Lift at the first instant

Problem

Lift at the first instant. A lift is at rest. At time $t=0$ it starts to rise. Discuss in five lines how the reading of the scale on which a person stands varies, over the three intervals: initial acceleration, constant-velocity phase, and arrival deceleration.

Solution

The scale reads the normal force $N$ that it exerts on the person’s feet. From the second law in the vertical direction (positive upward), $N - mg = m a$, i.e. $N = m(g + a)$: the reading depends on the sign of the acceleration.

• Initial acceleration (starts and rises). To set off upward the lift accelerates upward, $a > 0$: so $N = m(g+a) > mg$. The scale reads more than the true weight.
• Constant-velocity phase. With $a = 0$ the resultant is zero: $N = mg$. The scale reads exactly the true weight, as if the lift were stationary (uniform motion is indistinguishable from rest).
• Arrival deceleration. To stop at the top the rising lift slows down: the acceleration points downward, $a < 0$, so $N = m(g+a) < mg$. The scale reads less than the true weight.

$\ev{N > mg \;\text{(start)} \quad N = mg \;\text{(constant vel.)} \quad N < mg \;\text{(arrival)}}$

Links

Topics: Dynamics Concepts: Newton’s second law · Normal force Objects: Lift · Scale