# Why exactly does dot product not work if the axis are not 90°

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. 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.

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?

• Could you share the full text of the problem statement?
– user
Apr 19 at 8:54
• 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. Apr 19 at 9:13

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

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