An Atwood\'s machine consists of blocks of masses m 1 = 11.0 kg and m 2 = 17.0 k

Question

An Atwood's machine consists of blocks of masses

m1 = 11.0 kg

and

m2 = 17.0 kg

attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass

M = 9.30 kg

and radius

r = 0.200 m.

The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

(a) Why must the tension T2 be greater than the tension T1?
(b) What is the acceleration of the system, assuming the pulley axis is frictionless? (Give the magnitude of a)

(c) Find the tensions T1 and T2.

Explanation / Answer

a) as we see pulley it turns that means T2 have to be greater than T1.

so that torque due to T2 ( r x T) is greater than T1.

b) on pulley,
torque = I x alpha

r ( T2 - T1) = ( M r^2 /2 ) ( a/r)

T2 - T1 = 9.30a/2 = 4.65a ........... (i)

on bock m2,

17g - T2 = 17a .... (ii)

on block m1,

T1 - 11g = 11a ... (iii)

add i, ii and (iii).

17g - 11g = (4.65 + 17 +11)a

a = 1.80 m/s^2

c) using in (iii)

T1 - 11g = (11 x 1.80 )

T1 = 127.74 N

using in(ii)

17g - T2 =17(1.80)

T2 =136.17 N

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