### Example Problem on Pulley Systems#### Problem Statement:In the figure, two blocks are connected by a massless string over a pulley of radius \(2R\) and rotational inertia, \(I\). When this system is released from rest, the blocks (masses: \(m_1 = m\) and \(m_2 = 3m\)) are moving with a constant acceleration. (The friction constant between \(m_1\) and the surface is equal to \(\mu\)).##### Question:a) Draw the free body diagram and write the Newton’s equations for both masses and pulley.b) Find the angular velocity of the pulley in 5 seconds.#### Diagram Description:- The figure shows a pulley system with two blocks. - Block \(m_1 = m\) is on a horizontal surface experiencing friction, represented by the friction constant \(\mu\).- Block \(m_2 = 3m\) is hanging vertically.- The pulley is depicted with radius \(2R\).- A massless string connects both blocks over the pulley.### Solution:#### Part (a): Free Body Diagram and Newton's Equations1. **Free Body Diagram:** - For \(m_1\): - Tension (\(T\)) acts horizontally towards the pulley. - Friction (\(f = \mu m_1 g\)) acts in the opposite direction of motion. - Normal force (\(N\)) acts vertically upwards. - Gravity (\(m_1g\)) acts vertically downwards. - For \(m_2\): - Tension (\(T\)) acts upwards. - Gravity (\(3mg\)) acts downwards. - For the Pulley: - Tension forces (\(T\)) act tangentially at two points on the pulley.2. **Newton's Equations:** - For \(m_1\) (horizontal direction): \[ T - \mu m_1 g = m_1 a \] - For \(m_2\) (vertical direction): \[ 3mg - T = 3m a \] - For the pulley (rotational motion): \[ \tau = I \alpha \] where \(\tau\) is

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2) In figure, two blocks are connected by a massless string over a pulley of radius, 2R, and rotational inertia, I. When this system is released at rest, the blocks (masses: m₁ = m and m₂ = 3m) are moving with a constant acceleration. (The friction constant between my and the surface is equal to μ.) a) Draw the free body diagram and write the Newton's equations for both masses and pulley. b) Find the angular velocity of the pulley in 5 second. 1₂ m₂=3m m₁-m

2) In figure, two blocks are connected by a massless string over a pulley of radius, 2R, and rotational inertia, I. When this system is released at rest, the blocks (masses: m₁ = m and m₂ = 3m) are moving with a constant acceleration. (The friction constant between my and the surface is equal to μ.) a) Draw the free body diagram and write the Newton's equations for both masses and pulley. b) Find the angular velocity of the pulley in 5 second. 1₂ m₂=3m m₁-m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

### Example Problem on Pulley Systems

#### Problem Statement:
In the figure, two blocks are connected by a massless string over a pulley of radius \(2R\) and rotational inertia, \(I\). When this system is released from rest, the blocks (masses: \(m_1 = m\) and \(m_2 = 3m\)) are moving with a constant acceleration. (The friction constant between \(m_1\) and the surface is equal to \(\mu\)).

##### Question:
a) Draw the free body diagram and write the Newton’s equations for both masses and pulley.

b) Find the angular velocity of the pulley in 5 seconds.

#### Diagram Description:
- The figure shows a pulley system with two blocks.
- Block \(m_1 = m\) is on a horizontal surface experiencing friction, represented by the friction constant \(\mu\).
- Block \(m_2 = 3m\) is hanging vertically.
- The pulley is depicted with radius \(2R\).
- A massless string connects both blocks over the pulley.

### Solution:

#### Part (a): Free Body Diagram and Newton's Equations

1. **Free Body Diagram:**
- For \(m_1\):
- Tension (\(T\)) acts horizontally towards the pulley.
- Friction (\(f = \mu m_1 g\)) acts in the opposite direction of motion.
- Normal force (\(N\)) acts vertically upwards.
- Gravity (\(m_1g\)) acts vertically downwards.

- For \(m_2\):
- Tension (\(T\)) acts upwards.
- Gravity (\(3mg\)) acts downwards.

- For the Pulley:
- Tension forces (\(T\)) act tangentially at two points on the pulley.

2. **Newton's Equations:**
- For \(m_1\) (horizontal direction):
\[
T - \mu m_1 g = m_1 a
\]

- For \(m_2\) (vertical direction):
\[
3mg - T = 3m a
\]

- For the pulley (rotational motion):
\[
\tau = I \alpha
\]
where \(\tau\) is

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