Four (long?) vertex problems

**nirva** · July 7th 2006, 08:42 PM

Here are some vertex problems that I don't understant, if anyone does please teach me

1. V lies in the first quadrant of the graph and makes an angle pi/3 with the positive x-axis. And |v| = 4. Find V in component form

2.A clothesline tied between 2 poles 8m apart. When a wet shirt with a mass of 0.8kg is hung at the middle the midpoint is pulled down 8cm. Find the tension in each half of the clothesline.

3. A woman walks west on the deck of a ship at 3miles/hr. The ship is moving North at 22mi/hr. Find the speed and direction of the woman relative to the surface of water.

4. F1 and F2 with magnitude 10lb and 12lb act on an object at point P. Find the resulant force F acting at point P with the magnitude and direction.

For #4:

(The X is including F2's 30 degrees)

**earboth** · July 8th 2006, 12:48 AM

Originally Posted by nirva

Here are some vertex problems that I don't understant, if anyone does please teach me

1. V lies in the first quadrant of the graph and makes an angle pi/3 with the positive x-axis. And |v| = 4. Find V in component form
...

Hello, nirva,

let the angle between pos. x-Axis and the vector be $\alpha$ . Then you can express the vector as $\overrightarrow{v}=|\overrightarrow{v}| \cdot (\cos(\alpha),\sin(\alpha))$

Now plug in the values you know and you'll get:

$\overrightarrow{v}=4 \cdot \left(\cos\left(\frac{\pi}{3}\right),\sin\left( \frac{\pi}{3}\right) \right)$ = $\left(2,2\cdot\sqrt3\right)$

Greetings

EB

**earboth** · July 8th 2006, 04:21 AM

Originally Posted by nirva

Here are some vertex problems that I don't understant, if anyone does please teach me...
2.A clothesline tied between 2 poles 8m apart. When a wet shirt with a mass of 0.8kg is hung at the middle the midpoint is pulled down 8cm. Find the tension in each half of the clothesline.

...

Hello, nirva,

I've attached a diagram to show you what I've calculated:

1. Calculate the weight of the wet shirt:
$|\overrightarrow{F_{shirt}}|=0.8 kg\cdot 9.81\frac{m}{s^2}=7.848 N$

2. The two parts of the line have to pull up the shirt with this force. Calculate the angle between the vertical line and one part of the cloth-line:
$\alpha=\arctan\left( \frac{4 m}{0.08 m}\right)\approx 88.85^\circ$

3. The force pulling up the shirt has the same value as the weight of the shirt but the opposite direction:
$|\overrightarrow{F_{up}}|=7.848 N$

4. $\overrightarrow{F_{up}}$ and $\overrightarrow{F_{tension}}$ are legs of a right triangle:

$\cos(\alpha)=$ $\frac{0.5 \cdot |\overrightarrow{F_{up}}|} {|\overrightarrow{F_{tension}}|}$

That means: $F_{tension}=\frac{0.5 \cdot |\overrightarrow{F_{up}}|}{\cos(\alpha)}\approx 195.5 N$

that means, that both parts of the line have to pull with a force more than 20 times as large as the weight of the shirt. Amazing!

Greetings

EB

**Quick** · July 8th 2006, 05:05 AM

Originally Posted by nirva

3. A woman walks west on the deck of a ship at 3miles/hr. The ship is moving North at 22mi/hr. Find the speed and direction of the woman relative to the surface of water.

This is a good vector problem, you can see it drawn out at the bottom (pay no attention to the fact I used a graph program).

Anyway, looking at the diagram you can see that her movement in relation to the water is the hypotenuse of a triangle, the sides of course are the speed of her walking and the ship, now you can merely use the pythagorean theorum...

$a^2+b^2=c^2\quad\rightarrow\quad\sqrt{a^2+b^2}=c \quad\rightarrow\quad\sqrt{3^2+22^2}=c$

I'm trusting you can solve it from there...

**galactus** · July 8th 2006, 05:15 AM

Originally Posted by nirva

3. A woman walks west on the deck of a ship at 3miles/hr. The ship is moving North at 22mi/hr. Find the speed and direction of the woman relative to the surface of water.

Hello. first time poster. Hello, especially, to that ubiquitous Soroban.

The velocity of the woman relative to the ship's deck is -3i.

The velocity of the ship relative to the water is 22j.

So, the velocity of the woman relative to the water is -3i+22j

Use Pythagoras and

The direction is $tan^{-1}(\frac{3}{22})=7.765$

or about N8W

**Quick** · July 8th 2006, 05:22 AM

Originally Posted by galactus

The direction is $tan^{-1}(\frac{3}{22})=7.765$

or about N8W

It is a good thing you saw the part of direction (I completely overlooked it).

Hello. first time poster. Hello, especially, to that ubiquitous Soroban.

Welcome to MHF, I'm Quick and I'm the 14th highest poster on the site, (soon to be 13th

). You'll find the biggest posters on the site are ThePerfectHacker and Cap'n Black, Although Soroban is pretty high. Anyway, welcome!

**galactus** · July 8th 2006, 05:27 AM

Thank you, very much.

**earboth** · July 8th 2006, 07:40 AM

Originally Posted by nirva

Here are some vertex problems that I don't understant, if anyone does please teach me
...
4. F1 and F2 with magnitude 10lb and 12lb act on an object at point P. Find the resulant force F acting at point P with the magnitude and direction.

For #4:

(The X is including F2's 30 degrees)

Hello, nirva,

I've attached a digram to demonstrate what I've calculated:

You have to add two vectors. Use Cosine Rule. The angle between F1 and F2 ist 105°.

$(\overrightarrow{F})^2=(\overrightarrow{F_1})^2+( \overrightarrow{F_2})^2-2 \cdot (\overrightarrow{F_1}) \cdot (\overrightarrow{F_2}) \cdot \cos(75^\circ)$

Plug in the values you know:

$|\overrightarrow{F}|=\sqrt{10^2+12^2-2\cdot 10\cdot 12\cdot \cos(75^\circ)}\approx 13.486 N$

By the way: I don't understand your last remark about the X in your drawing. You find an approximate value of the angle between F1 and F.

Greetings

EB

**nirva** · July 8th 2006, 09:48 AM

Alright, I think I get it now, thanks a lot guys

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