Without the diagram we cannot see the relationships of the given facts. Where is point A?
Can you at least describe the figure?
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 ?
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 =]
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
In the right triangle with the box, the rope is
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.
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.