On a mass

The forces F1, and F2 with magnitudes of 40N act on a mass point M. Their resultant has a magnitude of 60N. Determine the angle that the forces F1 and F2 make.

Correct answer:

α =  97.1808 °

Step-by-step explanation:

$$\begin{gathered} F_1=40\text{N} \\ {F}_2=F_1=40\text{N} \\ F=60\text{N} \\ {F}^2=F_1^2+F_2^2-2\cdot F_1\cdot F_2\cdot \cos \alpha =F_1^2+F_2^2-2\cdot F_1\cdot F_2\cdot c \\ c=\frac{F_1^2+F_2^2-F^2}{2\cdot F_1\cdot F_2}=\frac{40^2+40^2-60^2}{2\cdot 40\cdot 40}=-\frac{1}{8}=-0.125 \\ c=\cos \alpha \\ A=\arccos c=\arccos ((-0.125))\doteq 1.6961\text{rad} \\ \alpha =A\to \text{°}=A\cdot \frac{180}{\pi }\text{°}=1.6961\cdot \frac{180}{\pi }\text{°}=97.181\text{°}=97.1808^{\circ }=97°10^{\prime }51" \end{gathered}$$

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