Math Is Fun Forum

jimi70

Member

Registered: 2012-04-04

Posts: 22

Pulley and string problem

| O | pulley
|    |
|    |
!    !
O  O
A B
Acceleration = 1.4 m s^-2 downwards from particle A

Particles A and B are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The particles are released from rest, with the string taut, and A and B at the same height above a horizontal floor (see terrible diagram). In the subsequent motion, A descends with acceleration 1.4 m s^2 and strikes the floor 0.8 s after being released. It is given that B never reaches the pulley and before A strikes the floor the tension in the string is 5.88 N.

(a) Calculate the mass of A and the mass of B.
(b) The pulley has mass 0.5 kg, and is held in a fixed position by a light vertical chain. Calculate the tension in the chain immediately before and after A strikes the floor.

(a) for A: mg - 5.88 = 1.4ma so m = 0.7 kg
     for B: 5.88 - mg = 1.4ma so m = 0.525 kg

I need help with (b). I would've thought we could just take the weight of the particles and the pulley, i.e. (0.5 + 0.7 + 0.525) x 9.8 = 16.905 N but this doesn't equate to the sum of the weight of the pulley and the tensions in the string, i.e. 0.5 x 9.8 + 2 x 5.88 = 16.66 N
Why is there a difference?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: Pulley and string problem

hi jimi70

I agree with your answers to part a.

Tension in chain before A hits the ground.

The pulley isn't moving so do the calculation without worrying about the particles. They aren't directly contributing to the forces on the pulley.

So your second calculation is the correct one:  Tc = T + T + 0.5g = 5.88 + 5.88 + 4.9 = 16.66 N

The difference is because the particles are moving.  You could perhaps think of it like this:  some of the weight is 'used up' in making the system accelerate, so not all can be considered as part of the static force on the pulley.  If you put a break on the pulley so it cannot turn, then 16.905N would be correct.

But you are also asked for the tension after A hits the ground.

A stays on the ground and B travels up slowing under gravity alone.  There's no tension now because the string will go slack. 

So at the pulley, there's just Tc acting up and the weight of the pulley acting down.  Tc = 0.5g = 4.9N

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jimi70

Member

Registered: 2012-04-04

Posts: 22

Re: Pulley and string problem

thanks once again bob