times 10 \times \frac{1}{2} \Rightarrow F=50 \mathrm{~N}$

Example 2: As shown in the figure block of mass m is placed on a smooth inclined plane which is accelerating with speed b. What should be the value of 'b' so that the block m remains stationary w.r.t. incline plane is:

1) $\frac{2}{\sqrt{3}} g$
2) $\frac{\sqrt{3}}{2} g$
3) $\sqrt{3 g}$
4) $\frac{g}{\sqrt{3}}$

Solution:

Acceleration of Block on Smooth Inclined Plane -

When an Inclined Plane is given Acceleration 'b'

$
\begin{aligned}
& R=m g \cos \theta+m b \sin \theta \\
& m a=m g \sin \theta-m b \cos \theta \\
& a=g \sin \theta-b \cos \theta
\end{aligned}
$

The condition of the body is to be at rest relative to the inclined plane.
$
\begin{aligned}
& a=g \sin \theta-b \cos \theta=0 \\
& b=g \tan \theta
\end{aligned}
$

On the basis of this,

for the block to be stationary $m b \cos \theta=m g \sin \theta$
$
b=g \tan \theta=\sqrt{3} g
$

Hence, the answer is option (3).

Example 3: Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B.

1) 4.9 ms ^-2 in the vertical direction

2) 4.9 ms ^-2 in the horizontal direction

3) 9.8 ms ^-2 in the vertical direction

4) Zero

Solution :

Let acceleration of block be 'a' as shown in the figure.

Along the incline-

$
m g \sin \theta=m a \Rightarrow a=g \sin \theta
$

Acceleration vector according to the coordinate system shown in the figure-
$
\vec{a}=g \sin \theta \cos \theta \hat{i}-g \sin ^2 \theta \hat{j}
$

Acceleration of $\mathrm{A}$ -
$
\overrightarrow{a_A}=g \sin 60^{\circ} \cos 60^0 \hat{i}-g \sin ^2 60^0 \hat{j}
$

Acceleration of $\mathrm{B}$ -
$
a_B=g \sin 30^{\circ} \cos 30^0 \hat{i}-g \sin ^2 30^{\circ} \hat{j}
$

Acceleration of $\mathrm{A}$ with respect to $\mathrm{B}$ -
$
a_{\overrightarrow{A B}}=\overrightarrow{a_A}-\overrightarrow{a_B} a_{A B}=-\left(\frac{3 g}{4}-\frac{g}{4}\right) \hat{j}\left|a_{\overrightarrow{A B}}\right|=\frac{g}{2}=4.9 \mathrm{~m} / \mathrm{s}^2
$

Hence, the answer is option (2).

Qu 4: A small mass m is slipping over a frictionless incline made of wood of mass M. The acceleration by which the incline should be pushed so that block m remains stationary w.r.t incline is :

1) $\frac{2}{\sqrt{3}} g$
2) $\frac{g}{\sqrt{3}}$
3) $\frac{g}{\sqrt{2}}$
4) $\frac{\sqrt{3}}{2} g$

Solution-

Let the acceleration of the triangular wedge be 'a', as shown in the figure.

As the block remains stationary with respect to the wedge, the block should be in equilibrium in the frame of reference of the wedge.

F.B.D in the frame of reference attached with the wedge along the incline-

Note- Since the frame of reference is accelerated (non-inertial), a pseudo force (ma) acts on the block towards the left.

Applying condition of equilibrium along the incline-

$\begin{aligned}
& \operatorname{macos} 30^{\circ}=m g \sin 30^{\circ} a=g \tan 30^{\circ} \Rightarrow a=\frac{g}{\sqrt{3}} \\
& m a \cos \theta=m g \sin \theta a=g \tan \theta=g \tan 30^{\circ}=\frac{g}{\sqrt{3}}
\end{aligned}$

Example 5: A block of mass $200 \mathrm{~g}$ is kept stationary on a smooth inclined plane by applying a minimum horizontal force $\mathrm{F}=\sqrt{\mathrm{xN}}$ as shown in the figure.

The value of $\mathrm{x=}$_________________.

1) 13

2) 12

3) 120

4) 130

Solution

$
\mathrm{N}=\mathrm{mg} \cos 60^{\circ}+\mathrm{F} \sin 60^{\circ}
$

For stationary block
$
\begin{aligned}
& \mathrm{F} \cos 60^{\circ}=m g \sin 60^{\circ} \\
& \sqrt{x}=2 \times \sqrt{3} \\
& x=12
\end{aligned}
$

Hence, the answer is option (2).

Summary

The acceleration of a block on a smooth inclined plane can be analyzed using free-body diagrams. When the plane is at rest, the block's acceleration along the incline is $a=g \sin \theta$. If the inclined plane is given an acceleration $b$, the block's acceleration relative to the plane is $a=g \sin \theta$ $b \cos \theta$. For the block to remain stationary relative to the accelerating plane, $b=g \tan \theta$. This concept is crucial for understanding motion in physics and is applicable in various competitive exams.