A man of mass 60 kg is in a lift and lift is accelerating upwards with acceleration `4 m//s^(2)`. Calculate effective weight of man in lift.

Answer

To calculate the effective weight of a man in a lift that is accelerating upwards, we can follow these steps:

### Step 1: Understand the forces acting on the man
When the lift accelerates upwards, two main forces act on the man:
- The gravitational force (weight) acting downwards, which is given by \( W = mg \), where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity.
- The normal force (effective weight) acting upwards, which we need to calculate.

### Step 2: Identify the values
Given:
- Mass of the man, \( m = 60 \, \text{kg} \)
- Acceleration of the lift, \( a = 4 \, \text{m/s}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (approximately)

### Step 3: Calculate the gravitational force
The gravitational force acting on the man is calculated using the formula:
\[ W = mg \]
Substituting the values:
\[ W = 60 \, \text{kg} \times 10 \, \text{m/s}^2 = 600 \, \text{N} \]

### Step 4: Calculate the effective weight
The effective weight (normal force \( N \)) when the lift is accelerating upwards can be calculated using the formula:
\[ N = mg + ma \]
Here, \( ma \) is the additional force due to the upward acceleration of the lift.

Substituting the values:
\[ N = mg + ma = 60 \, \text{kg} \times 10 \, \text{m/s}^2 + 60 \, \text{kg} \times 4 \, \text{m/s}^2 \]
\[ N = 600 \, \text{N} + 240 \, \text{N} = 840 \, \text{N} \]

### Step 5: Conclusion
Thus, the effective weight of the man in the lift is:
\[ \text{Effective Weight} = 840 \, \text{N} \]