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.

$\begin{array}{rl}& a=\frac{[m_2-m_1\sin \theta ]g}{m_1+m_2}\\ & T=\frac{m_1{m}_2(1+\sin \theta )g}{m_1+m_2}\end{array}$

Double-Inclined Plane with Different Angles

$\begin{array}{rl}& a=\frac{(m_2\sin {\theta }_2-m_1\sin {\theta }_1)g}{m_1+m_2}\\ & T=\frac{m_1{m}_2(\sin {\theta }_1+\sin {\theta }_2)g}{m_1+m_2}\end{array}$

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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 :
$\begin{array}{rl}& m_{2}g-T=m_{2}a\\ & T=m_2(g-a)\end{array}$

For $m_1$ block :
$\begin{array}{rl}& T=m_{1}g\sin {30}^{\circ }=m_{1}a\\ & T=m_1(a+g\sin {30}^{\circ })\end{array}$
equating above equations
$\begin{array}{rl}& m_1(a+g\sin {30}^{\circ })=m_2(g-a)\\ & \mathrm{a}=\frac{{\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_1\mathrm{g}\sin {30}^{\circ }}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ & \mathrm{a}=4\mathrm{m}/{\sec }^2\end{array}$

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

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

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

Adding (1) and (2)

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

Now

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

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 $\mathrm{g}=10{\mathrm{ms}}^{-2}]$

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

Solution:

For a system at rest

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

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

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:

$\begin{array}{rl}& T+f_1=m_{1}g\sin \alpha \\ & \text{For}T=0,f_1=m_{1}g\sin \alpha \\ & m_{1}g\sin \alpha ⩽{\mu }_1{m}_{1}g\cos \alpha \\ & \tan \alpha ⩽{\mu }_1\\ & T+f_2=m_{2}g\sin \beta \\ & T=0,m_{2}g\sin \beta =f_2\\ & m_{2}g\sin \beta ⩽{\mu }_2{m}_{2}g\cos \beta \\ & \tan \beta ⩽{\mu }_2\end{array}$

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$$\begin{array}{rl}\$& a=\frac{6g-5g\sin {30}^{\circ }}{6+5}\\ & \Rightarrow a=\frac{60-(50\times \frac{1}{2})}{11}=3.18{\mathrm{ms}}^{-2}\end{array}$$ $

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.