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I have worked through the following problem and I am not sure, why this can not be solved using a dot product:

Edit: Text of the problem: Split the forces $F_1$ and $F_2$ into their components along the $u$ and the $v$ axis.

enter image description here

The Question is: What part of $F_2$ is along the $v$ axis ? Given $$F_2=500N, \alpha_2=45°, \beta=70° $$ The problem can be solved using the law of sines: $$ \frac{F_2}{\sin\beta}=\frac{F_{2,v}}{\sin(180-\beta-\alpha_2)} $$ and gives you 483N .

However I first tried using:

$$ F_2\cdot\cos(\alpha_2) = 353.6N $$

This would be the procedure if the axis were in a $90°$ angle to each other.

I thought my wrong first result might be owed to the missing unitcircle in the background that is reasoning the classical mechanical projections of vectors:

$$ \vec{F}= |F|\cos(\alpha)\vec{u_x}+ |F|\sin(\alpha)\vec{u_y} $$

But then I thought about the dot product, using a unit vector of the $v$-axis: $$|F_2||u_v|\cdot\cos(\alpha_2) = F_{2,v} $$

Where I project $F_2$ on the axis of $v$ without scaling this projection. But this would give me my wrong result again.

Leading to my question:

What am I missing here: Why am I not able to use the last two methods to get to the correct result ?

My first uneducated guess: If you split a Vector into two components along a $90°$ axis, you receive a right triangle, the two components are in a $90°$ angle to each other.

In my example, the force has two components which are not in a $90°$ angle to each other then we can not treat them this way. Same goes for the dot product since I only receive a projection shorter than the vector lying along the $v$ axis.

Am I right here?

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    $\begingroup$ Could you share the full text of the problem statement? $\endgroup$
    – user
    Apr 19 at 8:54
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    $\begingroup$ You are projecting F2 perpendicularly onto the v-axis. It seems however that the problem (which you have not fully stated) expects you to split F2 into components along the u and v axes. In that case you would have to project F2 along the direction of the u-axis onto the v-axis. In that way you take away the u-component of F2 to leave you with the v-component. $\endgroup$
    – Jaap Scherphuis
    Apr 19 at 9:13

1 Answer 1

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The $u$ and the $v$ axis are not perpendicular then the component of $\vec{F_2}$ along $v$ according to the following sketch

enter image description here

with $F_2=\|\vec{F_2}\|$ is indeed given by

$$F_{2,v}=F_2\frac{\sin(180-\beta-\alpha_2)}{\sin\beta}$$

and, assuming $\hat u$ directed on the right

$$F_{2,u}=-F_2\frac{\sin \alpha_2}{\sin\beta}$$

in such way that

$$\vec{F_2}= F_{2,u} \;\hat u+F_{2,v} \;\hat v$$

This kind of decomposition is different from the projection of $\vec{F_2}$ along the direction $\hat v$ which is indeed given by

$$F^{\text{proj}}_{2,v}=\vec{F_2}\cdot \hat v = F_2 \cos \alpha_2$$

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