# I am so lost

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• Mar 1st 2006, 03:27 PM
jaycas21
I 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 PM
ThePerfectHacker
Quote:

Originally Posted by jaycas21
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?

1)Newton's second law states that $F=ma$ where $F$ is force in newtons, $m$ is mass in kilograms and $a$ is the acceleration in meters per second squared. In this case the weight of the object is its force due to gravity. In fact, weight is defined as the force of gravity. Since on earth the acceleration is $9.8\frac{m}{s^2}$ and it mass is $2 kg$ then its force is their product thus, $19.6 \frac{m\cdot kg}{s^2}$ just remember that the definition of a newton is $N=\frac{m\cdot kg}{s^2}$. Thus the force is $19.6 N$.
• Mar 1st 2006, 04:22 PM
ThePerfectHacker
Quote:

Originally Posted by jaycas21
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?

2)The ring is a equilibrium. Because it is motionless by being held by the strings. This means that the 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 $\vec i,\vec j$ form. Let $F_1$ 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 $\cos 30^o=\frac{x}{F_1}$ thus, $x=F_1\cos 30^o=F_1\frac{\sqrt{3}}{2}$because we know the cosine of this angle. Thus, the y-coordinate is also found through trigonometry, since it forms a right angle, $\sin 30^o=\frac{y}{F_1}$ thus, $y=F_1\sin 30^o=F_1\frac{1}{2}$. Thus, the vector $\vec F_1$ can be expressed as $F_1\frac{\sqrt{3}}{2}\vec i+F_1\frac{1}{2}\vec j$. By similar reasoning and the fact that $\alpha=45^o$. We arrived at $\vec F_2=-F_2\frac{\sqrt{2}}{2}\vec i+F_2\frac{\sqrt{2}}{2}\vec j$ (because $\cos 45^o=\sin 45^o=\frac{\sqrt{2}}{2}$ also important that the sign in front of $\vec i$ 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 $-19.6\vec j$ because it only goes down.
Now we state that $\sum \vec F_k=\vec 0$ meaning the vector sum. Which in this case is,
$F_1\frac{\sqrt{3}}{2}\vec i+F_1\frac{1}{2}\vec j
-F_2\frac{\sqrt{2}}{2}\vec i+F_2\frac{\sqrt{2}}{2}\vec j-19.6\vec j=\vec 0$

Organize them,
$\left(F_1\frac{\sqrt{3}}{2}-F_2\frac{\sqrt{2}}{2}\right)\vec i+\left(F_1\frac{1}{2}+F_2\frac{\sqrt{2}}{2}-19.6\right)\vec j=\vec 0$
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
$\left\{\begin{array}{cc}F_1\frac{\sqrt{3}}{2}-F_2\frac{\sqrt{2}}{2}&=0\\F_1\frac{1}{2}+F_2\frac{ \sqrt{2}}{2}-19.6&=0\end{array}\right$
just simplify the second one, to get,
$\left\{\begin{array}{cc}F_1\frac{\sqrt{3}}{2}-F_2\frac{\sqrt{2}}{2}&=0\\F_1\frac{1}{2}+F_2\frac{ \sqrt{2}}{2}&=19.6\end{array}\right$.

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 PM
topsquark
Quote:

Originally Posted by jaycas21
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?

In reference to the second problem I am going to slightly correct ThePerfectHacker... Newton's Second Law states that $\sum F=ma$ where F and a are vectors. So in problem 2 we have three forces acting on a single point. The first step is to set up a coordinate system; when in doubt choose +x to the right and +y upward, which is what I believe the book did. So now we break each force up into components in the x and y directions and then add them. They failed to specify an acceleration so I'm going to assume it is 0 m/s2. (That fits with the book's solution.) The components of a vector of zero length are zero so we can ignore components of a.

