# Thread: Working out resultant??

1. ## Working out resultant??

How do I work out the resultant of the following?? please see attachments hope someone can help. I know how to do them (kind of) but gone blank

2. Originally Posted by juangambino
How do I work out the resultant of the following?? please see attachments hope someone can help. I know how to do them (kind of) but gone blank
For the first one, you want to put everything in x and y forces, so:

$\displaystyle x = 100cos(30)-25cos(40)$
$\displaystyle y = 25sin(40)-100sin(30)$

Can you solve for x and y? To figure out the new angle, use tangent:

$\displaystyle tan(\frac{y}{x})=\theta$

3. ## thank you

thank you. I already worked them out in x and y then I didnt subtract them lol must have been asleep this morning. Thank you soo much

4. Originally Posted by juangambino
How do I work out the resultant of the following?? please see attachments hope someone can help. I know how to do them (kind of) but gone blank
Hello,

to #1:

1. I've added a sketch

2. use cosine rule: The resulting force divides the parallelogram of forces into 2 congruent triangles. 2 legs of the triangle are the known forces and the included angle is $\displaystyle \theta = 10^\circ$. Therefore the resulting force is:

$\displaystyle r = \sqrt{10^2+2.5^2-2 \cdot 10 \cdot 2.5 \cdot \cos(10^\circ) }\approx 7.55 \ kN$

3. to calculate the angle between the force of 10 kN and the resulting force $\displaystyle F_r$ use cosine rule again:

$\displaystyle \cos(\theta)=\frac{c^2-a^2-b^2}{-2 \cdot a \cdot b}$.......Plug in all values you know:

$\displaystyle \cos(\theta)=\frac{2.5^2-10^2-7.55^2}{-2 \cdot 10 \cdot 7.55} \approx 0.998361~\implies~\theta\approx 3.281^\circ$

In my drawing I've measured the angle between the x-axis and the resulting force.