Ask Me Help Desk - Tension of a Cord

Quote:

Originally Posted by cshirema1863

a frictionless pulley of unknown mass M and radius r = 0.200 m which is used to draw water from a well. A bucket of mass m = 1.50 kg is attached to a massless cord wrapped around the pulley. The bucket starts from rest at the top of the well and falls for t = 3.00 s before hitting the water h = 6.89 m below the top of the well.

What is the Tension of the Cord

My partner and I keep usinng the Tension Formula F= mass * acceleration. We found a an linear acceleration of 1.531 m/s2.

any help with the formula would be appreciated

Lets first figure out acceleration in the y direction

$y_f=\frac{at^2}{2}+v_{0y}t+y_0$

the final position is (yf) is 0, the initial position (y0) is 6.89, and the initial velocity(v0y) in the y direction is 0, and time (t) is 3.

$0=\frac{a(3)^2}{2}+6.89$
$a= -2.30 \frac{m}{s^2})$

This is OVERALL acceleration. We have two forces in play here causing this direction. Weight, and Tension which are acting in opposite directions.

$F_g=mg$
$T=ma$

Fg is the force due to gravity (weight) and T is tension. If the overall acceleration is -2.3 m/s^2 downwards, it would mean that there is an acceleration upward of (9.8-2.3) to give an overall acceleration of -2.3. Mass here is the mass of the bucket + mass of the rope (But the rope's pass is said to be negligible.

$T=(1.5)(9.8-2.3)$
$T=11.25N$

I can't figure out why they would mention that the mass of the pulley was unknown and that the radius of the pulley was 0.2. Perhaps I am missing something but it would seem to be extraneous information. (If the pulley had somehow become dislodged and attached to the bucket, then we would use its mass but again I don't understand why they tell us it's radius.)