Chapter 1, Problem 77P

(a)

To determine

To find: The two forces in unit vectors.

Answer to Problem 77P

The force applied by person P is $\left(35.36\text{lb}\right)\widehat{i}+\left(35.36\text{lb}\right)\widehat{j}$ and the force applied by Person Jis $-\left(53.24\text{lb}\right)\widehat{i}-\left(37.28\text{lb}\right)\widehat{j}$ .

Explanation of Solution

Given:

The force applied by person P is $50\text{lb}$ in northeast direction.

The force applied by person J is $65\text{lb}$ , $35°$ south of due west.

Formula used:

Write the expression for the horizontal component of force.

  $F_x=F\cos \theta$

Write the expression for the vertical component of force.

  $F_y=F\sin \theta$

Write the expression of force in terms of a unit vector.

  $F_R=F_x\widehat{i}+F_y\widehat{j}$ …… (1)

Substitute $F\cos \theta$ for $F_x$ and $F\sin \theta$ for $F_y$ in equation (1).

  $F_R=F\cos \theta \widehat{i}+F\sin \theta \widehat{j}$ …… (2)

Here, F is the magnitude of the force, $F_R$ is the force in vector form and $\theta$ is the angle.

Calculation:

For person P:

The direction of force is between the positive x-axis and the positive y-axis.

Substitute $50\text{lb}$ for F and $45°$ for $\theta$ in equation (2).

  $\begin{array}{c}{F}_1=\left(50\text{lb}\right)\cos 45°\widehat{i}+\left(50\text{lb}\right)\sin 45°\widehat{j}\\ =\left(35.36\text{lb}\right)\widehat{i}+\left(35.36\text{lb}\right)\widehat{j}\end{array}$

For Person J:

The direction of force is between the negative x-axis and the positive y-axis.

Substitute $65\text{lb}$ for F and $35°$ for $\theta$ in equation (2).

  $\begin{array}{c}{F}_2=\left(65\text{lb}\right)\cos 35°\widehat{i}-\left(65\text{lb}\right)\sin 35°\widehat{j}\\ =-\left(53.24\text{lb}\right)\widehat{i}-\left(37.28\text{lb}\right)\widehat{j}\end{array}$

Conclusion:

Thus, the force applied by person P is $\left(35.36\text{lb}\right)\widehat{i}+\left(35.36\text{lb}\right)\widehat{j}$ and the force applied by Person Jis $-\left(53.24\text{lb}\right)\widehat{i}-\left(37.28\text{lb}\right)\widehat{j}$ .

(b)

To determine

To write: The force applied by person C.

Answer to Problem 77P

The force applied by person C is $\left(17.88\text{lb}\right)\widehat{i}+\left(1.92\text{lb}\right)\widehat{j}$ ; magnitude of force $17.98\text{lb}$ at an angle of $6.13°$ north of due east.

Explanation of Solution

Given:

The resultant of three forces equal to zero.

Formula used:

The force applied by person C should be equal and opposite to the resultant sum of forces applied by person P and person J.

Write the expression for the force applied by person C.

  $F_3=-\left(F_1+F_2\right)$ …… (3)

Here, $F_3$ is the force applied by person C.

Calculation:

Substitute $\left(35.36\text{lb}\right)\widehat{i}+\left(35.36\text{lb}\right)\widehat{j}$ for $F_1$ and $-\left(53.24\text{lb}\right)\widehat{i}-\left(37.28\text{lb}\right)\widehat{j}$ for $F_2$ in equation (3).

  $\begin{array}{c}{F}_3=-\left[\left(35.36\text{lb}\right)\widehat{i}+\left(35.36\text{lb}\right)\widehat{j}-\left(53.24\text{lb}\right)\widehat{i}-\left(37.28\text{lb}\right)\widehat{j}\right]\\ =\left(17.88\text{lb}\right)\widehat{i}+\left(1.92\text{lb}\right)\widehat{j}\end{array}$

Calculate the magnitude of force $F_3$ .

  $\begin{array}{c}\left|F_3\right|=\sqrt{{\left(17.88\text{lb}\right)}^2+{\left(1.92\text{lb}\right)}^2}\\ =17.98\text{lb}\end{array}$

Calculate the angle of force.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.92\text{lb}}{17.88\text{lb}}\right)\\ =6.13°\end{array}$

Conclusion:

Thus, the force applied by person Cis $\left(17.88\text{lb}\right)\widehat{i}+\left(1.92\text{lb}\right)\widehat{j}$ ; the magnitude of force $17.98\text{lb}$ at an angle of $6.13°$ north of due east.