Find the resultant of the following forces: 3.0N east, 4.0N west, and 5.0N in a direction north 60° west.

Question

Find the resultant of the following forces: 3.0N east, 4.0N west, and 5.0N in a direction north 60° west.

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Farah 3 years 2021-08-30T01:45:28+00:00 1 Answers 9 views 0

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    2021-08-30T01:47:17+00:00

    Answer:

    5.5 N at 50.8° north of west.

    Explanation:

    To find the resultant of these forces, we have to resolve each force along the x- and y-direction, then find the components of the resultant force, and then calculate the resultant force.

    The three forces are:

    F_1=3.0 N (east)

    F_2=4.0 N (west)

    F_3=5.0 N (at 60° north of west)

    Taking east as positive x-direction and north as positive y-direction, the components of the forces along the 2 directions are:

    F_{1x}=3.0 N\\F_{1y}=0

    F_{2x}=-4.0 N\\F_{2y}=0

    F_{3x}=-(5.0)(cos 60^{\circ})=-2.5 N\\F_{3y}=(5.0)(sin 60^{\circ})=4.3 N

    Threfore, the components of the resultant force are:

    F_x=F_{1x}+F_{2x}+F_{3x}=3.0+(-4.0)+(-2.5)=-3.5 N\\F_y=F_{1y}+F_{2y}+F_{3y}=0+0+4.3=4.3 N

    Therefore, the magnitude of the resultant force is

    F=\sqrt{F_x^2+F_y^2}=\sqrt{(-3.5)^2+(4.3)^2}=5.5 N

    And the direction is:

    \theta=tan^{-1}(\frac{F_y}{|F_x|})=tan^{-1}(\frac{4.3}{3.5})=50.8^{\circ}

    And since the x-component is negative, it means that this angle is measured as north of west.

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