Question

. An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab’s speed is (a) increasing at a rate of 1.22 m/s2 and (b) decreasing at a rete of 1.22 m/s2?

Answer 1

Explanation:

It is given that, $F_{w}$ = 27.8 kN = $27.8 \times 10^{3} N$

Now, formula for the mass of elevator is as follows.

              $F_{w} = mg$

                m = $\frac{F_{w}}{g}$

                    = $\frac{27.8 \times 10^{3}}{9.8}$

                    = $2.84 \times 10^{3}$ kg

(a)   When acceleration is increasing at a rate of 1.22 $m/s^{2}$. Force according to the Newton's second law of motion is as follows.

               F = T - mg = ma

         $T - 27.8 \times 10^{3} = 2.84 \times 10^{3} \times 1.22$

            T = $31.27 \times 10^{3} N$

Hence, tension in the cable if the cab’s speed is increasing at a rate of 1.22 $m/s^{2}$ is $31.27 \times 10^{3} N$.

(b)   When acceleration is decreasing at a rate of 1.22 $m/s^{2}$. Force according to the Newton's second law of motion is as follows.

                      F = T - mg = ma

         $T - 27.8 \times 10^{3} = 2.84 \times 10^{3} \times (-1.22)$

            T = $24.34 \times 10^{3} N$

Hence, tension in the cable if the cab’s speed is decreasing at a rate of 1.22 $m/s^{2}$ is $24.34 \times 10^{3} N$.