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?

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.

To find the torque about the origin, we can use the equation:

τ = r x F

where 'x' represents the cross product. The torque vector will be perpendicular to both the position vector (r) and the force vector (F).

(a) For force 1:
Given: F1x = 2.70 N, F1y = F1z = 0

The position vector (r) is given by the particle's coordinates: (0, -2.24 m, 9.78 m)

Now, we can calculate the cross product of r and F1:

(0, -2.24 m, 9.78 m) x (2.70 N, 0, 0)

To perform the cross product, we use the following formula:

r x F = (r2 * F3 - r3 * F2, r3 * F1 - r1 * F3, r1 * F2 - r2 * F1)

By substituting the given values, we get:

(0 * 0 - (-2.24 m) * 0, 9.78 m * 2.70 N - 0 * 0, 0 * 0 - 9.78 m * 2.70 N)

Simplifying, we find:

(0, 26.436 N·m, 0)

Therefore, the torque about the origin due to force 1 is (0, 26.436 N·m, 0) in unit-vector notation.

(b) For force 2:
Given: F2x = 0, F2y = 2.92 N, F2z = 4.02 N

Using the same process as above, we can calculate the cross product:

(0, -2.24 m, 9.78 m) x (0, 2.92 N, 4.02 N)

By applying the cross product formula, we get:

(0 * 4.02 N - (-2.24 m) * 2.92 N, 9.78 m * 0 - 0 * 4.02 N, 0 * 2.92 N - 9.78 m * 0)

Simplifying further, we find:

(6.5408 N·m, 0, 0)

Therefore, the torque about the origin due to force 2 is (6.5408 N·m, 0, 0) in unit-vector notation.

In summary:
(a) The torque about the origin due to force 1 is (0, 26.436 N·m, 0).
(b) The torque about the origin due to force 2 is (6.5408 N·m, 0, 0).

To find the torque about the origin, we need to find the cross product of the position vector and the force vector for each case. The torque is given by the formula:

τ = r x F

where "x" represents the cross product.

For case (a) with force 1, the torque is given by:

τ1 = r x F1

To find the cross product, we need to represent the vectors in unit-vector notation. In this notation, the force and position vectors are represented as:

r = (x, y, z)
F = (Fx, Fy, Fz)

For case (a):

r1 = (0, -2.24, 9.78) m
F1 = (2.70, 0, 0) N

To find the cross product, we use the following formula:

a x b = (a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k

where i, j, and k are the unit vectors along the x, y, and z axes respectively.

Using this formula, we can now calculate the cross product:

r1 x F1 = [(0 × 0) - (0 × 0)]i + [(0 × 0) - (2.70 × 9.78)]j + [(2.70 × -2.24) - (0 × 0)]k
= [0 - 0]i + [0 - 26.446]j + [-6.048 - 0]k
= 0i - 26.446j - 6.048k

Therefore, the torque about the origin due to force 1 is:

τ1 = 0i - 26.446j - 6.048k

For case (b) with force 2, the torque is given by:

τ2 = r x F2

Similarly, we need to represent the vectors in unit-vector notation:

r2 = (0, -2.24, 9.78) m (same as before)
F2 = (0, 2.92, 4.02) N

Using the cross product formula:

r2 x F2 = [(0 × 4.02) - (0 × 2.92)]i + [(0 × 0) - (0 × 0)]j + [(0 × 2.92) - (0 × 4.02)]k
= [0 - 0]i + [0 - 0]j + [0 - 0]k
= 0i + 0j + 0k

Therefore, the torque about the origin due to force 2 is:

τ2 = 0i + 0j + 0k

In both cases, the torque about the origin is zero.