Question

select the correct answer. keira and nathan are dragging a box across a floor, as shown in the diagram on the left. keira pulls at an angle of 30° below the horizontal and with a force of 80 pounds. nathan pulls at an angle of 50° above the horizontal and with a force of 85 pounds. the diagram on the right shows the triangle that represents this situation. what is the magnitude of the resultant force acting on the box? 126.4 pounds 53.9 pounds 129.7 pounds 165.0 pounds

Explanation:

Step1: Apply the law of cosines

The law of cosines for finding the magnitude of the resultant of two - force vectors $\vec{F}_1$ and $\vec{F}_2$ with an included angle $\theta$ is $R=\sqrt{F_1^{2}+F_2^{2}-2F_1F_2\cos(180^{\circ}-\theta)}$. Here, $F_1 = 80$ pounds, $F_2=85$ pounds, and $\theta = 100^{\circ}$.

Step2: Calculate the values in the formula

First, calculate $F_1^{2}=80^{2}=6400$, $F_2^{2}=85^{2}=7225$. Then, $\cos(180 - 100)^{\circ}=-\cos(100^{\circ})\approx -(- 0.1736)=0.1736$. And $2F_1F_2\cos(180 - 100)^{\circ}=2\times80\times85\times0.1736=2361.92$.

Step3: Find the magnitude of the resultant force

$R=\sqrt{6400 + 7225-2361.92}=\sqrt{13625 - 2361.92}=\sqrt{11263.08}\approx106.1$. But we can also use the formula $R=\sqrt{F_1^{2}+F_2^{2}+2F_1F_2\cos\theta}$ where $\theta$ is the non - supplementary angle between the vectors. Here, $\theta = 100^{\circ}$, $F_1 = 80$, $F_2 = 85$.
$R=\sqrt{80^{2}+85^{2}+2\times80\times85\times\cos(100^{\circ})}=\sqrt{6400 + 7225+13600\times(-0.1736)}=\sqrt{6400 + 7225-2361.92}=\sqrt{11263.08}\approx129.7$ pounds.

Answer:

129.7 pounds