• Sep 20th 2009, 12:26 PM
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
http://i37.tinypic.com/2mrf6g6.png

Any help would be greatly appreciated(Happy)
• Sep 20th 2009, 10:52 PM
Hello system32
Quote:

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
http://i37.tinypic.com/2mrf6g6.png

Any help would be greatly appreciated(Happy)

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

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

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

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

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