# mechanics question, resultant forces.

• Feb 13th 2008, 09:21 AM
Oranges&Lemons
mechanics question, resultant forces.
Two forces P and Q act on a particle. The force P has magnitude 7N and acts due north. The resultant of P and Q is a force of magnitude 10N acting in a direction with bearing 120 degrees. Find

i) the magnitude of Q

any help would be much appreciated, thanks :)
• Feb 13th 2008, 09:44 AM
colby2152
Quote:

Originally Posted by Oranges&Lemons
Two forces P and Q act on a particle. The force P has magnitude 7N and acts due north. The resultant of P and Q is a force of magnitude 10N acting in a direction with bearing 120 degrees. Find

i) the magnitude of Q

any help would be much appreciated, thanks :)

Use some trig...

$7+sin(\theta)Q=10sin(120)$
$cos(\theta)Q=10cos(120)$

Solve for the system of equations. Basically, you want to split all the forces into x's (top equation) and y's (bottom equation). Use the sine of an angle multiplied by the force to get the vertical force; use cosine for horizontal directions.
• Feb 13th 2008, 10:18 AM
earboth
Quote:

Originally Posted by Oranges&Lemons
Two forces P and Q act on a particle. The force P has magnitude 7N and acts due north. The resultant of P and Q is a force of magnitude 10N acting in a direction with bearing 120 degrees. Find

i) the magnitude of Q

any help would be much appreciated, thanks :)

Here is another attempt:

1. draw a rough sketch

2. You can calculate the dotted green line by Cosine rule. The length of this line correspond with the magnitude of Q:

$|Q| = \sqrt{7^2+10^2-2 \cdot 7 \cdot 10 \cdot \cos(120^{\circ})} \approx 14.798648...\ N$

3. To calculate the bearing of Q you have to add the angle $\alpha_2 = \alpha$ to the 120° = bearing of the resultant force R:

$\cos(\alpha)=\frac{7^2-10^2-14.799^2}{-2\cdot 10 \cdot 14.799}$

This will give you $\alpha \approx 24.182^\circ$

And therefore the bearing of Q is 144.182°
• Feb 13th 2008, 12:37 PM
Oranges&Lemons
thankyou so much!