Work (in physics) is done whenever a force causes linear displacement. That is to say, work is the amount of force along the displacement vector. Work is usually measured in Joules (J).
Torque is the turning effect done by a force that causes angular displacement. Torque is usually measure in Newton-metres (N-m).
Outlined below is table outlining the different formulas used to determine Work, Areas of of Parallelograms, and Torque:
| Work | Area of Parallelogram | Torque |
|---|---|---|
| \(W = \vec{F}\cdot\vec{d} \) | \(A = l \cdot w \) | \(\vec{T} = \vec{r} \times \vec{F} \) |
| \(W = \vert\vec{F}\vert\vert\vec{d}\vert\cos{\theta} \) | \(A = \vert\vec{a}\vert\vert\vec{b}\vert\sin{\theta} \) | \(\vec{T} = \vert\vec{r}\vert\vert\vec{F}\vert\sin{\theta} \text{ } \hat{n} \) |
| \(A = \vert\vec{a} \times \vec{b}\vert \) |
First, we can determine the coordinates for\(\vec{PQ}\) by calculating the difference between \(Q\) and \(P\):
\(\vec{PQ} = (2, -3, -1)\)
Next, we can determine the coordinates for\(\vec{PR}\) by calculating the difference between \(R\) and \(P\):
\(\vec{PR} = (0, 1, 2) \)
Then, we can use the Area of a Parallelogram formula but divide the product by \(2\):
\( = \dfrac{\sqrt{5^2 + 4^2 + 2^2}}{2} = \dfrac{3\sqrt{5}}{2}\)
Therefore, we can determine that the area of the parallelogram is \(\boldsymbol{\dfrac{3\sqrt{5}}{2}}\) square units.
In order to determine the work done by the force of gravity, we can plug the appropriate values into the Work formula and solve:
\(W = 606 \; \text{[Nm] or [Joules]}\)
Therefore, we can determine that the work done by the force of gravity on the box is \(\boldsymbol{606\;[\textbf{J}]}\).
In order to determine the amount of torque produced by the cyclist, we can plug the appropriate values into the torque formula and solve:
\(\vec{T} = 18.12 \; [\text{N-m}] [\text{into the page}]\)
Therefore, we can determine the amount of torque produced by the cyclist is \(\boldsymbol{18.12 \; [\textbf{N-m}]}\).