What is the sum of the following set of three forces? express the resultant force in cartesian vector form, and give also the magnitude and direction of the resultant force. a (10 lb)+ (16 lb) j + (6 lb)k, b (3 lb)? + (2 lb)k, c is a force in the xz plane at an inclination of 45° to the positive x-axis and directed away from the origin with a magnitude of 25 lb. ans.: r (24.7 lb)i+ (16 lb)j + (25.7 lb)k; r- 39 lb, θ,-50.70, e,-65.80.4-48.80

R = ( 24.68 i^ + 16j^ + 25.68k^) lb

Explanation:

The way to work the vectors is to add each component independently of the others and when you have the resulting components the vector is constructed.

Let's write the competency of each vector

X axis

     Ax = 10 lb

     Bx = -3 lb

The vector C is given in the form of magnitude (C = 25 lb) and angle 45º so we will use trigonometry to find its components

      Cos θ = Cx / C

      Cx = C cos θ

      Cx = 25 cos 45

      Cx = 17.68 lb

Axis y

       Ay = 16 lb

       By = 0

       Cy = 0

Z axis

      Az = 6 lb

      Bz = 2 lb

      sin θ = Cz / C

      Cz = C sin θ

      Cz = 25 sin 45

      Cz = 17.68 lb

We calculate in the components of the resulting vector

      Rx = Ax + bx + Cx

      Rx = 10 -3 + 17.68

      Rx = 10 -3 + 17.68

      Rx = 24.68 lb

      Ry = Ay + By + Cy

      Ry = 16 + 0 + 0

      Ry = 16 lb

      Rz = Az + Bz + Cz

      Rz = 6 +2 + 17.68

      Rz = 25.68 lb

We build the resulting vector

      R = ( 24.68 i^ + 16j^ + 25.68k^) lb

      R = √ (Rrx² + Ry² + Rz²)

      R = √ (24.68² + 16² + 25.68²)

      R = 39.05 lb

Note that this is a three-dimensional system so we have angles between xy, xz and yz

Let us calculate each angle separately, for this we will use the concept of cosine directors

       Cos α = x / R

       Cos β = y / R

       Cos γ = z / R

       cos α = 24.68 / 39

       α = cos⁻¹ 0.632

       α = 50.8º

       Cos β = 16/39

       β = cos⁻¹ (04097)

       β = 68.8º

      cos γ = 25.68/39

      γ= cos⁻¹ (0.658)

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