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 N2

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

c) up

hope this