Question

Determine the magnitude, fda, acting along member da, due to the applied force f=10.0 iâ20.0 j+27.2 k n.

Answer 1

Force vector is given as

$F = 10\hat i + 20\hat j + 27.2\hat k$

now we need to find the magnitude of this force

as we know that

$F_x = 10$

$F_y = 20$

$F_z = 27.2$

so here we will find the net force

$F_{net} = \sqrt{F_x^2 + F_y^2 + F_z^2}$

$F_{net} = \sqrt{10^2 + 20^2 + 27.2^2}$

$F_{net} = 35.2 N$

so it is having magnitude as 35.2 N