A force, F = (2x + 3y) N, is applied to an object at a point whose position vector with respect to the pivot point is r = (5x + 4y + 2z) m. Calculate the torque (in units on Nm) created by the force about that pivot point.

To calculate the torque created by the force about the pivot point, we need to find the cross product of the force vector and the position vector.

The cross product formula for two vectors A and B is given by:

A x B = (A_y * B_z - A_z * B_y)i + (A_z * B_x - A_x * B_z)j + (A_x * B_y - A_y * B_x)k

Given:
Force vector, F = (2x + 3y) N
Position vector, r = (5x + 4y + 2z) m

We can break down the position vector as follows:
r_x = 5x, r_y = 4y, r_z = 2z

Now, let's calculate the cross product:

F x r = (2x + 3y) x (5x + 4y + 2z)
= (2*(4y) - 3*(2z))i + (2z - 2*(5x))j + ((5x)*(3y) - (2z)*(2*(4y)))k
= (8y - 6z)i + (2z - 10x)j + (15xy - 16yz)k

The torque vector is given as (8y - 6z)i + (2z - 10x)j + (15xy - 16yz)k.

Since torque is a vector quantity, it does not have units of Nm. It only represents the direction and magnitude of the torque.

To calculate the torque created by a force about a pivot point, we use the formula:

\[\text{Torque} (\tau) = \text{Force} (F) \times \text{Lever arm} (d)\]

The force vector given is \(\textbf{F} = (2x + 3y) \, \text{N}\), and the position vector (\(\textbf{r}\)) with respect to the pivot point is \(\textbf{r} = (5x + 4y + 2z) \, \text{m}\).

First, we find the lever arm vector (\(\textbf{d}\)) which is the vector from the pivot point to the point of application of the force. It is given by:

\[\textbf{d} = \textbf{r}\]

Now, we can calculate the torque using the formula:

\[\tau = |\textbf{F}| \times |\textbf{d}| \times \sin(\theta)\]

Where:
- \(|\textbf{F}|\) is the magnitude of the force vector.
- \(|\textbf{d}|\) is the magnitude of the position vector.
- \(\theta\) is the angle between the force vector and the position vector.

Considering the given values, we have:

\(|\textbf{F}| = \sqrt{(2x)^2 + (3y)^2} = \sqrt{4x^2 + 9y^2}\) (magnitude of the force vector)

\(|\textbf{d}| = \sqrt{(5x)^2 + (4y)^2 + (2z)^2} = \sqrt{25x^2 + 16y^2 + 4z^2}\) (magnitude of the position vector)

The angle between the force and position vectors (\(\theta\)) is 0 degrees because they are parallel, and the sine of 0 degrees is 0, so we can ignore it for now.

Finally, we can calculate the torque:

\(\tau = |\textbf{F}| \times |\textbf{d}| = \sqrt{4x^2 + 9y^2} \times \sqrt{25x^2 + 16y^2 + 4z^2} = \sqrt{(4x^2 + 9y^2)(25x^2 + 16y^2 + 4z^2)}\) (units in Nm)

Simplifying and expanding the expression inside the square root gives us the final torque in units of Nm.