Force f = (-6.0 n)ihat + (2.0 n) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
(a) what is the torque on the particle aboutthe origin?
1 n·m ihat+ 2 n·m jhat+ 3 n·m khat

(b) what is the angle between the directions of rand f? (if there is no torque, enter 0.)
4°

Explanation:

Given that.

Force acting on the particle,

Position of the particle,

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r,

Magnitude of F,

Using dot product formula,

Therefore, this is the required solution.

the answer is b

explanation: