Question

Find the sum of these four vector forces: 12.0 N to the right at 35.0° above the horizontal, 31.0 N to the left at 55.0° above the horizontal, 8.40 N to the left at 35.0° below the horizontal, and 24.0 N to the right at 55.0° below the horizontal. (Hint: Make a drawing of this situation and select the best axes for x and y so that you have the least number of components. Then add the vectors, using the component method.)

Answer 1

Answer:

R = 7,87N to the left at 82,26° above the horizontal

Explanation:

Graph1: situation drawing (attached image)

Graph2: Resulting from the forces on the x '- y' axes (attached image)

Fx' = (12 - 8,4)N = 3,6N to the right at 35° above the horizontal

Fy' = (31 - 24)N = 7N to the left at 55° above the horizontal

Graph 3: Resulting vector of forces (R) on x-y coordinate axes (attached image)

R = Rx i + Ry j

Rx = 3,6Cos(35°) - 7Cos(55°) = -1,06N

Rx = 3,6Sin(35°) + 7Sin(55°) = 7,8N

R = (1,06 -i + 7,8 j ) N

Magnitude of R:

$R = \sqrt{(-1,06)^2+(7,8)^2} = 7,87N$

$\alpha = tan^{-1} (\frac{7,8}{1,06} ) = 82,26 degrees$ to the left above the horizontal.

Answer 2

The sum of all the four forces using the component method is 29.67 N.

The given parameters;

• 12.0 N to the right at 35.0° above the horizontal,
• 31.0 N to the left at 55.0° above the horizontal,
• 8.40 N to the left at 35.0° below the horizontal
• 24 N to the right at 55.0° below the horizontal

The x and y component of each force is calculated as follows;

force ------------ angle ---------x - component ------ y - component

F                          θ               Fcosθ                         Fsinθ

12                      35                 9.83                         6.88

31                      55                -17.81                         25.39

8.4                    35                 -6.88                        -4.82

24                     55                 13.77                        -19.66

----------------------------------------------------------------------------------

 ∑F                                         -28.63                      7.79

-----------------------------------------------------------------------------------

The resultant force is calculated as follows;

$F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(-28.63)^2 + (7.79)^2} \\\\F = 29.67 \ N$

Thus, the sum of all the four forces is 29.67 N.

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