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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:

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.

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    Solution to what? The diagram does not make the actual problem clear. Are you calculating a reaction force, a turning moment or what? – 2017-01-14
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    Hi @Paul, the question reads `find the sum of the moments about P of the forces shown in the following diagram`. I'll add that to the post, apologies. – 2017-01-14
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    Do you know the definition of the turning moment or is your problem the trigonometry calcuations? – 2017-01-14
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    @Paul, I know the definitions but what I don't understand is why they've selected the values they have for the distances and degrees. – 2017-01-14

1 Answers 1

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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.

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    What I don't currently understand is why the force `3N` (bottom left) is omitted. In the anticlockwise direction `3 x 4 x Sin55` (why is it not Sin35?). It also doesn't seem like moments are taken about point `P`, there's still a distance from `P` to where the forces are taken? Thanks. – 2017-01-15
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    The force $3N$ has zero moment because $\alpha=180$ and $\sin180=0$, in fact $\vec r$ goes from $P$ to the application point of the force (the left vertex of the triangle), so the force (it goes from the left vertex towards $P$) is anti-parallel to $\vec r$. – 2017-01-15
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    In the anti clockwise direction we have: the force $3N$ is horizontal to the left, $\vec r$ is the vertical leg of the triangle, so $\vec r\perp\vec F$ and $\alpha =90$, hence $\sin90=1$. Now, how do we get $|\vec r|$? We have to apply trigonometry to the right triangle: $|\vec r|=4\cos55$. Thus we get $M=3\cdot(4\cos55)\cdot\sin90$ – 2017-01-15
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    Remember that you have chosen $P$ as pole, so you have to evaluate any moment respect to $P$. – 2017-01-15
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    Doesn't `(5 x 4 x sin55) - (3 x 4 x sin35 + 5 x 2 x sin55)` equal the same? I don't understand why the example shows the way it does. – 2017-01-15
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    Yes it is right: $clockwise - anti clockwise =1,31Nm$ and you can put $\sin35$ instead of $\cos55$ because $\sin35=\cos55$ ($35$ and $55$ are complementary angle). – 2017-01-15