Understanding Forces in an Accelerating Lift

When a passenger is inside a lift (elevator), there are primarily two forces acting on them:

• Gravity: The force exerted by the Earth pulling the passenger downwards. This force is equal to the passenger's mass multiplied by the acceleration due to gravity ($mg$).
• Normal Force: The force exerted by the floor of the lift pushing the passenger upwards. This force is often referred to as the apparent weight of the passenger.

According to Newton's second law of motion, the net force acting on an object is equal to its mass multiplied by its acceleration ($\Sigma F = ma$). In the case of the passenger in the lift, the net force is the vector sum of the normal force and the gravitational force.

Applying Newton's Second Law to a Downward Accelerating Lift

Let's consider the forces and acceleration. We can define the upward direction as positive and the downward direction as negative. The forces are:

• Normal force ($N$): Acting upwards (positive).
• Gravitational force ($mg$): Acting downwards (negative).

The net force ($\Sigma F$) acting on the passenger is $N - mg$.

The lift is accelerating downward at $0.81 \, m/s^2$. If we take upward as positive, the acceleration of the passenger ($a$) is $-0.81 \, m/s^2$.

Applying Newton's second law ($\Sigma F = ma$):

$\qquad N - mg = m(-a)$

We can rearrange this equation to solve for the normal force ($N$), which is the force exerted by the floor on the passenger:

$\qquad N = mg - ma$

$\qquad N = m(g - a)$

Calculating the Force Exerted by the Floor

We are given the following values:

• Mass of the passenger, $m = 80 \, kg$.
• Downward acceleration of the lift, $a = 0.81 \, m/s^2$.

We need to use the acceleration due to gravity, $g$. A commonly used value is $g = 9.8 \, m/s^2$. However, to match the given options precisely, it is likely that a value of $g = 9.81 \, m/s^2$ was used.

Using $g = 9.81 \, m/s^2$, substitute the values into the formula for $N$:

$\qquad N = m(g - a)$

$\qquad N = 80 \, kg \times (9.81 \, m/s^2 - 0.81 \, m/s^2)$

First, calculate the difference in accelerations:

$\qquad 9.81 \, m/s^2 - 0.81 \, m/s^2 = 9.00 \, m/s^2$

Now, multiply by the mass:

$\qquad N = 80 \, kg \times 9.00 \, m/s^2$

$\qquad N = 720 \, N$

The force exerted by the floor of the lift on the passenger is 720 N when the lift is accelerating downward at $0.81 \, m/s^2$. This force is less than the passenger's actual weight ($80 \, kg \times 9.81 \, m/s^2 \approx 785 \, N$), which is expected when the lift is accelerating downwards.