Statics of rigid bodies - understanding worked example

Question

I'm having some difficulty understanding how this solution was found.

The question reads:

Find the sum of the moments about P of the forces shown in the diagram. 

Diagram:

Solution:

clockwise: 5 x 4 x sin55 = 16.38...
anticlockwise: 3 x 4 x cos55 + 5 x 2 x sin55 = 15.07...
=> 1.31 Nm clockwise

I know the definitions, what I don't understand is the value selection for the distances and degrees.

Would it be possible for someone to add further information to the solution?

Any help is much appreciated.

Answer 1

We can evaluate the moment about $P$ by the following formula:

$$\vec M=\vec r\times \vec F$$

and $$|\vec M|=|\vec r|\cdot|\vec F|\sin\alpha\longrightarrow M=r\cdot F\sin\alpha$$

where $r$ is the distance between $P$ and the application point of the force $F$; $\alpha$ is the angle between the vectors $\vec r$ and $\vec F$.

So if you want to calculate the moment of the $5N-force$ on the left, for instance, we have: $F=5N$, $r$ is the base of the dashed right triangle and you can evaluate it by $r=4\sin 55$ (just an application of trigonometry), $\alpha=90$ ($\vec r$ and $\vec F$ are normal), so $\sin90=1$. Replacing in the formula:

$$M=5\cdot 4\sin55\cdot1=20\sin55$$

Similarly for the others forces.