Here is what I got:

**Spoiler:**

**Moderators:** gmalivuk, Moderators General, Prelates

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:**

Here is what I got:

I am imaginary, prove me wrong!

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

x' = -(y'/2)cos(θ)

[Mythbusters]Well there's your problem.[/Mythbusters]

Theta isn't constant with time.

This signature is Y2K compliant.

Last updated 6/29/108

Last updated 6/29/108

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.

- SWGlassPit
**Posts:**312**Joined:**Mon Feb 18, 2008 9:34 pm UTC**Location:**Houston, TX-
**Contact:**

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.

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

Erdös number: 5

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.

This signature is Y2K compliant.

Last updated 6/29/108

Last updated 6/29/108

Ah thank you very much letterX it makes sense now. I knew in the back of my head I had done something wrong!

I am imaginary, prove me wrong!

- SWGlassPit
**Posts:**312**Joined:**Mon Feb 18, 2008 9:34 pm UTC**Location:**Houston, TX-
**Contact:**

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

Erdös number: 5

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.

This signature is Y2K compliant.

Last updated 6/29/108

Last updated 6/29/108

Users browsing this forum: No registered users and 16 guests