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### Re: Hovercraft B/C

Posted: March 22nd, 2018, 5:40 pm
Ohh, I see.
$\frac12(\frac13m(0.4 m)^2)(3.5 \frac{m}{s} * \frac{1 rad}{0.4 m})^2 = mg(\frac12 * 0.4 m * cos \theta + \frac12 * 0.4 m)$
$\theta = 1.53 \textrm{radians}$
Not really sure since I still haven't quite wrapped my head around rotational stuff.

EDIT: Over-complicated things the first attempt

### Re: Hovercraft B/C

Posted: March 22nd, 2018, 6:11 pm
Cool (finally!)

A 1 lb point mass is held up by two ideal ropes. One rope is vertical and 55 cm long. The other is at an angle from the vertical rope. It is attached to the same "ceiling" as the other rope, so it looks something like this:

Code: Select all

--------------------- | / | / | / |/ O 
Imagine you suddenly decrease the length of the rope to the right by half of the length of the rope to the left. Find the tension in each rope (magnitude and direction) after the system reaches static equilibrium if the angle between them was originally
a) as close to 0 degrees as possible
b) 30 degrees
c) 45 degrees
d) 60 degrees
e) as close to 90 degrees as possible

EDIT: Unfortunately, I made an error when I first did this problem, and the algebra is more complicated than I thought (meaning very time-consuming). Thus, I will ask another problem :/

Three masses are held up to a ceiling by three ropes, like so:

Code: Select all

___ | O | O | O 
The ball on top has a mass of 5 kg, the ball in the middle 3 kg, the ball on the bottom 1 kg. Assuming the ropes are ideal, find the tension in each rope.

### Re: Hovercraft B/C

Posted: March 30th, 2018, 2:55 pm
Cool (finally!)

A 1 lb point mass is held up by two ideal ropes. One rope is vertical and 55 cm long. The other is at an angle from the vertical rope. It is attached to the same "ceiling" as the other rope, so it looks something like this:

Code: Select all

--------------------- | / | / | / |/ O 
Imagine you suddenly decrease the length of the rope to the right by half of the length of the rope to the left. Find the tension in each rope (magnitude and direction) after the system reaches static equilibrium if the angle between them was originally
a) as close to 0 degrees as possible
b) 30 degrees
c) 45 degrees
d) 60 degrees
e) as close to 90 degrees as possible

EDIT: Unfortunately, I made an error when I first did this problem, and the algebra is more complicated than I thought (meaning very time-consuming). Thus, I will ask another problem :/

Three masses are held up to a ceiling by three ropes, like so:

Code: Select all

___ | O | O | O 
The ball on top has a mass of 5 kg, the ball in the middle 3 kg, the ball on the bottom 1 kg. Assuming the ropes are ideal, find the tension in each rope.
Going from top to bottom, the tensions are as follows:

$Tension_1=(5+3+1)*9.8=88.2N$

$Tension_2=(3+1)*9.8=39.2N$

$Tension_3=(1)*9.8=9.8N$

Essentially, the tension in each rope is equal to the weight that it holds up.

### Re: Hovercraft B/C

Posted: March 30th, 2018, 2:56 pm

### Re: Hovercraft B/C

Posted: March 30th, 2018, 3:02 pm
Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?

### Re: Hovercraft B/C

Posted: March 30th, 2018, 3:14 pm
Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
32?

### Re: Hovercraft B/C

Posted: March 30th, 2018, 3:19 pm
Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
Assuming there is no change in lift (which affects friction) and the hovercraft gets to max speed instantly:
One penny roll is around the mass of 50 (new) pennies, 50 * 3.11 g or 0.156 kg.
Required speed is 1.8 m / 15 s = 0.12 m/s.
(2.25 kg)(0.2 m/s) = (2.25 kg + n * 0.156 kg)(0.12 m/s)
n = 9.6

10

### Re: Hovercraft B/C

Posted: March 30th, 2018, 5:38 pm
Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
Assuming there is no change in lift (which affects friction) and the hovercraft gets to max speed instantly:
One penny roll is around the mass of 50 (new) pennies, 50 * 3.11 g or 0.156 kg.
Required speed is 1.8 m / 15 s = 0.12 m/s.
(2.25 kg)(0.2 m/s) = (2.25 kg + n * 0.156 kg)(0.12 m/s)
n = 9.6

10
Oh sorry, I forgot to mention that you should assume that each roll of pennies has a mass of 125 grams (it's mentioned in the rules manual). However, your answer is technically correct, so you can go ahead with the next one!

### Re: Hovercraft B/C

Posted: March 31st, 2018, 6:16 pm
Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.

### Re: Hovercraft B/C

Posted: April 1st, 2018, 10:44 am
Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.
First, find the individual components of the 2D velocity vector:

$v_x = vcos\theta$
$v_y = vsin\theta$

The amount of time the ball will be in the air can be found using

$\Delta y = v_{y}t + \frac{at^2}{2}$

Since $\Delta y = 0$ and $a = -9.8$:

$0 = t(vsin\theta - 4.9t)$

$t = 0, \frac{vsin\theta}{4.9}$

To then get the total distance traveled, we simply use $v_{x}t = \Delta x$.

$vcos\theta*\frac{vsin\theta}{4.9} = \Delta x$

$\frac{v^2sin(2\theta)}{9.8} = \Delta x$ is the final equation.