Mechanics Question of a particle

[_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

(b) the direction of Q, giving your answer as a bearing.

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_

 

[nokia8650]

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

 

[_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_

 

[_Mayday_]

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

 

[_Mayday_]

Hey Nokia, is my answer correct?

 

[Hootenanny]

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_

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?

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

I'm assuming that you mean in the case of a right-triangle, in which case your angle in $\pi/2$.

 

[_Mayday_]

For P (0i + 7j)

For R (8.6i - 5j)

 

[Hootenanny]

Ahhh, I've just figured out what your doing! However, the angle

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

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

$a=2.79N$

Should be 120^o not 30 ^o. The angle that you take should be opposite the side you are trying to determine.

 

[_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!

 

[Hootenanny]

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!

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.

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 120^o 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.

 

[_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.

 

[_Mayday_]

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

 

[Hootenanny]

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.

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?

 

[_Mayday_]

Here we go.

 

[_Mayday_]

Is that correct?

 

[Hootenanny]

Is that correct?

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?

 

[_Mayday_]

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

 

[Hootenanny]

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

A pleasure as always _Mayday_

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

 

[_Mayday_]

!(Solved)!