An Atwood machine is constructed using a

hoop with spokes of negligible mass. The

2.1 kg mass of the pulley is concentrated on

its rim, which is a distance 22.1 cm from the

axle. The mass on the right is 1.56 kg and on

the left is 1.93 kg. What is the magnitude of the linear acceler-

ation of the hangingmasses? The acceleration

of gravity is 9.8 m/s2.

Answer in units of m/s2

a=1.03 m/s²

Explanation:

given required

dist=22.1 cm a=?

m1=2.1 kg

m2=1.56 kg

m3=1.93 kg

g=9.8 m/s²

solution

there are three forces on this system so we can solve by using newton's second law as follows

t1, t2, and weight of are the only forces acted on the system,

if we take the system on the left mass m3

-m3a=t2-m3g

-1.93 kg*a=t2-1.93 kg*9.8 m/s²

-1.93 kg*a=t2-18.91 N ... eq 1

take at the right mass m2

m2a=t2-m2*g

1.56 kg*a=t2-1.56 kg*9.8 m/s²

1.56 kg*a=t2-15.288 N ... 2

add the two equations simultaneously and then solve for a,

-1 * (-1.93 kg*a=t2-18.91 N) multiply by - 1 the first and add the two equations

1.56 kg*a=t2-15.288 N

3.49 kg*a=3.622 N

a=1.03 m/s²