Here is the problem I was given I have no clue how to get the answer's given.I have a test in three weeks on the first four chapters and I am lost like a rat in a maze.

Can someone help?

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- Mar 1st 2006, 03:27 PMjaycas21I am so lost
Here is the problem I was given I have no clue how to get the answer's given.I have a test in three weeks on the first four chapters and I am lost like a rat in a maze.

Can someone help? - Mar 1st 2006, 03:50 PMThePerfectHackerQuote:

Originally Posted by**jaycas21**

- Mar 1st 2006, 04:22 PMThePerfectHackerQuote:

Originally Posted by**jaycas21**

**Vector Sum of the Force Vectors is a Zero Vector**. This is a rule for equilibrium. Maybe you remember that you can convert any vector in a linear combination of unit vectors-meaning expressing them in $\displaystyle \vec i,\vec j$ form. Let $\displaystyle F_1$ represent the magnitude of this vector. Then its x-coordinate is is found throught trigonometry. Since the x-coordinate forms a right triangle we have that $\displaystyle \cos 30^o=\frac{x}{F_1}$ thus, $\displaystyle x=F_1\cos 30^o=F_1\frac{\sqrt{3}}{2}$because we know the cosine of this angle. Thus, the y-coordinate is also found through trigonometry, since it forms a right angle, $\displaystyle \sin 30^o=\frac{y}{F_1}$ thus, $\displaystyle y=F_1\sin 30^o=F_1\frac{1}{2}$. Thus, the vector $\displaystyle \vec F_1$ can be expressed as $\displaystyle F_1\frac{\sqrt{3}}{2}\vec i+F_1\frac{1}{2}\vec j$. By similar reasoning and the fact that $\displaystyle \alpha=45^o$. We arrived at $\displaystyle \vec F_2=-F_2\frac{\sqrt{2}}{2}\vec i+F_2\frac{\sqrt{2}}{2}\vec j$ (because $\displaystyle \cos 45^o=\sin 45^o=\frac{\sqrt{2}}{2}$ also important that the sign in front of $\displaystyle \vec i$ is negative because we are in the 2nd quadrant thus 135 degrees actually and the cosine has a negative sign there unlike the sine which remains positive). Lastly, the thrid vector is the weight of the ring which is $\displaystyle -19.6\vec j$ because it only goes down.

Now we state that $\displaystyle \sum \vec F_k=\vec 0$ meaning the vector sum. Which in this case is,

$\displaystyle F_1\frac{\sqrt{3}}{2}\vec i+F_1\frac{1}{2}\vec j

-F_2\frac{\sqrt{2}}{2}\vec i+F_2\frac{\sqrt{2}}{2}\vec j-19.6\vec j=\vec 0$

Organize them,

$\displaystyle \left(F_1\frac{\sqrt{3}}{2}-F_2\frac{\sqrt{2}}{2}\right)\vec i+\left(F_1\frac{1}{2}+F_2\frac{\sqrt{2}}{2}-19.6\right)\vec j=\vec 0$

Now, this is important that you can only express the zero vector when both the unit vector are equal to zero themselves.

Thus, a system of linear equations

$\displaystyle \left\{\begin{array}{cc}F_1\frac{\sqrt{3}}{2}-F_2\frac{\sqrt{2}}{2}&=0\\F_1\frac{1}{2}+F_2\frac{ \sqrt{2}}{2}-19.6&=0\end{array}\right$

just simplify the second one, to get,

$\displaystyle \left\{\begin{array}{cc}F_1\frac{\sqrt{3}}{2}-F_2\frac{\sqrt{2}}{2}&=0\\F_1\frac{1}{2}+F_2\frac{ \sqrt{2}}{2}&=19.6\end{array}\right$.

From here I think you can handle it, use any method you wish. I would recommend using determinants to solve this system. - Mar 1st 2006, 04:29 PMtopsquarkQuote:

Originally Posted by**jaycas21**

In the x direction:

mg is straight down, so has no x-component.

F1 is acting up and to the right at an angle of 30 degrees, so

F1x=F1*cos(30).

F2 is acting up and to the left, so F2x will be negative:

F2x=-F2*cos(45).

Thus $\displaystyle \sum F_x=F_1cos(30)-F_2cos(45)=0$.

In the y direction we have something similar. F1 and F2 are going to have positive y components and mg will be negative. So:

$\displaystyle \sum F_y=F_1sin(30)+F_2sin(45)-mg=0$, where mg was given in the first part.

Now we need to solve for F1 and F2. This is a system of 2 equations in 2 unknowns, so we can do it. In Physics the best way to start this is with the substitution method. Solve one of the equations for, say the first equation for F2 in terms of F1 and plug this into the second equation. Then you can solve for F1. etc.

-Dan - Mar 1st 2006, 06:34 PMThePerfectHackerQuote:

Originally Posted by**topsquark**

- Mar 1st 2006, 09:07 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

The gravitational force is $\displaystyle m.g$, where $\displaystyle g$ is the acceleration

due to gravity (it is independent of mass as Galileo tells us). At the earth's

surface $\displaystyle g \approx 9.81 \mathrm{m/s^2}$, so the gravitational force

(ie the weight) is $\displaystyle mg \approx 2 \times 9.81 =19.62\ \mathrm{Newtons}$

RonL - Mar 2nd 2006, 03:56 AMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

-Dan - Mar 2nd 2006, 03:58 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan - Mar 2nd 2006, 10:27 AMCaptainBlackQuote:

Originally Posted by**topsquark**

- Mar 2nd 2006, 10:39 AMCaptainBlackQuote:

Originally Posted by**topsquark**

In Hackers post he is using $\displaystyle \frac{m}{s^2}$ to represent metres per second, but he has already

introduced $\displaystyle m$ as the mass.

Now I believe the convention is the italic letters like $\displaystyle s$ represents a variable or parameter,

in which case the units should be non-italic letters like $\displaystyle \mathrm{s}$, or in this case we should

write metres per second as $\displaystyle \mathrm{m/s^2}$ or $\displaystyle \mathrm{ms^{-2}}$.

RonL - Mar 2nd 2006, 11:43 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan