Two forces, F1 vec and F2 vec, act at a point. F1 vec has a magnitude of 9.60 N and is directed at an angle of 55.0∘ above the negative x ax

Question

Two forces, F1 vec and F2 vec, act at a point. F1 vec has a magnitude of 9.60 N and is directed at an angle of 55.0∘ above the negative x axis in the second quadrant. F2 vec has a magnitude of 5.40 N and is directed at an angle of 53.1∘ below the negative x axis in the third quadrant. A. What is the x component of the resultant force?B. What is the y component of the resultant force?C. What is the magnitude of the resultant force?

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Doris 3 years 2021-08-26T13:45:25+00:00 1 Answers 14 views 0

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  1. RuslanHeatt
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    2021-08-26T13:46:47+00:00
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    Answer:

    A. F_x=-8.74N

    B. F_y=3.55N

    C. F=9.43N

    Explanation:

    First, we break down the two vectors F₁ and F₂ into their components. In order to do that, we have to use trigonometry:

    F_{1x}=F_1\cos\theta_1\implies F_{1x}=-9.60N\cos55.0\°=-5.50N\\\\F_{1y}=F_1\sin\theta_1\implies F_{1y}=9.60N\sin55.0\°=7.86N\\\\\\F_{2x}=F_2\cos\theta_2\implies F_{2x}=-5.40N\cos53.1\°=-3.24N\\\\F_{2y}=F_2\sin\theta_2\implies F_{2y}=-5.40N\sin53.1\°=-4.31N\\\\

    (Note: F_{1x}, F_{2x} and F_{2y} are negative because the vectors are in the second and third quadrant, respectively. So those components must be negative)

    Next, we sum the x components and the y components to get the components of the resultant force:

    F_x=F_{1x}+F_{2x}=-5.50N-3.24N=-8.74N\\\\F_y=F_{1y}+F_{2y}=7.86N-4.31N=3.55N

    So, the x component of the resultant force is -8.74N (A), and its y component is 3.55N (B).

    Finally, to get the magnitude of the resultant force, we have to use the Pythagorean Theorem:

    F=\sqrt{F_x^{2}+F_y^{2}}\\ \\F=\sqrt{(-8.74N)^{2}+(3.55N)^{2}}=9.43N

    In words, the magnitude of the resultant force is 9.43N (C).

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