# Mechanics Question of a particle

1. May 19, 2008

### _Mayday_

[SOLVED] Mechanics Question

Hey.

I have a quick question relating to what I think is resolving forces. I am stuck at the first hurdle, I have no idea how to approach it.

Question

The forces P and Q act on a particle. The force P has a magnitude of 7N and acts due north. The resultant of P and Q is a force of magnitude 10N acting in a direction with bearing 120. Find:

(a) the magnitude of Q

Attempt

No idea, but as there isn't going to be a right angle in this case, I would assume this question would involve the cosine rule? Could someone please give me a nudge in the right direction.

Thanks
_Mayday_

2. May 19, 2008

### nokia8650

Resolve the resultant into its components (i and j).

3. May 19, 2008

### _Mayday_

Okay, I have given it a bash. I don't know if someone has responded as I haven't refreshed the page.

Using the cosine rule.

$a^2=b^2+c^2-2bc\cosA$

$a^2=7^2+10^2-2(70)\cos30$

$a=2.79N$

Now for the second bit I just calculated the angle using trig and got 105.57 degrees. Now I simply subtract that from 360 which gives me 284 degrees.

Hope this is somewhat along the right lines!

_Mayday_

4. May 19, 2008

### _Mayday_

Probably a silly question, but in Trig, what happens if your opposite side is also the hypotenuse? What do you do?

5. May 19, 2008

### _Mayday_

Hey Nokia, is my answer correct?

6. May 19, 2008

### Hootenanny

Staff Emeritus
Hey _Mayday_,

Not sure what you're doing here, but you're certainly not following nokia's hints. Nokia suggested that you resolve the to two forces into perpendicular components. So for example,

$$\mathbold{P} = 0\mathbold{\hat{i}} + 7\mathbold{\hat{j}}$$

Can you do something similar for Q?
I'm assuming that you mean in the case of a right-triangle, in which case your angle in $\pi/2$.

7. May 19, 2008

### _Mayday_

For P (0i + 7j)

For R (8.6i - 5j)

8. May 19, 2008

### Hootenanny

Staff Emeritus
Ahhh, I've just figured out what your doing! However, the angle
Should be 120o not 30 o. The angle that you take should be opposite the side you are trying to determine.

9. May 19, 2008

### _Mayday_

I actually think it should be 60. 7N acts vertically upwards. The resultant acts at 180 - 60, so when I add that vector onto the one of 7N you get 60.

Correct me if I am wrong though! I just drew it out and it looks good to go!

10. May 19, 2008

### Hootenanny

Staff Emeritus
A bearing of 120 degree means, 120 degrees clockwise from North. So start by drawing a vertical line upwards, this is P. Now from the point where you started to draw P, draw a second line at 120o clockwise measured from your vertical line, this is the resultant vector. Now draw a third line joining the ends of the previous two vectors, this is Q.

11. May 19, 2008

### _Mayday_

If I put arrow heads on the forces, do they not need to follow on? The way you have said it means that the arrow heads will not follow on.

12. May 19, 2008

### _Mayday_

If the cosine rule works well, then I would prefer to do it this way.

13. May 19, 2008

### Hootenanny

Staff Emeritus
Yes the do follow on. P starts at the origin, Q starts from the tip of P and points in a generally SW direction. The resultant force then starts at the origin and points towards the tip of Q. Could you perhaps post your diagram?

14. May 19, 2008

### _Mayday_

Here we go.

#### Attached Files:

• ###### PHSO.bmp
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15. May 20, 2008

### _Mayday_

Is that correct?

16. May 20, 2008

### Hootenanny

Staff Emeritus
Does it look right to you? If you have two forces acting on a particle, one acting left and one acting upwards; how can the resultant force be acting downwards and rightwards?

Last edited: May 20, 2008
17. May 20, 2008

### _Mayday_

Bah, dammit I've got it now! Cheers Hoot.

18. May 20, 2008

### Hootenanny

Staff Emeritus
A pleasure as always _Mayday_

Don't forget to mark the thread as 'solved' when you're done.

19. May 20, 2008

!(Solved)!