Question

After coming to a stop, the elevator then accelerates at 3.0m/s? as it begins to move downward.Calculate the tension in the cable.

Answer 1

We know that

• The acceleration is 3.0 m/s^2.

Using Newton's Second Law, we have

$mg-T=ma$

The two forces involved are the weight of the elevator and the tension force. Let's solve got T.

$\begin{gathered} mg-ma=T \\ T=m(g-a) \end{gathered}$

The mass of the elevator is m = 5000kg, g = 9.8 m/s^2, and a = 3 m/s^2.

$\begin{gathered} T=5000\operatorname{kg}(9.8(\frac{m}{s^2})-3(\frac{m}{s^2})) \\ T=5000\operatorname{kg}\cdot6.8(\frac{m}{s^2}) \\ T=5000\operatorname{kg}\cdot7(\frac{m}{s^2}) \\ T=35,000N \end{gathered}$

Therefore, the tension in the cable is 35,000 N.