How to calulate angle?

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February 18th 2011, 03:14 PM
Barthayn

How to calulate angle?

I am working out how to solve for angles during vectors problems, and the only way I can solve for the angles is by trying all possibilities and seeing which ones is the most realistic way of solving for it. So are there any patterns or laws that one can know for knowing which angle to use to subtract it from 180^o? The problem that got me guessing is below:

6. Each of three brothers has tied a rope to a buoy floating on a lake. Paco pulls towards the east with a force of 40 N. Louis pills towards the southwest with a force of 30 N. What forces should Pepe exert to hold his brothers' efforts in equilibrium?

The answer is 28.3 N at 131^o. If you draw a diagram the angle that you need to use to subtract from 180^o is between the 40 N force and the 28.3 N force.
February 18th 2011, 05:45 PM
skeeter

Quote:

Originally Posted by Barthayn

I am working out how to solve for angles during vectors problems, and the only way I can solve for the angles is by trying all possibilities and seeing which ones is the most realistic way of solving for it. So are there any patterns or laws that one can know for knowing which angle to use to subtract it from 180^o? The problem that got me guessing is below:

6. Each of three brothers has tied a rope to a buoy floating on a lake. Paco pulls towards the east with a force of 40 N. Louis pills towards the southwest with a force of 30 N. What forces should Pepe exert to hold his brothers' efforts in equilibrium?

The answer is 28.3 N at 131^o. If you draw a diagram the angle that you need to use to subtract from 180^o is between the 40 N force and the 28.3 N force.

let $P_x$ and $P_y$ be the components of Pepe's force

$\sum{F_x} = 0$

$40 + 30\cos(-135^\circ) + P_x = 0$

$P_x = -30\cos(-135^\circ) - 40$

$\sum{F_y} = 0$

$0 + 30\sin(-135^\circ) + P_y = 0$

$P_y = -30\sin(-135^\circ)$

$|P| = \sqrt{P_x^2 + P_y^2}$

$\theta = \arctan\left(\dfrac{P_y}{P_x}\right) + 180^\circ$
February 18th 2011, 07:27 PM
Barthayn

Not sure what you did, however, since my class is a grade 12 university class, it can't be that hard. I was asking how does one know that correct angle within the triangle to subtract it from 180^o. Anyone know other than guess work?
February 18th 2011, 08:06 PM
Soroban

Hello, Barthayn!

Quote:

$\text{6. Each of three brothers has tied a rope to a buoy floating on a lake.}$
$\,A\text{ pulls towards the east with a force of 40 N.}$
$\,B\text{ pulls towards the southwest with a force of 30 N.}$
$\text{What force should }C\text{ exert to hold the buoy in equilibrium?}$

$\text{The answer is: }\, 28.3\,N\text{ at }131^o.$

Code:

C o * * * * o * * * * * o A * O 40 30 * * o B

We have: . $\begin{Bmatrix}\overrightarrow{OA} &=& \langle 40,0\rangle \\<br /> \overrightarrow{OB} &=& \langle\text{-}15\sqrt{2},\text{-}15\sqrt{2}\rangle \\ \overrightarrow{OC} &=& \langle x,y\rangle \end{Bmatrix}$

We want: . $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} \:=\:\overrightarrow{0}$

. . $\langle 40,0\rangle + \langle\text{-}15\sqrt{2},\text{-}15\sqrt{2}\rangle + \langle x,y\rangle \;=\;\langle 0,0,0\rangle$

. . $\langle x + 40 - 15\sqrt{2},\;y - 15\sqrt{2}\rangle \;=\;\langle 0,0\rangle$

$\text{Then: }\;\begin{Bmatrix} x+40 - 15\sqrt{2}\:=\:0 & \Rightarrow &x \:=\:15\sqrt{2}-40 \\ y - 15\sqrt{2} \:=\:0 & \Rightarrow & y \:=\:15\sqrt{2} \end{Bmatrix}$

$|\overrightarrow{OC}| \;=\;\sqrt{x^2+y^2} \;=\;\sqrt{\left(15\sqrt{2}-40\right)^2 + \left(15\sqrt{2}\right)^2}$

. . . $=\;28.33626167 \;\approx\;\boxed{28.3\text{ N}}$

$\tan\theta \:=\:\dfrac{y}{x} \;=\;\dfrac{15\sqrt{2}}{15\sqrt{2} - 40} \;=\;-1.129154902$

. . $\theta \;=\;131.5286818 \;\approx\;\boxed{131.5^o}$
February 18th 2011, 08:29 PM
pflo

Quote:

Originally Posted by Barthayn

Not sure what you did, however, since my class is a grade 12 university class, it can't be that hard. I was asking how does one know that correct angle within the triangle to subtract it from 180^o. Anyone know other than guess work?

All of Paco's force is pulling in the positive x direction (east). Part of Louis' force is in the negative x direction (southwest is half way between the negative y direction and the negative x direction, which is 135 degrees from the postive x direction). If you add these forces together (Louis' contribution is negative, so adding give you a difference) you get the total force along the east-west axis. Part of Louis' force is in the negative y direction and this is the only force along the north-south axis.

If you combine these two forces into the total force, Pepe's force must be equal in magnitude and in the opposite direction (your "subtract from 180" dilemma). Draw the component vectors (the ones along the axes) head-to-tail on a coordinate plane with the first starting at the origin, then draw a vector from the origin to where they end. Look at the trig ratios of this right triangle to figure out the angle. See Spider's reply to check out the equations and actual numbers involved.