Chapter 4, Problem 87QAP

To determine

(a)

The reading on the scale in an elevator when elevator accelerates at $\text{3}\text{.50}{\text{m/s}}^2$ downward

Answer to Problem 87QAP

The reading on the scale is $\text{=189}\text{N}$

Explanation of Solution

Givendata:

  $\begin{array}{l}{m}_{\text{dog}}=30.0\text{kg}\\ a_y=\text{3}\text{.50}{\text{m/s}}^2\end{array}$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

Calculation:

We are interested in calculating the reading on the scale for accelerationof the elevator, at $\text{3}\text{.50}{\text{m/s}}^2$ downwardwhich is equal to the acceleration of the scale and dog since they travel

together as one object.

The reading on the scale is equal in magnitude to the normalforce acting on the dog. Gravity also acts on the dog.

We can use Newton's second law tocalculate N as a function of the acceleration.

We'll define up to be positive y throughout ourcalculation.

Free-body diagram of the dog:

  

  $\begin{array}{l}\sum F_{\text{ext,y}}=n-w_{\text{dog}}=m_{\text{dog}}{a}_y\\ =>n=w_{\text{dog}}+m_{\text{dog}}{a}_y=m_{\text{dog}}g+m_{\text{dog}}{a}_y=m_{\text{dog}}(g+a_y)\end{array}$

  $=>n=m_{\text{dog}}(g+a_y)=(30.0\text{kg})\left(\left(\text{9}\text{.80}{\text{m/s}}^2\right)-\left(\text{3}\text{.50}{\text{m/s}}^2\right)\right)=189\text{N}$

Conclusion:

Thus, we have the reading on the scale in an elevator when elevator accelerates at $\text{3}\text{.50}{\text{m/s}}^2$ downward is $\text{=189}\text{N}$.

To determine

(b)

The reading on the scale in an elevator when elevator cruises down at a steady speed.

Answer to Problem 87QAP

The reading on the scale is $=294\text{N}$

Explanation of Solution

Given data:

  $m_{\text{dog}}=30.0\text{kg}$

  $a_y=\text{0}{\text{m/s}}^2$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

Calculation:

We are interested in calculating the reading on the scale for elevator cruises down at a steady speed., which is equal to the acceleration of the scale and dog since they traveltogether as one object.

The reading on the scale is equal in magnitude to the normal force acting on the dog. Gravity also acts on the dog.

We can use Newton's second law to calculate N as a function of the acceleration.

We'll define up to be positive y throughout our calculation.

Free-body diagram of the dog:

  

  $a_y=\text{0}{\text{m/s}}^2$ because of steady speed.

  $\begin{array}{l}\sum F_{\text{ext,y}}=n-w_{\text{dog}}=m_{\text{dog}}{a}_y\\ =>n=w_{\text{dog}}+m_{\text{dog}}{a}_y=m_{\text{dog}}g+m_{\text{dog}}{a}_y=m_{\text{dog}}(g+a_y)\end{array}$

  $=>n=m_{\text{dog}}(g+a_y)=(30.0\text{kg})\left(\left(\text{9}\text{.80}{\text{m/s}}^2\right)+\left(0\right)\right)=294\text{N}$

Conclusion:

Thus, we have the reading on the scale in an elevator $=294\text{N}$, when elevator cruises down at a steady speed.

To determine

(c)

The reading on the scale in an elevator when elevator accelerates at $4.00{\text{m/s}}^2$ upward

Answer to Problem 87QAP

The reading on the scale is $=414\text{N}$

Explanation of Solution

Given data:

  $m_{\text{dog}}=30.0\text{kg}$

  $a_y=4.00{\text{m/s}}^2$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

Calculation:

We are interested in calculating the reading on the scale for acceleration as $4.00{\text{m/s}}^2$ upward

of the elevator, which are equal to the accelerations of the scale and dog since they travel

together as one object.

The reading on the scale is equal in magnitude to the normal force acting on the dog. Gravity also acts on the dog.

We can use Newton's second law to calculate N as a function of the acceleration.

We'll define up to be positive y throughout our calculation.

Free-body diagram of the dog:

  

  $\begin{array}{l}\sum F_{\text{ext,y}}=n-w_{\text{dog}}=m_{\text{dog}}{a}_y\\ =>n=w_{\text{dog}}+m_{\text{dog}}{a}_y=m_{\text{dog}}g+m_{\text{dog}}{a}_y=m_{\text{dog}}(g+a_y)\end{array}$

  $=>n=m_{\text{dog}}(g+a_y)=(30.0\text{kg})\left(\left(\text{9}\text{.80}{\text{m/s}}^2\right)+\left(\text{4}\text{.00}{\text{m/s}}^2\right)\right)=414\text{N}$

Conclusion:

The reading on the scale in an elevator when elevator accelerates at $4.00{\text{m/s}}^2$ upward is $=414\text{N}$