the door of an almirah is 6 ft high, 1.5 ft wide and weights 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitude of the forces exerted by the hinges on the door are equal find this magnitude.

Explanation

To solve the problem, we need to find the magnitude of the forces exerted by the hinges on the door. Let's break down the solution step by step.

### Step 1: Understand the Setup
The door is 6 ft high, 1.5 ft wide, and weighs 8 kg. The hinges are located 1 ft from each end of the door. This means the distance between the two hinges is:
\[ \text{Distance between hinges} = 1.5 \, \text{ft} - 1 \, \text{ft} - 1 \, \text{ft} = 0.5 \, \text{ft} \]

### Step 2: Calculate the Weight of the Door
The weight of the door can be calculated using the formula:
\[ \text{Weight} = \text{mass} \times g \]
Where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \). First, we need to convert the mass from kg to Newtons:
\[ \text{Weight} = 8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N} \]

### Step 3: Set Up the Torque Equation
To find the forces exerted by the hinges, we can use the principle of moments (torque). The weight of the door acts at its center of gravity, which is at a distance of \( \frac{1.5}{2} = 0.75 \, \text{ft} \) from the left end. The distance from the left hinge to the center of gravity is:
\[ \text{Distance from left hinge to center of gravity} = 0.75 \, \text{ft} - 1 \, \text{ft} = -0.25 \, \text{ft} \]
(The negative sign indicates that the center of gravity is to the left of the left hinge.)

### Step 4: Calculate the Torque Due to the Weight
The torque due to the weight of the door about the left hinge is:
\[ \text{Torque} = \text{Weight} \times \text{Distance from hinge} \]
\[ \text{Torque} = 78.4 \, \text{N} \times 0.25 \, \text{ft} = 19.6 \, \text{N ft} \]

### Step 5: Set Up the Torque Balance Equation
Let \( F \) be the force exerted by each hinge. The torque due to the force exerted by the right hinge about the left hinge is:
\[ \text{Torque by right hinge} = F \times 0.5 \, \text{ft} \]

Setting the torques equal (since the door is in equilibrium):
\[ F \times 0.5 = 19.6 \]
\[ F = \frac{19.6}{0.5} = 39.2 \, \text{N} \]

### Step 6: Conclusion
The magnitude of the forces exerted by each hinge on the door is:
\[ F = 39.2 \, \text{N} \]