A force of 3 Nt is applied to the left side of an object, one of 4 N is applied from the bottom, and a force of 7 N is applied from an angle of π/4 to the horizontal. What is the resultant of forces applied to the object?

Question

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

Accepted Answer

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 [latex]\mathbf{F}_{1}= 3 \mathbf{i} ; \mathbf{F}_{2} = 4 \mathbf{j}[/latex]; and [latex]\mathbf{F}_{3}=-(7 / \sqrt{2})(\mathbf{i}+\mathbf{j})[/latex]. This last vector follows from the fact that the vector [latex]-(1 / \sqrt{2})(\mathbf{i}+\mathbf{j})[/latex] is a unit vector pointing toward the origin making an angle of [latex]\pi / 4[/latex] with the x-axis (see Example 1.1.10). Then the resultant is given by

[latex]\mathbf{F}=\mathbf{F}_{1}+\mathbf{F}_{2}+\mathbf{F}_{3}=\left(3-\frac{7}{\sqrt{2}} ight) \mathbf{i}+\left(4-\frac{7}{\sqrt{2}} ight) \mathbf{j}[/latex]

The magnitude of [latex]\mathbf{F}[/latex] is

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

The direction [latex] heta[/latex] can be calculated by first finding the unit vector in the direction of [latex]\mathrm{F}[/latex] :

[latex]\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 heta) \mathbf{i}+(\sin heta) \mathbf{j}.\end{aligned}[/latex]

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

Question 3.3: A force of 3 Nt is applied to the left side of an object, on...

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