Question

While an elevator of mass 2325 kg moves upward, the tension in the cable is 38.8 kN. Assume the elevator is supported by a single cable. Forces exerted by the guide rails and air resistance are negligible.What is the acceleration of the elevator? If the acceleration is in upward direction, enter a positive value and if the acceleration is in downward direction, enter a negative value.

Answer 1

Given

m: Mass

m = 2325 kg

T: Tension

T = 38.8 kN

Procedure

First, let's make a free body diagram for the elevator. There are two interacting forces, the weight which we will symbolize with the letter W and the tension which we will symbolize with the letter T.

We will perform the summation of forces on the Y-axis, following Newton's second law.

$\begin{gathered} \Sigma Fy=ma \\ T-mg=ma \\ a=\frac{T-mg}{m} \\ a=\frac{38.8kN-2325\operatorname{kg}\cdot9.8m/s^2}{2325\operatorname{kg}} \\ a=6.88m/s^2 \end{gathered}$

The acceleration of the elevator would be 6.88 m/s^2 in the upward direction, so it would be positive.