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Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables?
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Body A with weight 60N on a frictionless horizontal surface is connect...
**Solution:**
To solve this problem, we can use Newton's second law of motion and the principle of conservation of energy.
**1. Acceleration of the system:**
Let's assume that the acceleration of the system is 'a'. Since the pulley is frictionless, the tension in the cable connected to body A is equal to the tension in the cable connected to the hanging body.
The equation of motion for Body A can be written as:
\[T - W_A = m_A \cdot a\]
Where:
- T is the tension in the cable connected to Body A
- W_A is the weight of Body A
- m_A is the mass of Body A
The equation of motion for the hanging body can be written as:
\[W_B - T = m_B \cdot a\]
Where:
- W_B is the weight of the hanging body
- m_B is the mass of the hanging body
Since the hanging body is freely hanging, its weight is equal to the tension in the cable connected to it:
\[W_B = T\]
Substituting this into the equation of motion for the hanging body:
\[T - T = m_B \cdot a\]
\[0 = m_B \cdot a\]
Since the mass of the hanging body is non-zero, the acceleration of the system must be zero:
\[a = 0\]
Therefore, the system is in equilibrium and there is no acceleration.
**2. Tension in the cables:**
Since the system is in equilibrium, the tension in the cables can be calculated using the equation of motion for Body A:
\[T - W_A = 0\]
\[T = W_A\]
Substituting the values of the weights:
\[T = 60N\]
Therefore, the tension in the cables is 60N.
To solve this problem, we can use Newton's second law of motion and the principle of conservation of energy.
**1. Acceleration of the system:**
Let's assume that the acceleration of the system is 'a'. Since the pulley is frictionless, the tension in the cable connected to body A is equal to the tension in the cable connected to the hanging body.
The equation of motion for Body A can be written as:
\[T - W_A = m_A \cdot a\]
Where:
- T is the tension in the cable connected to Body A
- W_A is the weight of Body A
- m_A is the mass of Body A
The equation of motion for the hanging body can be written as:
\[W_B - T = m_B \cdot a\]
Where:
- W_B is the weight of the hanging body
- m_B is the mass of the hanging body
Since the hanging body is freely hanging, its weight is equal to the tension in the cable connected to it:
\[W_B = T\]
Substituting this into the equation of motion for the hanging body:
\[T - T = m_B \cdot a\]
\[0 = m_B \cdot a\]
Since the mass of the hanging body is non-zero, the acceleration of the system must be zero:
\[a = 0\]
Therefore, the system is in equilibrium and there is no acceleration.
**2. Tension in the cables:**
Since the system is in equilibrium, the tension in the cables can be calculated using the equation of motion for Body A:
\[T - W_A = 0\]
\[T = W_A\]
Substituting the values of the weights:
\[T = 60N\]
Therefore, the tension in the cables is 60N.
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Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables?
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Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables?.
Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body A with weight 60N on a frictionless horizontal surface is connected to pulley via cable with another body of weight 180N which hangs vertical freely . Given: mass moment of inertia I=0.2kgm^2 and R=0.5m. Find the acceleration and tension in the cables?.
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