Question

The cable supporting a 2125-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it give the elevator without breaking?

Short answer

The obtained value of the maximum upward acceleration without breaking is $a=0.425\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Step 1. Determine the forces acting on the elevator

In this problem, first, determine the number of forces acting on the elevator. Consider the upward direction to be positive, and use Newton’s second law to calculate the maximum upward acceleration of the elevator.

Given data:

The mass of the elevator is $m=2125\mathrm{kg}$.

The normal force is $N=21750\mathrm{N}$.

The free body diagram of the elevator is as follows:

The relation to calculate the vertical forces is given by:

$\begin{array}{c}\Sigma F_v=ma\\ N-W=ma\\ N-mg=ma\end{array}$

Here, a is the acceleration of the elevator, g is the gravitational acceleration, W is the weight, and N is the normal force.

Step 2. Determine the acceleration of the elevator

On plugging the values in the above relation, you get:

$\begin{array}{c}\left(21750\mathrm{N}\right)-\left(2125\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)=\left(2125\mathrm{kg}\right)a\\ a=0.425\mathrm{m}/{\mathrm{s}}^2\end{array}$

Here, g is the gravitational acceleration.

Thus, $a=0.425\mathrm{m}/{\mathrm{s}}^2$is the maximum upward acceleration of the elevator.