Chapter 4.2, Problem 21PP

(a)

To determine

The force when elevator moves upward at constant speed.

Answer to Problem 21PP

  $735\text{ N}$

Explanation of Solution

Formula used:

  $F=mg$

Calculation:

At constant speed, so $a=0{\text{ m/s}}^2$ ,

  $F_{\text{net}}=0\text{ N, }{F}_{\text{scale}}=F_{\text{net}}$

  $\begin{array}{l}F=mg\\ F=\left(75\times 9.8\right)\\ F=735\text{ N}\end{array}$

Conclusion:

The elevator moves upward at constant speed with $735\text{ N}$ force.

(b)

To determine

The force when it slows down at $2{\text{ m/s}}^2$ while moving downward.

Answer to Problem 21PP

  $885\text{ N}$

Explanation of Solution

Given:

  $a=2{\text{ m/s}}^2$

Formula used:

  $F_{\text{scale}}=F_{\text{net}}+F_{\text{g}}$

Calculation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=\left(75\times 2\right)+\left(75\times 9.8\right)\\ F_{\text{scale}}=150+735\\ F_{\text{scale}}=885\text{ N}\end{array}$

Conclusion:

The force acting is $885\text{ N}$ when it slows down at $2{\text{ m/s}}^2$ .

(c)

To determine

The force when it speed up at $2{\text{ m/s}}^2$ while moving downward.

Answer to Problem 21PP

  $585\text{ N}$

Explanation of Solution

Given:

  $a=-2{\text{ m/s}}^2$

Formula used:

  $F_{\text{scale}}=F_{\text{net}}+F_{\text{g}}$

Calculation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=\left(759\times \left(-2\right)\right)+\left(75\times 9.8\right)\\ F_{\text{scale}}=-150+735\\ F_{\text{scale}}=585\text{ N}\end{array}$

Conclusion:

The force acting is $585\text{ N}$ .

(d)

To determine

The force when it move down at constant speed.

Answer to Problem 21PP

  $735\text{ N}$

Explanation of Solution

Given:

  $a=0{\text{ m/s}}^2$

Formula used:

  $F_{\text{scale}}=F_{\text{net}}+F_{\text{g}}$

Calculation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=\left(759\times \left(0\right)\right)+\left(75\times 9.8\right)\\ F_{\text{scale}}=735\text{ N}\end{array}$

Conclusion:

The force acting is $735\text{ N}$ .

(e)

To determine

The direction of force as the elevator slows to stop as it is moving down.

Answer to Problem 21PP

Upwards

Explanation of Solution

If acceleration is acting on a body is upward then net force is also upward.

Conclusion:

The direction of force in in upward direction.