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Chapter 1

Q. 3.3

A force of 3 \mathrm{~N}^{t}is applied to the left side of an object, one of 4 \mathrm{~N} is applied from the bottom, and a force of 7 \mathrm{~N} is applied from an angle of \pi/4 to the horizontal. What is the resultant of forces applied to the object?

Step-by-Step

Verified Solution

The forces are indicated in Figure 3 . We write each force as a magnitude times a unit vector in the indicated direction. For convenience we can think of the center of the object as being at the origin. Then \mathbf{F}_{1}= 3 \mathbf{i} ; \mathbf{F}_{2} = 4 \mathbf{j}; and \mathbf{F}_{3}=-(7 / \sqrt{2})(\mathbf{i}+\mathbf{j}). This last vector follows from the fact that the vector -(1 / \sqrt{2})(\mathbf{i}+\mathbf{j}) is a unit vector pointing toward the origin making an angle of \pi / 4 with the x-axis (see Example 1.1.10). Then the resultant is given by

\mathbf{F}=\mathbf{F}_{1}+\mathbf{F}_{2}+\mathbf{F}_{3}=\left(3-\frac{7}{\sqrt{2}}\right) \mathbf{i}+\left(4-\frac{7}{\sqrt{2}}\right) \mathbf{j}

The magnitude of \mathbf{F} is

|\mathbf{F}|=\sqrt{\left(3-\frac{7}{\sqrt{2}}\right)^{2}+\left(4-\frac{7}{\sqrt{2}}\right)^{2}}=\sqrt{74-\frac{98}{\sqrt{2}}} \approx 2.17 \mathrm{~N}

The direction \theta can be calculated by first finding the unit vector in the direction of \mathrm{F} :

\begin{aligned}\frac{\mathbf{F}}{|\mathbf{F}|} &=\frac{3-(7 / \sqrt{2})}{\sqrt{74-(98 / \sqrt{2})}} \mathbf{i}+\frac{4-(7 / \sqrt{2})}{\sqrt{74-(98 / \sqrt{2})}} \mathbf{j} \\&=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}.\end{aligned}

Then
\cos \theta=\frac{3-(7 / \sqrt{2})}{\sqrt{74-(98 / \sqrt{2})}} \approx-0.8990 \quad and \quad \sin \theta=\frac{4-(7 / \sqrt{2})}{\sqrt{74-(98 / \sqrt{2})}} \approx-0.4379
This means that \theta is in the third quadrant, and \theta \approx 3.5949 \approx 206^{\circ} (or -154^{\circ} ). This is illustrated in Figure 4 .