Question

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an angle of ααalpha = 56.0 ∘∘ above the negative x axis in the second quadrant. F⃗ 2F→2F_2_vec has a magnitude of 7.00 NN and is directed at an angle of ββbeta = 52.8 ∘∘ below the negative x axis in the third quadrant.A. What is the x component FxFxF_x of the resultant force?B. What is the y component FyFyF_y of the resultant force?C. What is the magnitude FFF of the resultant force?D.What is the angle γγgamma that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative xx axis.

Answer 1

Answer:

• Fx = -9.15 N
• Fy = 1.72 N
• F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

• Fx = -9.15 N
• Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°