Practice Problems part 2 Problem 5 An engineering problem fr

Practice Problems -part 2: Problem 5: An engineering problem from Mechanics of Materials You have learned that a moment (or torque) about a point is just the force times the perpendicular distance for two-dimensional cases However, this definition becomes very tricky when using three-dimensional force/position vectors. For three-dimensional cases. the moment M is defined as the cross product of the position vector r and the force vector F. Note that the position from the point of interest o the beginning of the vector. Given The following position and force vectors: ra m 0.3i+0.4j (m): Fa 20k (N) rb 06i+0.2j (m); Fb 10i+5k (N)

Solution

ra = [0.3 0.4 0.0];
Fa = [0 0 -20];
rb = [0.6 0.2 0.0];
Fb = [0 10 0.5];
rc = [-0.9 0.0 0.5];
Fc = [-4 8 0.0];
Ma = cross(ra, Fa)
Mb = cross(rb, Fb)
Mc = cross(rc, Fc)
Mt = Ma + Mb + Mc
Mt_mag = sqrt(Mt(1)^2 + Mt(2)^2 + Mt(3)^2])
unit_vec = Mt/Mt_magnitude