1. ## Resultant Force Problem

Hello, I have a trig/physics problem from my foundations of engineering class. If this is in the wrong forum please let me know. Here it is.

Two forces P and Q with respective magnitudes 100N and 200N are are applied to the upper right corner of a crate. The sum of the two forces is a horizontal force with magnitude 250N. The the angles that P and Q make with their sum. In the picture force P is on top and force Q is below the resultant horizontal force.

So far I have taken the scalar form of the two forces and summed them to their resultant. Im calling the angles p, and q for their respective forces.

100N*COS(p)i + 100N*SIN(p)j

+200N*COS(q)i - 200N*SIN(q)j

= 250Ni

From here I am not sure where to proceed. I tried summing the individual components but I did not make any progress. Any help would be appreciated.

2. You have two equation in two unknowns:

$\displaystyle 100 \cos(p) + 200 \cos(q) = 250$
$\displaystyle 100 \sin(p) - 200 \sin(q) = 0$

From the 2nd equaton:

$\displaystyle \sin(p) = 2 \sin(q)$, so
$\displaystyle \cos(p) = \sqrt{1 - \sin^2p} = \sqrt{1 - 4 \sin^2q}$

Put that back into the first, and solve for cos(q):

$\displaystyle 100 \sqrt {1-4 \sin^2q} = 250- 200 \cos (q)$

Square both sides, and rearrange:

$\displaystyle 1-4 \sin^2q = 2.5^2 - 10 \cos(q) + 4 \cos^2q$
$\displaystyle 0 = 5.25 - 10 \cos(q) + 4(cos^2q + sin^2q) = 9.25 - 10 \cos(q)$
$\displaystyle q = Arccos (0.925)$

3. Thank You!! That was very helpful. I thought there may have been a identity lurking in there. Your a life saver!

4. Hello, repete!

$\displaystyle \text{Two forces }P\text{ and }Q\text{ with respective magnitudes 100N and 200N}$
. . $\displaystyle \text{are applied to a point.}$

$\displaystyle \text{The sum of the two forces is a horizontal force with magnitude 250N.}$

$\displaystyle \text}Find the angles that }P\text{ and }Q\text{ make with their sum.}$

$\displaystyle \text{(In the diagram, force }P\text{ is over and force }Q\text{ is under}$
. . $\displaystyle \text{the resultant horizontal force.)}$
Code:
                  P
o
*    *
100  *         *   200
*              *
*                   *
*  p                     *
O o - - - - - - - - - - - - - - o R
*  q      250            *
*                   *
*              *
200   *         *  100
*    *
o
Q

We have: .$\displaystyle \begin{array}{cccccccccc} |\overrightarrow{OP}| \;=\;100 && p \:=\:\angle POR \\ \\[-4mm] |\overrightarrow{OQ}| \;=\;200 && q \;=\;\angle QOR \\ \\[-4mm] |\overrightarrow{OR}| \;=\;250 \end{array}$

$\displaystyle \text{In }\Delta OPR\!:\;\; \cos p \:=\:\dfrac{100^2 + 250^2 - 200^2}{2(100)(250)} \;=\; 0.65$

Hence: .$\displaystyle \angle p \;\approx\;49.46^o$

$\displaystyle \text{In }\Delta OQR\!:\;\;\cos q \:=\:\dfrac{200^2 + 250^2 - 100^2}{2(200)(250)} \;=\;0.925$

Hence:. . $\displaystyle \anglen q \;\approx\;22.33^o$