Question

Four forces are exerted on the object shown in FIGURE P3.44. (Forces are measured in newtons, abbreviated N.) The net force on the object is${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3+{\overrightarrow{F}}_4=4.0\widehat{ı}\text{N}$ . What are (a)${\overrightarrow{F}}_3$ and (b) ${\overrightarrow{F}}_4$? Give your answers in component form.

Short answer

The forces${\text{F}}_3$ and ${\text{F}}_4$are ${\overrightarrow{F}}_3=\left(2.9\text{\hspace{0.05em}N}\right)\widehat{j}$ and${\overrightarrow{F}}_4=\left(1.64\text{\hspace{0.05em}N}\right)\widehat{i}$

Step-by-step solution

Step 1. Given information

The net force

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3+{\overrightarrow{F}}_4=4\widehat{i}$

Step 2. Explanation

If a vector F is inclined by an angle $\theta$with respect to the x- axis, and the x-component is $\text{F}\cos \theta$and its y - component is $\text{F}\sin \theta$.

Resolving$F_2$ into two components, along x-axis and along the y-axis as shown in the following figure:

$\begin{array}{l}{\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3+{\overrightarrow{F}}_{4}4\widehat{i}+0\widehat{j}\\ {\overrightarrow{F}}_{net}=\left[5\right(-\widehat{j}\left)\right]+\left[6\cos {20}^{°}\widehat{i}+6\sin {20}^{°}\widehat{j}\right]+F_3\widehat{j}+\left[F_4(-\widehat{i})\right]4\widehat{i}+0\widehat{j}\\ 4\widehat{i}=\left(6\cos {20}^{°}-F_4\right)\widehat{i}+\left(6\sin {20}^{°}+F_3-5\right)\widehat{j}\mathrm{.....}\left(1\right)\end{array}$

Part (a)

Comparing both y-component of expression (1)

$\begin{array}{l}0\widehat{j}=\left(6\sin {20}^{°}+F_3-5\right)\widehat{j}\\ \Rightarrow {\overrightarrow{F}}_3=\left(2.9\text{\hspace{0.05em}N}\right)\widehat{j}\end{array}$

Part (b)

Comparing x-component of expression (1)

$\begin{array}{l}4\widehat{i}=\left(6\cos {20}^{°}-F_4\right)\widehat{i}\\ {\overrightarrow{F}}_4=(6\cos {20}^{°}-4)\widehat{i}\\ {\overrightarrow{F}}_4=\left(1.64\text{\hspace{0.05em}N}\right)\widehat{i}\end{array}$