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 9.20 NN and is directed at an angle of 62.0 ∘∘ above the negative x axis in the second quadrant. F⃗ 2F→2F_2_vec has a magnitude of 5.00 NN and is directed at an angle of 53.6 ∘∘ below the negative x axis in the third quadrant. Part A What is the x component of the resultant force?

Answers

  1. Answer:

    The x component of the resultant force is -7.27N.

    Explanation:

    To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

    |F_{1x}|=|F_1|\cos\theta_1=9.20N\cos62.0\°=4.31N \\|F_{2x}|=|F_2|\cos\theta_2=5.00N\cos53.6\°=2.96N

    Since both vectors are in the left side of the y-axis, they have a negative x component. So:

    F_{1x}=-4.31N;\\F_{2x}=-2.96N

    Finally, we sum both components to obtain the component of the resultant force:

    F_{Rx}=-4.31N-2.96N=-7.27N

    In words, the x component of the resultant force is -7.27N.

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