Solve Related Rates: More Homework Questions

[Physics197]

Homework Statement

2) One end of a rope is tied to a box. The other end is passed over a pulley 5 m above the floor and tied at a level 1 m above the floor to the back of a truck. The rope is L meters long. If the rope is taut and the truck moves at ½ m/s:
a. How fast is the box rising when the truck is 3 m from point directly below the pulley?
b. How far will the truck have to move to raise the box from the floor to a height of 2 m?

Homework Equations

Known:
dx/dt = 0.5 m/s
x^2 + 16 = z^2 (length of rope from pulley to truck)
z + (5-y) = L (length of rope)

The Attempt at a Solution

for a:
2x dx/dt = 2z dz/dt

and dz/dt - dy/dt = 0, therefore dz/dt = dy/dt

dy/dt = x/z dx/dt
z = sqr (x^2 + 16)

dy/dt = x/[sqr (x^2 + 16)]* 0.5, where x=3

dy/dt = 0.3 m/s <---- someone verify?

For part B: Not sure, just know that I need to find x, giving y.

 

[mighty2000]

Hi there! Your attempt at solving part a looks correct. I would just add that you can also use the Pythagorean theorem to find x (length of the rope from the pulley to the box) when the truck is 3 m from the point directly below the pulley. It would look something like this:

x^2 + (5-y)^2 = L^2
x^2 + (5-1)^2 = L^2 (since the height of the truck is 1 m above the floor)
x^2 + 16 = L^2
x = sqr (L^2 - 16)

So when the truck is 3 m from the point directly below the pulley, x = sqr (L^2 - 16) = sqr (L^2 - 4^2).

For part b, you can use the same equation above, but this time you know the final height (2 m) and you need to find the distance the truck needs to move (x). So the equation would look like this:

x^2 + (5-y)^2 = L^2
x^2 + (5-2)^2 = L^2 (since the final height is 2 m)
x^2 + 9 = L^2
x = sqr (L^2 - 9)

Hope this helps! Let me know if you have any other questions.