Two forces F1=(2^i–5^j−6^k) N and F2=(−^i+2^j−^k) N are acting on a body at the points (1,1,0) and (0,1,2) respectively. Find resultant torque acting on the body about point (−1,0,1) in (N-m).
Let the position vector of (1,1,0) and (0,1,2) w.r.t point (−1,0,1) be r1 and r2 .
Therefore, r1=(^i+^j)−(−^i+^k)=2^i+^j−^k
r2=(^j+2^k)−(−^i+^k)=^i+^j+^k
Then, net torque is given by
τ=r1×F1+r2×F2
Now, r1×F1=∣∣
∣
∣∣^i^j^k21−12−5−6∣∣
∣
∣∣=−11^i+10^j−12^k
and
r2×F2=∣∣
∣
∣∣^i^j^k111−12−1∣∣
∣
∣∣=−3^i+3^k
So, resultant torque =(−11^i+10^j−12^k)+(−3^i+3^k)
=−14^i+10^j−9^k N-m