Q44P - University Physics with Modern Physics

Question

A loaded elevator with very worn cables has a total mass of 2200 kg, and the cables can withstand a maximum tension of 28000 N. (a) Draw the free-body force diagram for the elevator. In terms of the forces on your diagram, what is the net force on the elevator? Apply Newton’s second law to the elevator and find the maximum upward acceleration for the elevator if the cables are not to break. (b) What would be the answer to part (a) if the elevator were on the moon, where $\mathbf{g}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{62}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

Step-by-Step Solution

Answer

(a) .   

The net force on the elevator is T-mg.

The maximum upward acceleration on the earth is $2.92 \mathrm{m}/{\mathrm{s}}^2$.

(b) The maximum upward acceleration on the moon is $11.1\mathrm{m}/{\mathrm{s}}^2$.

Step 1: Identification of the given data

The given data is listed below as:

 

• The total mass of the elevator is, m=2200 kg
• The maximum amount of tension the cable can withstand is, T=28000 N
• The acceleration due to gravity on the moon is, $g=1.62 \mathrm{m}/{\mathrm{s}}^2$

Step 2: Significance of the net force

The net force is referred to as the sum of the forces that acts on a particular object. The net force mainly represents the effect of the forces of the motion of a particle.

Step 3: (a) Determination of the free body diagram, the net force on the elevator, and the maximum upward acceleration

The free-body diagram of the elevator has been drawn below:

In the above diagram, the weight w of the cables and the elevator is acting in the downward direction that is the product of the mass m and the acceleration due to gravity g. The acceleration $a_y$ is acting in the upward direction and the tension T is also acting in the upward direction.

 

According to the above diagram, the equation of the net force can be expressed as:

 

$F=T-w$ 

 

Here,T is the tension and W is the weight of the elevator.

 

Substitute mg for w in the above equation.

 

$F=T-mg$

 

The equation of the upward acceleration can be expressed as:

 

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

 

Here, a is the maximum upward acceleration, T is the maximum tension,m is the total mass of the elevator and g is the acceleration due to gravity on earth.

 

Substitute all the values in the above equation.

$$\begin{gathered} a=\frac{28000\mathrm{N}-\left(2200kg\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200kg\right)} \\ =\frac{28000\mathrm{N}-\left(21560kg\cdot \mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200kg\right)} \\ =\frac{28000\mathrm{N}\times \left(\frac{1kg\cdot \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)-\left(21560kg\cdot \mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200kg\right)} \\ =\frac{28000kg\cdot \mathrm{m}/{\mathrm{s}}^2-\left(21560kg\cdot \mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200kg\right)} \end{gathered}$$ 

 

Hence, further as:

$$\begin{gathered} a=\frac{28000 kg.m/s^2-\left(21560kg.m/s^2\right)}{\left(2200 kg\right)} \\ =\frac{6400 kg.m/s^2}{\left(2200 kg\right)} \\ =2.92 m/s^2 \end{gathered}$$ 

 

Thus, the net force on the elevator is T-mg, and the maximum upward acceleration on the earth is $2.92\mathrm{m}/{\mathrm{s}}^2$.

Step 6: (b) Determination of the maximum upward acceleration in the moon

Substitute all the values in equation (i).

$$\begin{gathered} a=\frac{28000\mathrm{N}-\left(2200\mathrm{kg}\right)\left(1.62\mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200\mathrm{kg}\right)} \\ =\frac{28000\mathrm{N}-\left(3564\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200\mathrm{kg}\right)} \\ =\frac{28000\mathrm{N}\times \left(\frac{1\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)-\left(3564\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200\mathrm{kg}\right)} \\ =\frac{28000\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2-\left(3564\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200\mathrm{kg}\right)} \end{gathered}$$ 

 

 

Hence, further as:

$$\begin{gathered} \mathrm{a}=\frac{28000\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2-\left(3564 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)}{\left(2200 \mathrm{kg}\right)} \\ =\frac{24436 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{\left(2200 \mathrm{kg}\right)} \\ =11.1 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

Thus, the maximum upward acceleration on the moon is $11.1 \mathrm{m}/{\mathrm{s}}^2$.