Physics197
Nov15-09, 11:11 AM
1. The problem statement, all variables and given/known data
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?
2. Relevant 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)
3. 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.
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?
2. Relevant 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)
3. 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.