Question
A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N.
a) As the elevator moves up, the scale reading increases to 935 N, then decreases back to 836 N. Find the acceleration of the elevator.
b) As the elevator approaches the 74th floor, the scale reading drops as low as 782 N. What is the acceleration of the elevator?
Solution
We need to start by analyzing the forces involved in each scenario:
Part (a):
When the elevator is at rest, the reading on the scale is 836 N. This is the normal force N acting on the student, which is equal to the gravitational force mg:
Hence, N = mg
Since N = 836 N, we will now have
836 = mg
To find the mass (m), we need to make m subject of the formula from the above equation. Thus:
m = 836/g
and g = 9.8 m/s2
Thus, m = 836/9.8 = 85.31 kg
Next, as the elevator accelerates upward, the scale reading increases to 935 N. The new normal force N′ can be expressed as:
N′ = m (g + a)
Where a is the acceleration of the elevator. Inserting in the values:
935 N = 85.31 kg × (9.8 m/s2 + a)
Therefore, solving for a, 935N = 85.31 kg × (9.8m/s2 + a)
a = 1.17 m/s2
Therefore, the acceleration of the elevator when it is moving up is approximately 1.17 m/s2.
Part (b):
As the elevator approaches the 74th floor, the scale reading drops to 782 N. This time, the normal force N′′can be expressed as:
N′′ = m (g − a)
Where aa is now the deceleration of the elevator. When we need plug in the values:
782 N = 85.31 kg × (9.8 m/s2 − a)
By making a subject of the formula and solving the problem, we will now end up with:
a = 53.04/85.31 = 0.62 m/s2
Therefore, the acceleration of the elevator as it decelerates when approaching the 74th floor is approximately 0.62 m/s2 downwards.