Home
Class 11
PHYSICS
A force F=(2hati+3hatj-2hatk)N is acting...

A force `F=(2hati+3hatj-2hatk)N` is acting on a body at point (2m, `4m,-2m)`. Find torque of this force about origin.

Text Solution

Verified by Experts

`r=(2hati+4hatj+2hatk)`
`tau=rxxF`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ROTATIONAL MECHANICS

    DC PANDEY|Exercise Exercise 12.4|11 Videos
  • ROTATIONAL MECHANICS

    DC PANDEY|Exercise Exercise 12.5|7 Videos
  • ROTATIONAL MECHANICS

    DC PANDEY|Exercise Exercise 12.2|4 Videos
  • ROTATION

    DC PANDEY|Exercise (C) Chapter Exercises|39 Videos
  • ROTATIONAL MOTION

    DC PANDEY|Exercise Integer Type Questions|17 Videos

Similar Questions

Explore conceptually related problems

A force F=(2hati+3hatj+4hatk)N is acting at point P(2m,-3m,6m) find torque of this force about a point O whose position vector is (2hati-5hatj+3hatk) m.

A force vecF = (hati+hatj-hatk)N acts at a point P(3 m, 6 m, 9 m). The torque exerted by this force about a point Q(2 m, 7 m, 8 m) is

Knowledge Check

  • A force of vecF=2xhati+2hatj+3z^2hatk N is acting on a particle. Find the work done by this force in displacing the body from (1, 2, 3)m to (3, 6, 1)m .

    A
    (a) `-10J`
    B
    (b) `100J`
    C
    (c) `10J`
    D
    (d) `1J`
  • A force vecF=4hati-5hatj+3hatk N is acting on a point vecr_1=2hati+4hatj+3hatk m. The torque acting about a point vecr_2=4hati-3hatk m is

    A
    Zero
    B
    `42hati-30hatj+6hatk`Nm
    C
    `42hati+30hatj+6hatk`Nm`
    D
    `42hati+30hatj-6hatk`Nm
  • Three forces hati+2hatj-3hatk,2hati+3hatj+4hatk and hati-hatj+hatk acting on a particle at the point (0,1,2) the magnitude of the moment of the forces about the point (1,-2,0) is

    A
    `2sqrt(35)`
    B
    `6sqrt(10)`
    C
    `4sqrt(7)`
    D
    none of these
  • Similar Questions

    Explore conceptually related problems

    A force F=(2hati+3hatj-hatk)N is acting on a body at a position r=(6hati-3hatj-2hatk) . Calculate the torque about the origin

    A force vecF=(3hati+4hatj)N is acting on a point mass m=((1)/(2))kg at a point A(2m,2m) . Find the angular acceleration of the line OA at this instant.

    A force vecF=2hati +hatj-hatk acts at point A whose position vector is 2hati-hatj. Find the moment of force vecF about the origin.

    A force F= 2hati+3hatj-hatk acts at a point (2,-3,1). Then magnitude of torque about point (0,0,2) will be

    The torque of force F =(2hati-3hatj+4hatk) newton acting at the point r=(3hati+2hatj+3hatk) metre about origin is (in N-m)