Incline With Mass And Pulley

As we shall study, the acceleration of an object is the change in its velocity in each unit of time. In case the change in velocity in each unit of time is constant, the object is said to be moving with constant acceleration and such a motion is called uniformly accelerated motion. On the other hand, if the change in velocity in each unit of time is not constant, the object is said to be moving with variable acceleration and such a motion is called non-uniformly accelerated motion.

In this article, we will cover the concept of incline with Mass And Pulley. This topic falls under the broader category of laws of motion, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. In the last ten years of the JEE main exam, 2-3 questions have been asked but no direct questions have been asked in NEET.

Let's read this entire article to gain an in-depth understanding of Incline With Mass And Pulley.

When the Block is Hanging From the Incline

When One Block is Hanging, the Other is on the Incline Plane.

$\begin{aligned}
& a=\frac{\left[m_2-m_1 \sin \theta\right] g}{m_1+m_2} \\
& T=\frac{m_1 m_2(1+\sin \theta) g}{m_1+m_2}
\end{aligned}$

Double-Inclined Plane with Different Angles

$\begin{aligned}
& a=\frac{\left(m_2 \sin \theta_2-m_1 \sin \theta_1\right) g}{m_1+m_2} \\
& T=\frac{m_1 m_2\left(\sin \theta_1+\sin \theta_2\right) g}{m_1+m_2}
\end{aligned}$

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Solved Examples Based on Incline With Mass And Pulley

Example 1: A block of mass $m_1=2 \mathrm{Kg}$ on a smooth inclined plane at an angle $30^{\circ}$ is connected to the second block of mass $m_2=3 \mathrm{Kg}$ by a cord passing over as a frictionless pulley as shown in the figure How much time (in seconds) will be taken by block B to reach the ground if they are released from the rest:

1) 2

2) 5

3) 4

4) 3

Solution :

let tension be $\mathrm{T}$ and acceleration a of the blocks For $m_2$ block :
$
\begin{aligned}
& m_2 g-T=m_2 a \\
& T=m_2(g-a)
\end{aligned}
$

For $m_1$ block :
$
\begin{aligned}
& T=m_1 g \sin 30^{\circ}=m_1 a \\
& T=m_1\left(a+g \sin 30^{\circ}\right)
\end{aligned}
$
equating above equations
$
\begin{aligned}
& m_1\left(a+g \sin 30^{\circ}\right)=m_2(g-a) \\
& \mathrm{a}=\frac{\mathrm{m}_2 \mathrm{~g}-\mathrm{m}_1 \mathrm{~g} \sin 30^{\circ}}{\mathrm{m}_1+\mathrm{m}_2} \\
& \mathrm{a}=4 \mathrm{~m} / \mathrm{sec}^2
\end{aligned}
$

Now, Time taken by block $m_2$ to reach the ground (starting from rest $\mathrm{u}=0$ )
$
\begin{aligned}
& S=u t+\frac{1}{2} a t^2 \\
& 8=0+\frac{1}{2} \times 4 \times t^2 \\
& \therefore t=2 \mathrm{sec}
\end{aligned}
$

Hence, the answer is option (1).

Example 2: Two blocks of masses 50 Kg and 30 Kg connected by a massless string pass over a tight frictionless, pulley and rest on two smooth planes inclined at angles and respectively with horizontal as shown in the figure. If the system is released from rest then find the time taken by 30 Kg block to reach the ground

1) 20 sec

2) 30 sec

3) 10 sec

4) 50 sec

Solution:

Double-inclined plane with different angles

$\begin{gathered}
T-50 g \sin 30=50 a \ldots(1) \\
30 g \sin 60-T=30 a \ldots .(2)
\end{gathered}$

Adding (1) and (2)

$a=\frac{30 g \sin 60-50 g \sin 30}{80}=0.12 \mathrm{~m} / \mathrm{s}$

Now

$\begin{aligned}
& s=u t+1 / 2 a t^2 \\
& 6=(1 / 2) \times 0.12 \times t^2 \\
& t^2=\frac{6 \times 2}{0.12}=100 \\
& t=10 \mathrm{sec}
\end{aligned}$

Hence, the answer is option (3).

Example 3: Two bodies of masses $\mathrm{m}_1=5 \mathrm{~kg}$ and $\mathrm{m}_2=3 \mathrm{~kg}$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass $\mathrm{m}_1$ will be : [Take $\mathrm{g}=10 \mathrm{~ms}^{-2} \mathrm{]}$

1) $30 \mathrm{~N}$
2) $40 \mathrm{~N}$
3) $50 \mathrm{~N}$
4) $60 \mathrm{~N}$

Solution:

For a system at rest

$\begin{aligned}
& \mathrm{m}_1 \mathrm{~g} \cos \theta=\mathrm{N} \rightarrow(1) \\
& \text { ( } \mathrm{N} \rightarrow \text { Force by inclined plane on the block) } \\
& \mathrm{T}=\mathrm{m}_2 \mathrm{~g}=30 \mathrm{~N} \rightarrow \text { (2) } \\
& \mathrm{m}_1 \mathrm{~g} \sin \theta=\mathrm{T} \\
& 50 \sin \theta=30 \\
& \sin \theta=\frac{3}{5} \\
& \theta=37^{\circ} \\
& \mathrm{N}=\mathrm{m}_1 \mathrm{~g} \cos \theta \\
& =50 \times \cos 37^{\circ} \\
& \mathrm{N}=40 \mathrm{~N} \\
&
\end{aligned}$

$\text { The force exerted by the inclined plane on the body of mass } \mathrm{m}_1 \text { will be } 40 \mathrm{~N}$

Hence, the answer is option (2).

Example 4: Two blocks of masses $\mathrm{m}_1$ and $\mathrm{m}_2$ connected by a string are placed gently over a fixed inclined plane, such that the tension in the connecting string is initially zero. The coefficient of the firction between $\mathrm{m}_1$ and the inclined plane is $\mu_1$, between $\mathrm{m}_2$ and the inclined plane $\mu_2$. The tension in the string shall continue to remain zero if -

1) $\mu_1>\tan \alpha$ and $\mu_2<\tan \beta$
2) $\mu_1<\tan \alpha$ and $\mu_2>\tan \beta$
3) $\mu_1>\tan \alpha$ and $\mu_2>\tan \beta$
4) $\mu_1<\tan \alpha$ and $\mu_2<\tan \beta$

Solution:

$\begin{aligned}
& T+f_1=m_1 g \sin \alpha \\
& \text { For } T=0, f_1=m_1 g \sin \alpha \\
& m_1 g \sin \alpha \leqslant \mu_1 m_1 g \cos \alpha \\
& \tan \alpha \leqslant \mu_1 \\
& T+f_2=m_2 g \sin \beta \\
& T=0, m_2 g \sin \beta=f_2 \\
& m_2 g \sin \beta \leqslant \mu_2 m_2 g \cos \beta \\
& \tan \beta \leqslant \mu_2
\end{aligned}$

Hence, the answer is option (2).

Example 5: A block of mass 5 kg is there on a smooth inclined plane which is making 30 degrees angle with the horizontal, And a hanging mass of 6 kg is attached with it by a string which passes through a pulley; the acceleration of the system is:

1) $4 \mathrm{~ms}^{-2}$
2) $3.18 \mathrm{~ms}^{-2}$
3) $4.5 \mathrm{~ms}^{-2}$
4) $5 \mathrm{~ms}^{-2}$

Solution:

Let us see the FBD for the system


For acceleration, we can write the equation as\begin{aligned}
$ & a=\frac{6 g-5 g \sin 30^{\circ}}{6+5} \\
& \Rightarrow a=\frac{60-\left(50 \times \frac{1}{2}\right)}{11}=3.18 \mathrm{~ms}^{-2}
\end{aligned} $

Summary

A diagram for each body of the system depicting all the forces on the body by the remaining part of the system is called the free body diagram. From the free-body diagrams of different bodies, the equations of motion are obtained for each body.