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## Tension * Tension is the force exerted by a string, rope, cable or similar object on another object. * Tension is always a pulling force. * Tension is always directed along the string. ### Tension in a String Passing Over a Pulley If the string and pulley are massless and frictionless, t...

## Tension * Tension is the force exerted by a string, rope, cable or similar object on another object. * Tension is always a pulling force. * Tension is always directed along the string. ### Tension in a String Passing Over a Pulley If the string and pulley are massless and frictionless, then the tension in the string is the same on both sides of the pulley. ### Example A block of mass $m_1 = 5 kg$ rests on a frictionless table. A string is attached to the block and runs over a massless, frictionless pulley. A second block of mass $m_2 = 10 kg$ is attached to the other end of the string. Find the acceleration of the blocks and the tension in the string. ### Solution **Block 1:** $\sum F_x = T = m_1 a$ $\sum F_y = N - m_1 g = 0$ **Block 2:** $\sum F_y = m_2 g - T = m_2 a$ **Substituting** $T = m_1 a$ **into the second equation:** $m_2 g - m_1 a = m_2 a$ $m_2 g = (m_1 + m_2) a$ $a = \frac{m_2 g}{m_1 + m_2} = \frac{(10 kg)(9.8 m/s^2)}{5 kg + 10 kg} = 6.53 m/s^2$ $T = m_1 a = (5 kg)(6.53 m/s^2) = 32.7 N$ ### Example A block of mass $m_1 = 5 kg$ rests on a table. The coefficient of kinetic friction between the block and the table is $\mu_k = 0.2$. A string is attached to the block and runs over a massless, frictionless pulley. A second block of mass $m_2 = 10 kg$ is attached to the other end of the string. Find the acceleration of the blocks and the tension in the string. ### Solution **Block 1:** $\sum F_x = T - f_k = m_1 a$ $\sum F_y = N - m_1 g = 0$ $f_k = \mu_k N = \mu_k m_1 g$ $T - \mu_k m_1 g = m_1 a$ **Block 2:** $\sum F_y = m_2 g - T = m_2 a$ **Substituting** $T = m_2 g - m_2 a$ **into the first equation:** $m_2 g - m_2 a - \mu_k m_1 g = m_1 a$ $m_2 g - \mu_k m_1 g = (m_1 + m_2) a$ $a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2} = \frac{(10 kg)(9.8 m/s^2) - (0.2)(5 kg)(9.8 m/s^2)}{5 kg + 10 kg} = 5.87 m/s^2$ $T = m_2 g - m_2 a = (10 kg)(9.8 m/s^2) - (10 kg)(5.87 m/s^2) = 39.3 N$ ### Inclined Plane Example A block of mass m1 rests on a frictionless inclined plane that makes an angle of $\theta = 30°$ with the horizontal. A string is attached to the block and runs over a massless, frictionless pulley. A second block of mass $m_2$ is attached to the other end of the string. What must be the value of $m_2$ if the system is to remain at rest? ### Solution In order for the system to remain at rest, the acceleration must be zero. **Block 1:** $\sum F_x = T - m_1 g sin\theta = 0$ $\sum F_y = N - m_1 g cos\theta = 0$ $T = m_1 g sin\theta$ **Block 2:** $\sum F_y = m_2 g - T = 0$ $T = m_2 g$ $m_2 g = m_1 g sin\theta$ $m_2 = m_1 sin\theta = (5 kg) sin(30°) = 2.5 kg$