1. ## finding the angle

I'm having trouble with this question. I was thinking of using Lami's theorem, but its not helping too much.

A body of mass 5 kg is in equilibrium when it is acted upon by three concurrent coplanar forces P, Q and R as shown in the diagram below. P = 10 newtons and Q = 20 newtons. The angle between P and Q is 170 degrees and the angle between P and R is $\theta$ degrees

Find $\theta$.

2. Originally Posted by scorpion007
I'm having trouble with this question. I was thinking of using Lami's theorem, but its not helping too much.

A body of mass 5 kg is in equilibrium when it is acted upon by three concurrent coplanar forces P, Q and R as shown in the diagram below. P = 10 newtons and Q = 20 newtons. The angle between P and Q is 170 degrees and the angle between P and R is $\theta$ degrees

Find $\theta$.
Resolve all forces into components in the direction of Q and normal to Q
(with lets say positive upwards). Then as the body is in equilibrium the
sum of each components is zero. so

Put phi = theta-10 degrees

p_1 = P(parrallel to Q) = -10 cos(10)
p_2 = P(orthog to Q) = -10 sin(10)

r_1 = R(parrallel to Q) = -R cos(phi)
r_2 = R(orthog to Q) = R sin(phi)

then:

p_1 + r_1 = Q = -20
p_2 + r_2 = 0.

Which is a pair of simultaneous non-linear equationswhich you will have to
solve for phi and R.

RonL

3. hmm, what i have so far is these equations:

$10.15 - R\cos(\phi) = 0$
$-1.74 - R\sin(\phi) = 0$

I'm not too sure how to eliminate one of the pronumerals from here...
Also, im not 100% i got up to here correctly. Anyone care to check? lol sorry.

4. Originally Posted by scorpion007
hmm, what i have so far is these equations:

$10.15 - R\cos(\phi) = 0$
$-1.74 - R\sin(\phi) = 0$

I'm not too sure how to eliminate one of the pronumerals from here...
Also, im not 100% i got up to here correctly. Anyone care to check? lol sorry.
Two comments. Following CaptainBlack's notation the second equation should be:
$-1.74 + R\sin(\phi) = 0$

The second comment is that would keep a few extra digits on your numbers to stave off rounding errors, but that you otherwise calculated them correctly.

Solve the equations thusly:
$R\cos(\phi) = 10.15$

$R\sin(\phi) = 1.74$

Now divide the second by the first:

$\frac{R sin( \phi )}{R cos( \phi )} = \frac{1.75}{10.15}$

$tan( \phi ) = 0.17241$

Which you can solve for $\phi$.

You can now choose any way you like to get R, but the method I recommend is:

$\left ( R sin( \phi ) \right )^2 + \left ( R cos( \phi ) \right )^2 = 1.74^2 + 10.15^2$

$R^2 sin^2( \phi ) + R^2 cos^2( \phi ) = 106.0501$

$R^2 (sin^2( \phi ) + cos^2( \phi ) ) = 106.0501$

$R^2 = 106.0501$

Now you can solve for R.

(The reason I choose this method over the others is that this equation uses fewer values that have been calculated (as opposed to given), which reduces rounding error.)

-Dan

5. ahh, thank you very much!

i was concerned with some values for a sec, but i realised i made some stupid arithmetic mistakes.

EDIT: Although, i do have a minor question, CaptainBlack, why did you have p_1 and p_2 negative? shouldn't the sum of the (parallel to Q) components of R and P = 20?

6. Originally Posted by scorpion007
ahh, thank you very much!

i was concerned with some values for a sec, but i realised i made some stupid arithmetic mistakes.

EDIT: Although, i do have a minor question, CaptainBlack, why did you have p_1 and p_2 negative? shouldn't the sum of the (parallel to Q) components of R and P = 20?
The way the axes are chosen Q's in these axes is (+20,0), the other forces
components parallel to Q are in the opposite direction so the - signs.

RonL

7. Hmm, but if i translate your first equation, wouldnt this be it?

$-29.848 - R\cos(\phi) = 0$ ?

which is different to what I got.

8. Originally Posted by scorpion007
Hmm, but if i translate your first equation, wouldnt this be it?

$-29.848 - R\cos(\phi) = 0$ ?

which is different to what I got.
Mistake on my part should equal -20

That is the sum of the components of the forces with the correct signs should be 0 for equilibrium to prevail.

RonL

9. that's how i understand it too. Which is what i did. Thanks for the clarification anyway.