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

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

$\sum F_y=F_1sin(30)+F_2sin(45)-mg=0$, 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 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
In reference to the second problem I am going to slightly correct ThePerfectHacker... Newton's Second Law states that $\sum F=ma$ where F and a are vectors. So in problem 2 we have three forces acting on a single point. The first step is to set up a coordinate system; when in doubt choose +x to the right and +y upward, which is what I believe the book did. So now we break each force up into components in the x and y directions and then add them. They failed to specify an acceleration so I'm going to assume it is 0 m/s2. (That fits with the book's solution.) The components of a vector of zero length are zero so we can ignore components of a.

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

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

$\sum F_y=F_1sin(30)+F_2sin(45)-mg=0$, 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

Setting $\sum \vec F_x,\sum \vec F_y=0$ is the same as doing what I did throught unit vectors $\vec i,\vec j$ because they represent the x and y components.
• Mar 1st 2006, 09:07 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
1)Newton's second law states that $F=ma$ where $F$ is force in newtons, $m$ is mass in kilograms and $a$ is the acceleration in meters per second squared. In this case the weight of the object is its force due to gravity. In fact, weight is defined as the force of gravity. Since on earth the acceleration is $9.8\frac{m}{s^2}$ and it mass is $2 kg$ then its force is their product thus, $19.6 \frac{m\cdot kg}{s^2}$ just remember that the definition of a newton is $N=\frac{m\cdot kg}{s^2}$. Thus the force is $19.6 N$.

What is $s$?

The gravitational force is $m.g$, where $g$ is the acceleration
due to gravity (it is independent of mass as Galileo tells us). At the earth's
surface $g \approx 9.81 \mathrm{m/s^2}$, so the gravitational force
(ie the weight) is $mg \approx 2 \times 9.81 =19.62\ \mathrm{Newtons}$

RonL
• Mar 2nd 2006, 03:56 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Setting $\sum \vec F_x,\sum \vec F_y=0$ is the same as doing what I did throught unit vectors $\vec i,\vec j$ because they represent the x and y components.

Sorry, ThePerfectHacker, we posted a new reply at more or less the same time and I didn't note that you had made a second reply. My comment was in reference to your first post where you stated that F=ma was Newton's Second Law. I should have included the quote when I made the comment. My apologies for the confusion! :eek:

-Dan
• Mar 2nd 2006, 03:58 AM
topsquark
Quote:

Originally Posted by CaptainBlack
What is $s$?

$g \approx 9.81 \mathrm{m/s}$

An acceleration is defined as $a=\frac{d^2x}{dt^2}$ where x is a displacement and t is a time. Thus the standard (MKS) unit for an acceleration will be $m/s^2$, not m/s.

-Dan
• Mar 2nd 2006, 10:27 AM
CaptainBlack
Quote:

Originally Posted by topsquark
An acceleration is defined as $a=\frac{d^2x}{dt^2}$ where x is a displacement and t is a time. Thus the standard (MKS) unit for an acceleration will be $m/s^2$, not m/s.

-Dan

Opps, typo
• Mar 2nd 2006, 10:39 AM
CaptainBlack
Quote:

Originally Posted by topsquark
An acceleration is defined as $a=\frac{d^2x}{dt^2}$ where x is a displacement and t is a time. Thus the standard (MKS) unit for an acceleration will be $m/s^2$, not m/s.

-Dan

Interesting point here, (I believe its also what originally mislead me about $s$).

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

Now I believe the convention is the italic letters like $s$ represents a variable or parameter,
in which case the units should be non-italic letters like $\mathrm{s}$, or in this case we should
write metres per second as $\mathrm{m/s^2}$ or $\mathrm{ms^{-2}}$.

RonL
• Mar 2nd 2006, 11:43 AM
topsquark
Quote:

Originally Posted by CaptainBlack
Interesting point here, (I believe its also what originally mislead me about $s$).

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

Now I believe the convention is the italic letters like $s$ represents a variable or parameter,
in which case the units should be non-italic letters like $\mathrm{s}$, or in this case we should
write metres per second as $\mathrm{m/s^2}$ or $\mathrm{ms^{-2}}$.

RonL

Good point. Now that you mention it $\frac{m}{s^2}$ DOES rather look equation-like. I'll have to remember that myself in future posts.

-Dan