example: a cantilever beam with a force f at the free end. find m_a due to force f using cross - product

Explanation:

Step1: Resolve the force into components

The force $\vec{F}$ can be resolved into its $x$ - and $y$ - components. The magnitude of the force is $F = 10$ lbs and the angle with the $x$ - axis is $\theta=30^{\circ}$. The $y$ - component of the force is $F_y=F\sin\theta$ and the $x$ - component is $F_x = F\cos\theta$. So, $F_y=10\sin30^{\circ}=5$ lbs and $F_x = 10\cos30^{\circ}=5\sqrt{3}$ lbs.

Step2: Determine the moment - arm for each component

The moment about point $A$ is calculated using the cross - product $\vec{M}=\vec{r}\times\vec{F}$. For a two - dimensional case, $M = rF_{\perp}$, where $r$ is the distance from the point of interest to the line of action of the force and $F_{\perp}$ is the perpendicular component of the force. The distance from point $A$ to the point of application of the force is $r = 5$ ft. The $x$ - component of the force does not create a moment about point $A$ because its line of action passes through point $A$ (moment - arm for $F_x$ is $0$). The moment - arm for $F_y$ is $r = 5$ ft.

Step3: Calculate the moment about point $A$

The moment about point $A$, $M_A$ is given by $M_A=rF_y$. Substituting $r = 5$ ft and $F_y=5$ lbs, we get $M_A=(5)(5)=25$ lb - ft.

Answer:

$25$ lb - ft