The force of 20\hat{j}~\text{N}20 j ^ N is applied at \vec{r} =(4.0\hat{i}-2.0\hat{j})~\text{m} r ⃗ =(4.0 i ^ −2.0 j ^ ) m. What is the torque of this force about the origin?

Explanation:

Torque is given by

Note that this is a cross product of vectors. To perform a cross product, we need to express the vectors in three components.

The torque is then

(I'm not sure if the matrix will appear well because I'm constrained with the device I'm using but what I'm showing is the determinant of a matrix with i, j, k in the first row, the coefficients of F in the second and the coefficients of r in the third)

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.