Question

Derive formula from : Mass suspended over pulley from another on an inclined plane??

Answer 1

Answer...

For mass m1 : m1g - T = m1a.

For mass m2: T-M2g sin$\theta{}$=m2a.

acceleration a = (m1-m2sin$\theta{}$.

T=m1m2(1+sin$\theta{}$ / (m1+m2) g

hope it helps u✌

Answer 2

For mass m1 : m1g - T - m1a

For mass m2 : T - M2g sin0 = m2a

Acceleration a = (m1-m2sin0

T= m1m2(1+ sin0 / (m1m2) a

Hope it will help you