In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0, -2.24 m, 9.78 m) due to (a) force 1 with components F1x = 2.70 N and F1y = F1z = 0, and (b) force 2 with components F2x = 0, F2y = 2.92 N, F2z = 4.02 N?

Answer 1

We can find the torque using the cross product:

τ1 = r1 × F1

where τ1 is the torque due to force 1, r1 is the position vector of the particle, and F1 is the force vector.

For force 1, the position vector r1 is given by:

r1 = (0, -2.24, 9.78)

And the force vector F1 is given by:

F1 = (2.70, 0, 0)

Now we can find the cross product:

τ1 = r1 × F1 = (0 * 0 - 2.24 * 0, -2.24 * 0 - 9.78 * 2.70, -2.24 * 2.70 - 0 * 0)
τ1 = (0, -2.24 * 2.70 - 0, -2.24 * 2.70)
τ1 = (0, -6.048, -6.048)

That is the torque due to force 1. Now we need to find the torque due to force 2:

τ2 = r2 × F2

where τ2 is the torque due to force 2, r2 is the position vector of the particle, and F2 is the force vector.

For force 2, the position vector r2 is the same:

r2 = (0, -2.24, 9.78)

And the force vector F2 is given by:

F2 = (0, 2.92, 4.02)

Now we can find the cross product:

τ2 = r2 × F2 = ( - 2.24 * 4.02 - 9.78 * 2.92, 9.78 * 0 - 0 * 4.02, 0 * 2.92 - (-2.24) * 0)
τ2 = ( - 9.0048 - 28.5756, 0, 0)
τ2 = (-37.5804, 0, 0)

Now we have both torque components:

τ1 = (0, -6.048, -6.048)
τ2 = (-37.5804, 0, 0)

So the total torque on the particle is the sum of these two:

τ_total = τ1 + τ2 = (0 - 37.5804, -6.048 + 0, -6.048 + 0)
τ_total = (-37.5804, -6.048, -6.048)

In unit-vector notation, the torque due to force 1 is (0, -6.048, -6.048) Nm and the torque due to force 2 is (-37.5804, 0, 0) Nm.