Find the resultant of the following forces by component method: F1 =12N,south; F2 =24N, 30° north of west; F3 = 15N, 75° south of west; and

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Find the resultant of the following forces by component method: F1 =12N,south; F2 =24N, 30° north of west; F3 = 15N, 75° south of west; and F4 = 32 N, 50° south of east.​

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Bơ 5 years 2021-07-27T07:32:38+00:00 1 Answers 1193 views 3

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  1. Giakhanh
    -1
    2021-07-27T07:33:47+00:00
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    Answer:

    The resultant of given forces is \vec R = -16.099\,\hat{i} -27.002\,\hat{j}\,\,[N].

    Explanation:

    The component method consists in summing each component of known vectors in rectangular form to get the resultant. That is:

    \vec R = \left(\Sigma_{i=1}^{m} x_{i}\right)\,\hat{i}+\left(\Sigma_{i=1}^{m} y_{i}\right)\,\hat{j}

    Where:

    \vec R – Resultant, measured in newtons.

    x_{i} – i-th x-Component, measured in newtons.

    y_{i} – i-th y-Component, measured in newtons.

    We describe each known vector below:

    \|\vec F_{1}\| = 12\,N, South:

    \vec F_{1} = -12\,\hat{i}\,\,[N]

    \|\vec F_{2}\| = 24\,N, \angle = 30^{\circ} North of west:

    \vec F_{2} = 24\cdot (-\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})\,[N]

    \vec F_{2} = -20.785\,\hat{i}+12\,\hat{j}\,\,[N]

    \|\vec F_{3}\| = 15\,N, \angle = 75^{\circ} South of west:

    \vec F_{3} = 15\cdot (-\cos 75^{\circ}\,\hat{i}-\sin 75^{\circ}\,\hat{j})\,\,[N]

    \vec F_{3} = -3.883\,\hat{i}-14.489\,\hat{j}\,\,[N]

    \|\vec F_{4}\| = 32\,N, \angle = 50^{\circ} South of east:

    \vec F_{4} = 32\cdot (\cos 50^{\circ}\,\hat{i}-\sin 50^{\circ}\,\hat{j})\,\,[N]

    \vec F_{4} = 20.569\,\hat{i} -24.513\,\hat{j}\,\,[N]

    We find the resultant by vectorial sum:

    \vec R = \vec F_{1}+\vec F_{2}+\vec F_{3}+\vec F_{4}

    \vec R = (-12-20.785-3.883+20.569)\,\hat{i}+(12-14.489-24.513)\,\hat{j}\,\,[N]

    \vec R = -16.099\,\hat{i} -27.002\,\hat{j}\,\,[N]

    The resultant of given forces is \vec R = -16.099\,\hat{i} -27.002\,\hat{j}\,\,[N].

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