Complete question:
A particle located at the position vector r = (i + j) m has a force F = (2i + 3j) N acting on it. The torque about the origin is
Answer:
The torque about the origin is (5k) N.m
Explanation:
The torque about the origin is the vector or cross product of the two vectors.
τ = r x F (N.m)
Where;
τ is the torque about the origin
τ = r x F
τ = (i + j) x (2i + 3j)
For cross product;
i x j = k
i x k = j
j x k = i
i x i = 0
j x j = 0
k x k = 0
τ = (i + j) x (2i + 3j)
τ = (i x 2i) + (i x 3j) + (j x 2i) + (j x 3j)
τ = (0) + (3k)+ (2k) + 0
τ = (5k) N.m
Therefore, the torque about the origin is (5k) N.m
