Two masses 26kg and 24kg are attached to the ends of a string which passes over a frictionless pulley. The 26kg mass is lying over a smooth horizontal table while the 24kg mass is... Two masses 26kg and 24kg are attached to the ends of a string which passes over a frictionless pulley. The 26kg mass is lying over a smooth horizontal table while the 24kg mass is moving vertically downward. Find the tension in the string and the acceleration of the bodies.
Understand the Problem
The question is asking for the tension in the string and the acceleration of the two masses connected by the string over a pulley. We are dealing with two masses, one on a frictionless horizontal surface and the other hanging vertically. To solve this, we need to apply Newton's second law of motion to both masses to find the acceleration and then use it to find the tension in the string.
Answer
The tension in the string is given by $T = \frac{m_1 m_2 g}{m_1 + m_2}$ and the acceleration is $a = \frac{m_2 g}{m_1 + m_2}$.
Answer for screen readers
The tension in the string is given by $$ T = \frac{m_1 m_2 g}{m_1 + m_2} $$ and the acceleration of the masses is $$ a = \frac{m_2 g}{m_1 + m_2} $$
Steps to Solve
- Identify the forces acting on the masses
Consider the two masses: mass $m_1$ on the frictionless horizontal surface and mass $m_2$ hanging vertically. The forces acting on $m_1$ are the tension $T$ in the string, which pulls it to the right. For mass $m_2$, the forces are its weight $m_2 g$ acting downwards and the tension $T$ acting upwards.
- Set up the equations of motion for each mass
For the mass on the surface ($m_1$), applying Newton's second law ($F = ma$): $$ T = m_1 a $$
For the hanging mass ($m_2$), we apply Newton's second law again: $$ m_2 g - T = m_2 a $$
- Solve for acceleration (a)
Now we have two equations:
- $T = m_1 a$
- $m_2 g - T = m_2 a$
Substituting the first equation into the second to eliminate $T$: $$ m_2 g - m_1 a = m_2 a $$ Rearranging gives us: $$ m_2 g = m_1 a + m_2 a $$ $$ m_2 g = (m_1 + m_2) a $$ So, the acceleration $a$ can be found by: $$ a = \frac{m_2 g}{m_1 + m_2} $$
- Solve for tension (T)
Now, using the value of $a$ we obtained, we can substitute back into the equation for tension: $$ T = m_1 a $$ Substituting $a$ gives: $$ T = m_1 \left(\frac{m_2 g}{m_1 + m_2}\right) $$ This simplifies to: $$ T = \frac{m_1 m_2 g}{m_1 + m_2} $$
The tension in the string is given by $$ T = \frac{m_1 m_2 g}{m_1 + m_2} $$ and the acceleration of the masses is $$ a = \frac{m_2 g}{m_1 + m_2} $$
More Information
This problem explores the dynamics of two connected masses, demonstrating how the forces act on each mass through a string. It's a common problem in physics, illustrating fundamental principles of mechanics and Newton's laws.
Tips
- Forgetting to account for the direction of the forces can lead to incorrect equations. Always ensure to understand which way each force is acting.
- Not substituting correctly between equations may result in errors in calculating tension or acceleration.
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