7.1 Pulley

  Laws of Motion

Pulley is a device, which facilitates the smooth movement of string.

For general problem on pulleys, we consider pulley to be smooth and frictionless.

Problem based on pulleys becomes very simple using concept of pulling force method.

Let us consider two unequal masses $m_1$ and $m_2$ attached to the ends of a light inextensible string, which passes over a smooth massless pulley.

From the FBD, we can write, $$\begin{equation} \begin{aligned} T - {m_1}g = {m_1}a\quad ...(i) \\ {m_2}g - T = {m_2}a\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(ii)$ we get, $$a = \left( {\frac{{{m_2} - {m_1}}}{{{m_1} + {m_2}}}} \right)g\quad ...(iii)$$

From equation $(i)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} T - {m_1}g = {m_1}\left( {\frac{{{m_2} - {m_1}}}{{{m_1} + {m_2}}}} \right)g \\ T = \frac{{2{m_1}{m_2}g}}{{{m_1} + {m_2}}}\quad ...(iv) \\\end{aligned} \end{equation} $$

Tension at the hinge of the pulley, $$H = 2T$$

Question 5. Find the acceleration of the system, tension in the string and hinge reaction due to the pulley from the diagram as shown below.

Solution: From the FBD we can write, $$\begin{equation} \begin{aligned} T = {m_1}a\quad ...(i) \\ {m_2}g - T = {m_2}a\quad ...(ii) \\\end{aligned} \end{equation} $$

Adding equation $(i)$ and $(ii)$ we get, $$a = \frac{{{m_2}g}}{{\left( {{m_1} + {m_2}} \right)}}\quad ...(iii)$$

From equation $(i)$ and $(iii)$ we get, $$T = {m_1}a = \frac{{{m_1}{m_2}g}}{{\left( {{m_1} + {m_2}} \right)}}$$

Hinge reaction $(R)$, $$\begin{equation} \begin{aligned} \overrightarrow R = - T\widehat i - T\widehat j \\ \left| {\overrightarrow R } \right| = \sqrt {{T^2} + {T^2}} \\ R = \sqrt 2 T \\\end{aligned} \end{equation} $$

Question 6. All surface are assumed to be frictionless. Find the acceleration of the system. Also find the reaction at pulley hinge point.

Solution: When the acting force is not along the direction of motion, we resolve the components of force along the direction of motion.

From FBD we can write, $$\begin{equation} \begin{aligned} T = {m_2}a\quad ...(i) \\ {m_1}g\sin \theta - T = {m_1}a\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(ii)$ we get, $$a = \frac{{{m_1}g\sin \theta }}{{{m_1} + {m_2}}}$$

Hinge reaction $(R)$, $$\begin{equation} \begin{aligned} R = \sqrt {{T^2} + {T^2} + 2(T)(T)\cos (\pi - \theta )} \\ R = 2T\sin \frac{\theta }{2} \\\end{aligned} \end{equation} $$

Direction of the force, $$\begin{equation} \begin{aligned} \tan \phi = \frac{{T\sin \left( {\pi - \theta } \right)}}{{T + T\cos \left( {\pi - \theta } \right)}} \\ \tan \phi = \frac{{\sin \theta }}{{1 - \cos \theta }} = \cot \frac{\theta }{2} \\ \tan \phi = \tan \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) \\ \phi = \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) \\\end{aligned} \end{equation} $$

Question 7. Find the acceleration of the system. Also find the tension in all the strings.

(All surface are frictionless and wedge is fixed, Take $g = 10m/{s^2}$).

Solution: From FBD we can write, $$\begin{equation} \begin{aligned} {T_1} - 4g\sin 30^\circ = 4a\quad ...(i) \\ 2g\sin 60^\circ - {T_2} = 2a\quad ...(ii) \\ 6g\sin 60^\circ + {T_2} - {T_1} = 6a\quad ...(iii) \\\end{aligned} \end{equation} $$

Adding equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} a = \left( {\frac{{8g\sin 60^\circ - 4g\sin 30^\circ }}{{12}}} \right) \\ a = \left( {\frac{{2\sin 60^\circ - \sin 30^\circ }}{{12}}} \right)g \\ a = 4.10m/{s^2}\quad ...(iv) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(iv)$ we get, $$\begin{equation} \begin{aligned} {T_1} = 4\left( {a + g\sin 30^\circ } \right) \\ {T_1} = 4\left( {4.10 + 10 \times \frac{1}{2}} \right) \\ {T_1} = 36.4N \\\end{aligned} \end{equation} $$

From equation $(ii)$ and $(iv)$ we get, $$\begin{equation} \begin{aligned} {T_2} = 2\left( {g\sin 60^\circ - a} \right) \\ {T_2} = 2\left( {10 \times \frac{{\sqrt 3 }}{2} - 4.10} \right) \\ {T_2} = 9.12N \\\end{aligned} \end{equation} $$