## Dynamics Question

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iroZn
Posts: 33
Joined: Sat Oct 30, 2010 12:49 pm UTC

### Dynamics Question

Hey guys, I just finished a test with this question in it, I wanted to see what you guys get for it.

Here is what I got:

Spoiler:
let l be the length of the cable.
let d be the distance between A and B.
let y be the distance from the top to A.
let x be the horizontal distance from A to B.

l = y + 2d
d = l/2 - y/2

x = d*cos(θ)

x = (l/2)cos(θ) - (y/2)cos(θ)

x' = -(y'/2)cos(θ)

y' = 1 m/s

x' = -(1/2)cos(θ)

then just sub in the values for θ. The time allotted for it was 35 minutes and I got it in 5... so I'm wondering if I oversimplified it or something :s
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jmorgan3
Posts: 710
Joined: Sat Jan 26, 2008 12:22 am UTC

### Re: Dynamics Question

iroZn wrote:x = (l/2)cos(θ) - (y/2)cos(θ)

x' = -(y'/2)cos(θ)

Theta isn't constant with time.
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letterX
Posts: 535
Joined: Fri Feb 22, 2008 4:00 am UTC
Location: Ithaca, NY

### Re: Dynamics Question

iroZn wrote:x = (l/2)cos(θ) - (y/2)cos(θ)

x' = -(y'/2)cos(θ)

When you do this line, you're differentiating with respect to time. You differentiate x and y, but forget that [imath]\theta[/imath] is also a function of time, so you also need to apply the chain rule inside the [imath]\cos(\theta)[/imath]. However, doing it this way, you get a bunch of [imath]\theta'[/imath] terms running around, and we have no idea how fast [imath]\theta[/imath] changes wrt time.

Ok: typed out the answer because it's not homework, and you've already got a partial solution, but really don't look at it until you've thought about it for a while.

Spoiler:
x, d, l as before. y the vertical distance between A and B. Then we know that [imath]x^2 + y^2 = d^2[/imath]. Furthermore, since A is descending at 1 m/s, we know that we are eating up 1 meter of cable per second, so [imath]d' = -1/2[/imath]. Thus, we can take the derivative of the pythagorean equation I just wrote to get
$2xx' + 2yy' = 2dd'$

Since [imath]d' = -1/2[/imath] and [imath]y' = -1[/imath], we solve to get
$x' = \frac{2y - d}{2x} = \frac{y}{x} - \frac{d}{2x} = \tan(\theta) - \frac{1}{2}\sec(\theta)\$

It's still not a 35 minute problem, but it helps if you've seen similar before. The key is to remember to always remember which things are functions of time, and always remember to differentiate by them. Also, use the pythagorean theorem to turn things which are trig functions into lengths (which are much easier to work with).

SWGlassPit
Posts: 312
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Location: Houston, TX
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### Re: Dynamics Question

I like how the spring is a completely spurious addition to the problem. The cable could just as easily be replaced with a rigid bar with no change to the fundamental problem.
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jmorgan3
Posts: 710
Joined: Sat Jan 26, 2008 12:22 am UTC

### Re: Dynamics Question

SWGlassPit wrote:I like how the spring is a completely spurious addition to the problem. The cable could just as easily be replaced with a rigid bar with no change to the fundamental problem.

The point of the problem is that the hypotenuse between the two blocks changes as the position of a changes. A rigid bar would not behave that way.
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iroZn
Posts: 33
Joined: Sat Oct 30, 2010 12:49 pm UTC

### Re: Dynamics Question

Ah thank you very much letterX it makes sense now. I knew in the back of my head I had done something wrong!
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SWGlassPit
Posts: 312
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Location: Houston, TX
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### Re: Dynamics Question

jmorgan3 wrote:
SWGlassPit wrote:I like how the spring is a completely spurious addition to the problem. The cable could just as easily be replaced with a rigid bar with no change to the fundamental problem.

The point of the problem is that the hypotenuse between the two blocks changes as the position of a changes. A rigid bar would not behave that way.

Maybe it's just the scan, but I'm not seeing that.
Up in space is a laboratory the size of a football field zipping along at 7 km/s. It's my job to keep it safe.

Erdös number: 5

jmorgan3
Posts: 710
Joined: Sat Jan 26, 2008 12:22 am UTC

### Re: Dynamics Question

The cable is attached at the horizontal surface in the upper-left corner. It runs down to A and then around the pulley at A, then to pulley B and back to A, where the other end is secured. Its total length is therefore (Distance from horizontal surface to A)+(Two times distance from A to B). As A moves down, the first quantity becomes larger, so the second quantity becomes smaller.
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