Chapter 9, Problem 73P

A high-speed lifting mechanism supports an 800.-kg object with a steel cable that is 25.0 m long and 4.00 cm² in cross-sectional area. (a) Determine the elongation of the cable. (b) By what additional amount does the cable increase in length if the object is accelerated upward at a rate of 3.0 m/s²? (c) What is the greatest mass that can be accelerated upward at 3.0 m/s² if the stress in the fable is not to exceed the elastic limit of the cable, which is 2.2 × 108 Pa?

To determine

The elongation of the steel cable.

Answer to Problem 73P

The elongation of the steel cable is $2.5\text{mm}$.

Explanation of Solution

The definition of Young’s modulus is used for the elongation of the steel cable that is $Y=\left(F/A\right)/\left(\Delta L/L_0\right)\to \Delta L=\left(mg{L}_0\right)/\left(AY\right)$.

Given info: The mass of the object is $800\text{kg}$, acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, cross-sectional area of the steel cable is $4.00{\text{cm}}^{\text{2}}$, length of the cable is $25.0\text{m}$, and the Young’s modulus of the cable is $20.0\times {10}^{10}\text{Pa}$.

The formula for the elongation of the steel cable is,

$\Delta L=\frac{mg{L}_0}{AY}$

• $m$ is mass of the object.
• $g$ is acceleration due to gravity.
• $L_0$ is length of the cable.
• $A$ is area of the steel cable.
• $Y$ is Young’s modulus of the cable.

Substitute $800\text{kg}$ for $m$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $4.00{\text{cm}}^{\text{2}}$ for $A$, $25.0\text{m}$ for $L_0$, and $20.0\times {10}^{10}\text{Pa}$ for $Y$ to find $\Delta L$.

$\begin{array}{c}\Delta L=\frac{\left(800\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\left(25.0\text{m}\right)}{\left(4.00{\text{cm}}^{\text{2}}\right){\left(\frac{0.01\text{m}}{1\text{cm}}\right)}^2\left(20.0\times {10}^{10}\text{Pa}\right)}\\ =2.5\text{mm}\end{array}$

Thus, the elongation of the steel cable is $2.5\text{mm}$.

Conclusion:

Therefore, the elongation of the steel cable is $2.5\text{mm}$.

To determine

The additional the cable increases in length if the object is accelerated upward.

Answer to Problem 73P

The additional the cable increases in length if the object is accelerated upward is $0.70\text{mm}$.

Explanation of Solution

The sum of vertical forces on the object would be $F-mg=m{a}_y\Rightarrow F=m\left(g+a_y\right)$ and the change in length of the cable is $\Delta L=\left[m\left(g+a_y\right)L_0\right]/\left(AY\right)$.

Given info: The mass of the object is $800\text{kg}$, acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, the vertical acceleration of the object is $3.0{\text{m/s}}^{\text{2}}$, cross-sectional area of the steel cable is $4.00{\text{cm}}^{\text{2}}$, length of the cable is $25.0\text{m}$, and the Young’s modulus of the cable is $20.0\times {10}^{10}\text{Pa}$.

The formula for the elongation of the steel cable is,

$\Delta L=\frac{m\left(g+a_y\right)L_0}{AY}$

• $m$ is mass of the object.
• $g$ is acceleration due to gravity.
• $L_0$ is length of the cable.
• $A$ is area of the steel cable.
• $Y$ is Young’s modulus of the cable.
• $a_y$ is vertical acceleration of the cable.

Substitute $800\text{kg}$ for $m$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $3.0{\text{m/s}}^{\text{2}}$ for $a_y$, $4.00{\text{cm}}^{\text{2}}$ for $A$, $25.0\text{m}$ for $L_0$, and $20.0\times {10}^{10}\text{Pa}$ for $Y$ to find $\Delta L$.

$\begin{array}{c}\Delta L=\frac{\left(800\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}+3.0{\text{m/s}}^{\text{2}}\right)\left(25.0\text{m}\right)}{\left(4.00{\text{cm}}^{\text{2}}\right){\left(\frac{0.01\text{m}}{1\text{cm}}\right)}^2\left(20.0\times {10}^{10}\text{Pa}\right)}\\ =3.2\text{mm}\end{array}$

Thus, after the upward acceleration of the object, the increase elongation of the cable is $\Delta L-{\left(\Delta L\right)}_{\text{initial}}=3.2\text{mm}-2.5\text{mm}=0.70\text{mm}$.

Conclusion:

Therefore, the additional the cable increases in length if the object is accelerated upward is $0.70\text{mm}$.

To determine

The greatest mass that can be accelerated upward if the stress in the cable is not exceeding the classical limit of the cable.

Answer to Problem 73P

The greatest mass that can be accelerated upward if the stress in the cable is not exceeding the classical limit of the cable is $6.9\times {10}^3\text{kg}$.

Explanation of Solution

Given info: Given info: The acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, the vertical acceleration of the object is $3.0{\text{m/s}}^{\text{2}}$, and the maximum stress in the cable is $8.8\times {10}^4\text{N}$.

The formula for the mass of the maximum load is,

$m_{\max }=\frac{F_{\max }}{g+a_y}$

• $F_{\max }$ is the maximum stress in the cable.
• $g$ is acceleration due to gravity.
• $a_y$ is the vertical acceleration of the load.

Substitute $8.8\times {10}^4\text{N}$ for $F_{\max }$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, and $3.0{\text{m/s}}^{\text{2}}$ for $a_y$ to find $m_{\max }$.

$\begin{array}{c}{m}_{\max }=\frac{8.8\times {10}^4\text{N}}{9.80{\text{m/s}}^{\text{2}}+3.0{\text{m/s}}^{\text{2}}}\\ =6.9\times {10}^3\text{kg}\end{array}$

Thus, the greatest mass that can be accelerated upward if the stress in the cable is not exceeding the classical limit of the cable is $6.9\times {10}^3\text{kg}$.

Conclusion:

Therefore, the greatest mass that can be accelerated upward if the stress in the cable is not exceeding the classical limit of the cable is $6.9\times {10}^3\text{kg}$.