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Two masses \(8 \mathrm{~kg}\) and \(12 \mathrm{~kg}\) are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Short Answer

Expert verified
Acceleration is 1.96 m/s², and tension is 94.48 N.

Step by step solution

01

Understand the System

We have a system with two masses, 8 kg and 12 kg, connected by a string over a frictionless pulley. The pulley ensures the tension is the same on both sides. We need to find the acceleration of the system and the tension in the string.
02

Apply Newton's Second Law to 12 kg Mass

For the 12 kg mass moving downward, the forces acting are gravity and tension. Apply Newton's second law: \( m_1g - T = m_1a \), where \( m_1 = 12 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( T \) is the tension, and \( a \) is the acceleration.
03

Apply Newton's Second Law to 8 kg Mass

For the 8 kg mass moving upward, the forces are tension and gravity. Apply Newton's second law: \( T - m_2g = m_2a \), where \( m_2 = 8 \text{ kg} \).
04

Set Up Equations

Develop the equations based on the previous two steps: 1. \( 12g - T = 12a \)2. \( T - 8g = 8a \)Here, \( g = 9.8 \text{ m/s}^2 \).
05

Solve the Equations Simultaneously

Add the two equations to eliminate \( T \):\[12g - T + T - 8g = 12a + 8a\]This simplifies to:\[ 4g = 20a \]Solve for \( a \):\[ a = \frac{4g}{20} = \frac{4 \times 9.8}{20} = 1.96 \text{ m/s}^2 \]
06

Calculate Tension

Substitute \( a = 1.96 \text{ m/s}^2 \) back into either equation (e.g., \( T - 8g = 8a \)):\[T - 8 \times 9.8 = 8 \times 1.96\]Solve for \( T \):\[T = 8 \times 9.8 + 15.68 = 94.48 \text{ N}\]
07

Final Step: Result Summary

The acceleration of the system is \( 1.96 \text{ m/s}^2 \) and the tension in the string is \( 94.48 \text{ N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
When studying physics, especially when dealing with connected objects like masses and pulleys, understanding acceleration is key. Acceleration refers to the rate of change of velocity of an object. In this exercise, we have two masses, 12 kg and 8 kg, connected by a string over a frictionless pulley.The acceleration of the system is determined by the net force acting on the masses and the total mass of the system. Since the pulley is frictionless, it doesn't add any resistance to the motion. Therefore, the acceleration is purely due to the gravitational force acting on the masses.
  • For the 12 kg mass, gravity acts downwards.
  • For the 8 kg mass, gravity acts upwards since the mass moves up.
To find the acceleration, we apply Newton's second law, which is: \[ F = ma \]We have two forces to consider: one on the 12 kg mass caused by gravity minus tension, and one on the 8 kg mass caused by tension minus gravity. By setting up the equations and solving them for acceleration, we find that the acceleration of this system is \( 1.96\; \text{m/s}^2 \). This value tells us how quickly both masses increase their speed as they move.
Tension in the String
Tension is a force exerted along the length of a string, rope, or in this case, the inextensible string connecting the two masses. Tension is crucial in understanding how the masses move in relation to one another over the pulley. It's what keeps the system in equilibrium until released and what governs their motion once released.In this scenario, the tension is the same throughout the entire string since we assume the string is light and the pulley is frictionless. This simplifies calculations as the tension does not vary on different sides of the pulley.To find the tension, after determining the acceleration of the system, we plug the known value of the acceleration into either of the forces' equations derived from Newton's second law:\[T - 8g = 8a\]Solving this equation, we substitute the acceleration and gravitational acceleration \( (g = 9.8 \; \text{m/s}^2) \) to get the tension value:\[T = 94.48\; \text{N}\]This tension value balances out the force of gravity on the 8 kg mass while also pulling the mass upwards against gravity.
Frictionless Pulley
The concept of a frictionless pulley is an idealization used to simplify calculations in mechanics problems. A frictionless pulley allows the string to move smoothly over it without any resistance. It means there’s no energy loss due to friction when the string moves, making calculations straightforward. In our exercise, the frictionless nature of the pulley ensures:
  • The tension in the string is the same on both sides of the pulley.
  • There's no rotational resistance that would otherwise need to be factored in.
This assumption allows us to focus solely on the forces and motion of the masses. With a real pulley, friction would reduce the net force and thus the acceleration would be lesser than idealized. It also keeps the solution within straightforward algebra, without requiring calculus to compute energy losses or rotational inertia adjustments. Comprehending how the frictionless pulley assists in simplifying dynamic problems is vital for effective problem-solving in physics. It allows students to understand and predict the system’s behavior under ideal conditions before tackling more complex real-world scenarios.

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Most popular questions from this chapter

A batsman deflects a ball by an angle of \(45^{\circ}\) without changing its initial speed which is equal to \(54 \mathrm{~km} / \mathrm{h}\). What is the impulse imparted to the ball? (Mass of the ball is \(0.15\) kg.)

A thin circular loop of radius \(R\) rotates about its vertical diameter with an angular frequency \(\omega\). Show that a small bead on the wire loop remains at its lowermost point for \(\omega \leq \sqrt{g / R} .\) What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for \(\omega=\sqrt{2 g / R} ?\) Neglect friction.

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