# pulley problem

• Oct 26th 2007, 09:26 PM
winterwyrm
pulley problem
A worker sets up a pulley 12 meters high,
to move a box with a rope of length 28 meters. If the
worker moves to the left at 1
2 meter per second, then
at what rate does the box move along the ground
when the worker is 5 meters from point A ?
[IMG]file:///C:/Users/WINTER%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG]
Thanks in advance for any help. Btw, as a hint, it's a triangle problem, there's a diagram, but it won't copy/paste properly.

s/n kk, got the diagram attached, thanks again =]
• Oct 26th 2007, 09:57 PM
ticbol
Without the diagram we cannot see the relationships of the given facts. Where is point A?
Can you at least describe the figure?
• Oct 26th 2007, 10:51 PM
CaptainBlack
Quote:

Originally Posted by winterwyrm
A worker sets up a pulley 12 meters high,
to move a box with a rope of length 28 meters. If the
worker moves to the left at 1
2 meter per second, then
at what rate does the box move along the ground
when the worker is 5 meters from point A ?
[IMG]file:///C:/Users/WINTER%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG]
Thanks in advance for any help. Btw, as a hint, it's a triangle problem, there's a diagram, but it won't copy/paste properly.

Use the manage attachments buttion below the message entry box (use
uploaded to am image hosting service.

RonL
• Oct 27th 2007, 11:17 AM
ticbol
So that is the figure. As I thought, point A is vertically below the pulley.

Since the worker and the box are assumed to be points on the horizontal ground, we can solve the Problem if we assign distances from point A for the worker and the box.
Let x = distance from worker to point A.
And y = distance from box to point A.

Two right triangles are formed.

In the right triangle with the worker, the length of the rope is
sqrt(x^2 +12^2)

In the right triangle with the box, the rope is
sqrt(y^2 +12^2)

Then,
sqrt(x^2 +144) +sqrt(y^2 +144) = 28 ------(i)

The Problem needs to find rate of movement of the box along the ground, and the worker is given (1/2) m/sec rate, so we need to differentiate both sides of (i) with respect to time t to find dy/dt.

[x /sqrt(x^2 +144)](dx/dt) +[y/sqrt(y^2 +144)](dy/dt) = 0 -----(ii)

At the instant x = 5m, what is y?

At that instant, the length of the rope with the worlker is
sqrt(5^2 +144) = sqrt(169) = 13m

So, in the right triangle with the box,
the length of the rope is 28 -13 = 15m.
hence,
y = sqrt(15^2 -144) = sqrt(81) = 9m

Substitute those int (ii),
[x /sqrt(x^2 +144)](dx/dt) +[y/sqrt(y^2 +144)](dy/dt) = 0
[5 /sqrt(5^2 +144)](1/2) +[9 /sqrt(9^2 +144)](dy/dt) = 0
[5/13](1/2) +[9/15](dy/dt) = 0
dy/dt = -[5/26]/[9/15]
dy/dt = -0.3205 m/sec -----the negative sign only means that y is decreasing.

Therefore, at that instant, the box is moving towards point A at the rate of 0.3205 m/sec. -----------------answer.