Chapter 4, Problem 93QAP

To determine

(a)

The tension in the rope.

Answer to Problem 93QAP

The tension in the rope is $644\text{N}$

Explanation of Solution

Givendata:

Distance, $\Delta y=2.00\text{m}$

Time, $t=10.0\text{s}$

Mass of Sue, $m_{\text{Sue}}=66.0\text{kg}$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

  $v=ut+\frac{1}{2}a{t}^2$

Calculation:

  

We'll use two different but related coordinate systems for the two people.

For Sue, positive y will point upward.

For Paul, the axes will be parallel and perpendicular to the inclined plane, where up the ramp and out of the ramp are positive.

Tension from the rope pulling up and gravity pulling down are the only forces acting on

Sue.

Assuming her acceleration is constant, we can use the constant acceleration equations

and Newton's second law to calculate the magnitude of the tension.

Sue's acceleration:

  $\begin{array}{l}y=y_0+v_0{}_{y}t+\frac{1}{2}{a}_y{t}^2=y_0+0+\frac{1}{2}{a}_y{t}^2\\ =>a_y=\frac{2(\Delta y)}{t^2}=\frac{2(-2.00\text{m)}}{{(10.0\text{s)}}^2}=-0.0400{\text{m/s}}^{\text{2}}\end{array}$

Free-body diagram of Sue:

  

Newton's second law for Sue:

  $\begin{array}{l}\sum F_{\text{ext,y}}=T-w_{\text{Sue}}=T-m_{\text{Sue}}g=m_{\text{Sue}}{a}_y\\ T=m_{\text{Sue}}(g+a_y)=(66.0\text{kg})\left(\left(9.80{\text{m/s}}^{\text{2}}\right)+\left(-0.0400{\text{m/s}}^{\text{2}}\right)\right)=644\text{N}\end{array}$

Conclusion:

Thus, from theNewton's second law for Sue we have the tension in the rope joining them as $644\text{N}$

To determine

(b)

Mass of Paul

Answer to Problem 93QAP

Mass of Paul is $92.4\text{kg}$

Explanation of Solution

Given data:

Distance, $\Delta y=2.00\text{m}$

Time, $t=10.0\text{s}$

Mass of Sue, $m_{\text{Sue}}=66.0\text{kg}$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

  $v=ut+\frac{1}{2}a{t}^2$

Calculation:

  

We'll use two different but relatedcoordinate systems for the two people.

For Sue,positive y will point upward.

For Paul, the axes willbe parallel and perpendicular to the inclined plane,where up the ramp and out of the ramp are positive.

Tension from the rope pulling up and gravity pulling down are the only forces acting on

Sue.

Assuming her acceleration is constant, we can use the constant acceleration equations

and Newton's second law to calculate the magnitude of the tension.

Since Paul and Sue aretethered to one another, the magnitudes of their accelerations are equal.

The tension in therope and gravity are the only forces acting on Paul that have components that are parallel tothe face of the glacier. We can then solve the parallel component of Newton's second law forPaul's mass.

Sue's acceleration:

  $\begin{array}{l}y=y_0+v_0{}_{y}t+\frac{1}{2}{a}_y{t}^2=y_0+0+\frac{1}{2}{a}_y{t}^2\\ =>a_y=\frac{2(\Delta y)}{t^2}=\frac{2(-2.00\text{m)}}{{(10.0\text{s)}}^2}=-0.0400{\text{m/s}}^{\text{2}}\end{array}$

Free-body diagram of Sue:

  

Newton's second law for Sue:

  $\begin{array}{l}\sum F_{\text{ext,y}}=T-w_{\text{Sue}}=T-m_{\text{Sue}}g=m_{\text{Sue}}{a}_y\\ T=m_{\text{Sue}}(g+a_y)=(66.0\text{kg})\left(\left(9.80{\text{m/s}}^{\text{2}}\right)+\left(-0.0400{\text{m/s}}^{\text{2}}\right)\right)=644\text{N}\end{array}$

Free-body diagram of Paul:

  

Newton's second law for Paul:

  $\begin{array}{l}\sum F_{\text{ext,parallel}}=T-w_{\text{Paul}}=T-m_{\text{Paul}}g\sin ({45.0}^{\circ })=m_{\text{Paul}}{a}_{\text{parallel}}\\ =>m_{\text{Paul}}=\frac{T}{a_{\text{parallel}}+g\sin ({45.0}^{\circ })}=\frac{644\text{N}}{\left(0.0400{\text{m/s}}^{\text{2}}\right)+\left(9.80{\text{m/s}}^{\text{2}}\right)\sin ({45.0}^{\circ })}=92.4\text{kg}\end{array}$

Conclusion:

Thus, by Newton's second law for Paul mass of Paul is $92.4\text{kg}$