Question

Use a graphing utility to find the magnitude and direction angles of the resultant of forces F1\mathbf{F}_1 and F2\mathbf{F}_2 with initial points at the origin. The magnitude and terminal point of each vector are given.

Vector Magnitude Terminal Point
F1\mathbf{F}_1 50lb50 \mathrm{lb} (10,5,3)(10,5,3)
F2\mathbf{F}_2 80lb80 \mathrm{lb} (12,7,5)(12,7,-5)

Solution

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Answered 2 years ago
Answered 2 years ago
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(1) Let M1M_1 be the magnitude per unit vector for F1F_1.

F1=M110,5,3F_1 = M_1\left\langle 10, 5, 3 \right\rangle

M1=M1F1=50lb1102+52+32=50lb11344.319lbM_1 = M* \frac{1}{\left\|F_1\right\|} = 50 lb* \frac{1}{\sqrt{10^2 + 5^2 + 3^2}}=50 lb * \frac{1}{\sqrt{134}} \approx 4.319 lb per 1 unit scalar.

Let M2M_2 be the magnitude per unit vector for F2F_2.

F2=M212,7,5F_2 = M_2\left\langle 12, 7, -5 \right\rangle

M2=M1F2=80lb1122+72+52=80lb12185.418lbM_2 = M* \frac{1}{\left\|F_2\right\|} = 80 lb* \frac{1}{\sqrt{12^2 + 7^2 + 5^2}}=80 lb * \frac{1}{\sqrt{218}} \approx 5.418 lb per 1 unit scalar.

F1=4.31910,5,3=43.19,21.59,12.95F_1 = 4.319 \left\langle 10, 5, 3 \right\rangle = \left\langle 43.19, 21.59, 12.95\right\rangle

F2=M212,7,5=5.41812,7,5=65.02,37.93,27.09F_2 = M_2\left\langle 12, 7, -5 \right\rangle = 5.418 \left\langle 12, 7, -5 \right\rangle = \left\langle 65.02 ,37.93 ,- 27.09\right\rangle

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