Four blocks of the same mass m connected by cords are pulled by a force F on a smooth horizontal surface as shown in Determine the tensions `T_1 ,T_2` and `T_3` in the cords.

Options

A. `T_(1)=(1)/(4)F,T_(2)=(3)/(2)F,T_(3)=(1)/(4)F`
B. `T_(1)=(1)/(4)F,T_(2)=(1)/(2)F,T_(3)=(1)/(2)F`
C. `T_(1)=(3)/(4)F,T_(2)=(1)/(2)F,T_(3)=(1)/(4)F`
D. `T_(1)=(3)/(4)F,T_(2)=(1)/(2)F,T_(3)=(1)/(2)F`

Answer

To solve the problem of determining the tensions \( T_1, T_2, \) and \( T_3 \) in the cords connecting four blocks of mass \( m \) being pulled by a force \( F \) on a smooth horizontal surface, we will follow these steps:

### Step 1: Determine the acceleration of the system
The total mass of the system is \( 4m \) (since there are four blocks, each of mass \( m \)). According to Newton's second law, the net force acting on the system is equal to the total mass times the acceleration:

\[ F = (4m) \cdot a \]

From this, we can solve for the acceleration \( a \):

\[ a = \frac{F}{4m} \]

### Step 2: Analyze the forces on the last block
For the last block (Block 4), the only force acting on it is the tension \( T_3 \) in the cord connected to it. According to Newton's second law:

\[ T_3 = m \cdot a \]

Substituting the value of \( a \):

\[ T_3 = m \cdot \frac{F}{4m} = \frac{F}{4} \]

### Step 3: Analyze the forces on the third block
For the third block (Block 3), the forces acting on it are the tension \( T_2 \) pulling it forward and the tension \( T_3 \) pulling it backward. Thus, we can write:

\[ T_2 - T_3 = m \cdot a \]

Substituting the value of \( a \) and \( T_3 \):

\[ T_2 - \frac{F}{4} = m \cdot \frac{F}{4m} \]

This simplifies to:

\[ T_2 - \frac{F}{4} = \frac{F}{4} \]

Solving for \( T_2 \):

\[ T_2 = \frac{F}{4} + \frac{F}{4} = \frac{F}{2} \]

### Step 4: Analyze the forces on the second block
For the second block (Block 2), the forces acting on it are the tension \( T_1 \) pulling it forward and the tension \( T_2 \) pulling it backward. Thus, we can write:

\[ T_1 - T_2 = m \cdot a \]

Substituting the value of \( a \) and \( T_2 \):

\[ T_1 - \frac{F}{2} = m \cdot \frac{F}{4m} \]

This simplifies to:

\[ T_1 - \frac{F}{2} = \frac{F}{4} \]

Solving for \( T_1 \):

\[ T_1 = \frac{F}{2} + \frac{F}{4} = \frac{2F}{4} + \frac{F}{4} = \frac{3F}{4} \]

### Final Results
The tensions in the cords are:
- \( T_1 = \frac{3F}{4} \)
- \( T_2 = \frac{F}{2} \)
- \( T_3 = \frac{F}{4} \)