Three masses m1=4 kg, m2=4 kg and m3=6 kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of T1/T2 is . (take g=10 m/s2) [figure]
- 5/3
- 2/3
- 3/5
- 2/5
Answer
(1) 5/3
Solution
Right side (m2+m3=10 kg) is heavier than left side (m1=4 kg) ⇒ right side goes down, m1 goes up with the same acceleration a.
a= (m2+m3-m1)gm1+m2+m3= (10-4) 1014= 307 m/s2
[figure]
For m1: T1-m1g=m1a ⇒ T1=4 (10+ 307 )= 4007 N
For m3: m3g-T2=m3a ⇒ T2=6 (10- 307 )= 2407 N
∴ T1T2= 400240= 53
Answer: (1)