Aidan Sharpe - Homework 1

\vec{F} has magnitude $800$N and makes an angle of 35^\circ with the y-axis in the second quadrant.

\vec{F}_x = \lVert \vec{F} \rVert \sin(-35^\circ) = 800 \times -0.5736 = -458.861

\vec{F}_y = \lVert \vec{F} \rVert \cos(-35^\circ) = 800 \times 0.8192 = 655.322

\vec{F} = -458.861\hat{i} + 655.322\hat{j}
The force, \vec{F}, has magnitude $6.6$kN and slope is -\frac{5}{12}.

\theta = \arctan \left(-\frac{5}{12} \right) = -0.3948 = -22.620^\circ

\vec{F}_x = 6600\cos(-22.620^\circ) = 6600 \times 0.9231 = 6092.308

\vec{F}_y = 6600\sin(-22.620^\circ) = 6600 \times -0.3846 = -2538.462

\vec{F} = 6092.308\hat{i} - 2538.462\hat{j}
$F_1 = 500$N and $F_2 = 350$-N. F_1 is in the direction of the x-axis, and F_2 makes an angle 60^\circ with the x-axis.

\vec{R} = \vec{F}_1 + \vec{F}_2

\vec{F}_1 = 500\hat{i} + 0\hat{j}

\vec{F}_2 = 350\cos(60^\circ)\hat{i} + 350\sin(60^\circ)\hat{j} = 175\hat{i} + 303.109\hat{j}

\vec{R} = 675\hat{i} + 303.109\hat{j}

\theta_R = \arctan(\frac{303.109}{675}) = 0.4221 = 24.182^\circ
$F_y = 70$lbs. The slope of \vec{F} is \frac{12}{5}.

\frac{70}{F_x} = \frac{12}{5}

70 \times 5 = 12 F_x

F_x = \frac{70 \times 5}{12} = 29.166

\vec{F} = F_x\hat{i} + F_y\hat{j}

$\lVert \vec{F} \rVert = \sqrt{F_x^2 + F_y^2} = \sqrt{29.166^2 + 70^2} = 75.833$N
\vec{F}_1 has magnitude $2$kN and makes an angle 30^\circ with the x-axis. \vec{F}_2 has magnitude $3$kN and has slope -\frac{4}{3}.

\vec{F}_1 = 2000\cos(30^\circ)\hat{i} + 2000\sin(30^\circ)\hat{j} = 1732.051\hat{i} + 1000\hat{j}

\theta_{F_2} = \arctan(-\frac{4}{3}) = -0.9273 = -53.130^\circ

\vec{F}_2 = 3000\cos(\theta_{F_2})\hat{i} + 3000\sin(\theta_{F_2})\hat{j} = 1800\hat{i} - 2400\hat{j}

\vec{R} = \vec{F}_1 + \vec{F}_2 = (1732.051 + 1800)\hat{i} + (1000 - 2400)\hat{j} = 3532.051\hat{i} - 1400\hat{j}

\theta = \arctan(-\frac{1400}{3532.051}) = -0.3774 = -21.622^\circ

\lVert \vec{R} \rVert = \sqrt{3532.051^2 + (-1400)^2} = 3799.393

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Aidan Sharpe - Homework 1

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