A man whose mass is 80kg stands on a spring weighing machine inside an elevator.
What is the reading of the weighing machine when:
the elevator is coming to rest with a retardation of 4.0ms
a is in units of m/s^2, not ms
F = m(g+a) = 80(9.81 + 4.0)
is it stopping at the top or stopping at the bottom ?
at bottom stop, g+a
at top stop, g-a
It's actually deceleration not acceleration
To find the reading of the weighing machine, we need to consider the forces acting on the man inside the elevator.
When the elevator is coming to rest with a retardation of 4.0 m/s^2, there are two forces acting on the man: his weight (mg) and the pseudo force (ma_p). The pseudo force is equal to the mass of the man (m) multiplied by the acceleration of the elevator (a).
Since the elevator is coming to rest, the acceleration of the elevator is equal to the acceleration due to gravity (g), but in the opposite direction. So the pseudo force is equal to the mass of the man (m) multiplied by the acceleration due to gravity (g).
The total force acting on the man can be calculated using Newton's second law (F = ma), where the mass (m) is the mass of the man and the acceleration (a) is the retardation of the elevator (4.0 m/s^2).
F_total = m * a
F_total = 80 kg * (-4.0 m/s^2)
F_total = -320 N
The negative sign indicates that the forces are acting in the opposite direction. Now, the reading on the weighing machine will be equal to the total force acting on the man.
Reading of the weighing machine = F_total
Reading of the weighing machine = -320 N
Therefore, the reading of the weighing machine when the elevator is coming to rest with a retardation of 4.0 m/s^2 is -320 N.