The force acting on a particle vecr=(4,6,12)m of a rigid body is vecF=(6,8,10)N. Find the magniude of the torque producing the rotational motion. Axis of rotation is along the unit vector (1)/(sqrt(3))(1,1,1).

The force acting on a particle vecr=(4,6,12)m of a rigid body is vecF=

1.	The force acting on a particle vecr=(4,6,12)m of a rigid body is vecF=(6,8,10)N. Find the magniude of the torque producing the rotational motion. Axis of rotation is along the unit vector (1)/(sqrt(3))(1,1,1).
Answer» SOLUTION :`vectau=vecrxxvecF` The magnitude of the torque with respect to the axis on which the unit VECTOR is `HATN` is `vec(tau_(n))=(vecrxxvecF).hatn` `THEREFORE vecrxxvecF=\|(hati,hatj,HATK),(4,6,12),(6,8,10)\|` `therefore vecrxxvecF=hati(60-96)-hatj(40-72)+hatk(32-36)` `=-36hati+32hatj-4hatk=(-36hati+32hatj-4hatk)Nm` `therefore` Magnitude of the torque responsible for rotational motion Now, `(vecrxxvecF).hatn=(-36,32,-4).(1)/(sqrt(3))(1,1,1)` `=(1)/(sqrt(3))(-36+32-4)` `therefore vec(tau_(n)).hatn=-(8)/(sqrt(3))Nm` `therefore tau=(8)/(sqrt(3))Nm`