I am in a Applied Mechanics for Engineering Course and need some help with vectors. I am struggling with this question and would like some help. How do I solve this? I am also very confused with how the Force of 68N shows an angle of (x=15, y=8) and not the angle. Can someone please explain this to me? I have drawn a picture in paint:

Question:
Determine the resultant of the force system shown

Any help would be greatly appreciated

2. Hello system32
Originally Posted by system32
I am in a Applied Mechanics for Engineering Course and need some help with vectors. I am struggling with this question and would like some help. How do I solve this? I am also very confused with how the Force of 68N shows an angle of (x=15, y=8) and not the angle. Can someone please explain this to me? I have drawn a picture in paint:

Question:
Determine the resultant of the force system shown

Any help would be greatly appreciated
First, the meaning of the $(x=15, y=8)$ direction is that the force of 68 N makes an angle $\arctan(\tfrac{8}{15})$ with the $x$-direction. You can work out this angle in degrees if you like, but you don't need to. The hypotenuse of the triangle is $\sqrt{8^2+15^2}=17$ exactly. So the sine of the angle is $\frac{8}{17}$ and its cosine is $\frac{15}{17}$.

Now, to find the resultant force, add up their $x-$ and $y-$components, taking $x$ as positive to the right, and $y$ positive upwards, using sines and cosines as appropriate. I'll do the $x$ one for you:

$X=-80\cos65 -50\sin15+68\times\frac{15}{17}=13.25$, where $X$ is the $x-$component of the resultant force.

Do the same for $Y$, the sum of the $y$-components.

Then, using a right-angled triangle with sides $X$ and $Y$ work out the magnitude of the resultant, using $\sqrt{X^2+Y^2}$ and the angle the resultant makes with the $x$-direction, which is $\arctan\Big(\frac{Y}{X}\Big)$.

3. thank you