Question

A force F=2i+3j-5k N acts at a point r1= 2i+4j+7k m The torque of the force about the point r2 =i+2j+3k m is

Answer 1

Hi , there !!!

here is your answer !!!

r1 = 2i + 4j + 7k

r2= i+ 2j + 3k

r2 -r1 = 2i-i + 4j - 2j + 7k - 3k

rf = i + 2j + 4k



torque => force . displacement

( 2i + 3j -5k ) (i+ 2j + 4k ) .


=> 2 + 6 -20

- 12 j





hope it helps you !!!!


thanks !!!!


Ranjankumar

Answer 2

To find the torque by the force F= (2i+3j-5k) acting on r1= (2i+4j+7k) m about r2= (i+2j+3k) m  

We will find the torque at r1 and subtract it from the torque at r2 as individually they are torques about the origin. Using determinant method,

τ1 =r1 × F  = i (21+20) + j (-10 -14) + k (8-6) = (41i -24j + 2k)

τ2 = r2 × F=  = i (9+10) + j(-5-6) + k(4-3) = (19i -11j + k)

finally torque about r2 is  

τ2 – τ2 =  19i – 41i -11j+24j +k-2k = (13j -22 I – k)