Question

A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector r$\rightarrow$ = (2.00 m) - (3.00 m) + (2.00 m) , the force is F$\rightarrow$ = Fx + (7.00 N) - (6.00 N) , and the corresponding torque about the origin is T$\rightarrow$ = (4.00 N·m) + (10.0 N·m) + (11.0 N·m) . Determine Fx.

Answer 1

here,

the position vector , r = ( 2 i - 3 j + 2 k) m

the applied force , F = ( Fx i - 6 j) N

the torque applied , T = r X F

(4 i + 10 j + 11 k) N.m = (2i - 3j + 2 k) X ( Fx i + 7 j - 6 k)

(4 i + 10 j + 11 k) N.m = (4 i + (2 Fx + 12) j + (14 + 3 Fx) k)

on compairing

we get

Fx = - 1 N

the value of Fx is (-1 N)