How to show force R^2=(Q-24)^2 + 7^2

[physicshawk]

A particle is in equilibrium under the action of three horizontal forces P, Q and R. The magnitudes of P, Q and R are 25 N, Q N, and R N respectively, and the cosine of the angle between P and Q is -0.96.

Show that R^2 = (Q - 24)^2 + 7^2.

Hence find:

(a) the least possible value of R,

(b) the corresponding angle between Q and R.

My working:
R^2 = P^2 + Q^2 - 2(P)(Q) Cos (theta)
R^2 = 25^2 + Q^2 - 2(25) (Q) (-0.96)
R^2= 25^2 + Q^2 + 48 Q
R^2 = Q^2 + 48Q + 25^2
Now, I was trying completing the squares but... i don't know how to proceed. Please Help.

 

[WillemBouwer]

If you have an equation in mathematics: ax^2 + bx +c = 0 Then how do we simplify to get the roots? do you remember the term that c = (b/2)^2 for it to be the same roots. (x+c_new)^2... Try doing this... if you get the new c put it into the equation but remember to subtract it again... Really trying hard not to give the answer...
so ax^2 +bx +c_old = R^2 becomes ax^2 +bx +c_new -c_new +c_old = R^2... Try this and I'll help you further...

 

[ehild]

A particle is in equilibrium under the action of three horizontal forces P, Q and R. The magnitudes of P, Q and R are 25 N, Q N, and R N respectively, and the cosine of the angle between P and Q is -0.96.

Show that R^2 = (Q - 24)^2 + 7^2.

Hence find:

(a) the least possible value of R,

(b) the corresponding angle between Q and R.

My working:
R^2 = P^2 + Q^2 - 2(P)(Q) Cos (theta)

The "minus" in red should be "+".

ehild

 

[physicshawk]

If you have an equation in mathematics: ax^2 + bx +c = 0 Then how do we simplify to get the roots? do you remember the term that c = (b/2)^2 for it to be the same roots. (x+c_new)^2... Try doing this... if you get the new c put it into the equation but remember to subtract it again... Really trying hard not to give the answer...
so ax^2 +bx +c_old = R^2 becomes ax^2 +bx +c_new -c_new +c_old = R^2... Try this and I'll help you further...

Thanks. So now we are doing this.

R^2= (Q)^2 + 2(Q)24 + (24)^2 - (24)^2 + 625
R^2= (Q + 24)^2 -576 + 625
R^2 = (Q + 24)^2 + 49
R^2 = (Q + 24)^2 + 7^2
it looks like we are almost there. But why is it (Q-24) in the required proof
and not (Q + 24) as i have got in my proof.
Am i making some mistake? please suggest. Thanks

 

[ehild]

Read #3.

ehild

 

[physicshawk]

Read #3.

ehild

I looked up in 2 sites, "wikipedia" and "math is fun" for Cosine Law. There it is saying
c^2=a^2 + b^2 -2ab cos(theta)

Am i missing out on something? Please suggest...

 

[ehild]

See the figure in #3. You applied the Law of Cosines to the triangle with sides P and Q and enclosed angle theta, but the third side of that triangle is not R.
In the triangle with sides of length P, Q, R, the angle enclosed by P and Q is 180-theta.

ehild

 

[physicshawk]

See the figure in #3. You applied the Law of Cosines to the triangle with sides P and Q and enclosed angle theta, but the third side of that triangle is not R.
In the triangle with sides of length P, Q, R, the angle enclosed by P and Q is 180-theta.

ehild

But this is an equilibrium problem. I understand by this that the vector triangle will be closed. Which means that the heads of two arrows will not meet. But in the diagram you have given, the two arrows' heads are meeting. Is there an error? Please suggest!

 

[ehild]

I don't quite follow you. I have shown the parallelogram method of addition. The sum of vectors R and Q is -P. R+Q +P=-P+P=0. You have a triangle (dotted) with sides P, Q, R and angle 180-θ, opposite to the side R.
ehild