1. m1 m2 Derive a general formula for the acceleration of the system and tension in the ropes

2. m1 m2 Derive a general formula for the acceleration of the system and tension in the ropes m1 m2 Draw FBDs and set your sign conventions: Up is positive down is negative

3. m1 m2 Derive a general formula for the acceleration of the system and tension in the ropes m1 m2 Draw FBDs and set your sign conventions: If up is positive down is negative on the right, the opposite is happening on the other side of the pulley. -T +m 1 g +T -m 2 g Then, do ΣF= ma to solve for a if possible.

4. m1 m2 Derive a general formula for the acceleration of the system and tension in the ropes m1 m2 Σ F 1 = - T + m 1 g = m 1 a -T +m 1 g +T -m 2 g Σ F 2 = T – m 2 g = m 2 a To find acceleration I must eliminate T: so solve one equation for T and then substitute it into the other equation (This is called “solving simultaneous equations” )

5. Σ F 1 = - T + m 1 g = m 1 a Solving this for T T = m 1 g - m 1 a Σ F 2 = T – m 2 g = m 2 a Σ F 2 = m 1 g - m 1 a – m 2 g = m 2 a Now get all the a variables on one side and solve for a.

6. m 1 g - m 1 a – m 2 g = m 2 a m 1 g – m 2 g = m 2 a + m 1 a Factor out g on left and a on right g(m 1 – m 2 ) = a (m 2 + m 1 ) Divide this away (m 1 – m 2 ) (m2 + m1) g = a

7. m1 m2 Now you can find the tension formula by eliminating a. Just plug the new a formula either one of the original force equations. m1 Σ F 1 = - T + m 1 g = m 1 a -T +m 1 g Σ F 2 = T – m 2 g = m 1 a (m 1 – m 2 ) (m2 + m1) g = a