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Tension is equally distributed in a rope and can be bidirectional

Tension is equally distributed in a rope and can be bidirectional. Draw FBD block Draw FBD hand. 30N. Tension does not always equal weight:. Derive a general formula for the acceleration of the system and tension in the ropes. m1. m2.

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Tension is equally distributed in a rope and can be bidirectional

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  1. Tensionis equally distributed in a rope and can be bidirectional Draw FBD block Draw FBD hand 30N Tension does not always equal weight:

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

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

  4. Derive a general formula for the acceleration of the system and tension in the ropes 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 +m1g +T -m2g m1 m2 m1 m2 Then, do ΣF= ma to solve for a if possible.

  5. Derive a general formula for the acceleration of the system and tension in the ropes -T +m1g +T -m2g m1 m2 m1 m2 Σ F2 = T – m2g = m2a Σ F1 = - T + m1g = m1a 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” )

  6. Σ F1 = - T + m1g = m1a Solving this for T T = m1g - m1 a Σ F2 = T – m2g = m2a Σ F2 = m1g - m1 a – m2g = m2 a Now get all the a variables on one side and solve for a.

  7. m1g - m1 a – m2g = m2 a m1g – m2g = m2 a + m1 a Factor out g on left and a on right g(m1– m2) = a (m2 + m1) Divide this away (m1– m2) g = a (m2 + m1)

  8. Now you can find the tension formula by eliminating a. Just plug the new a formula either one of the original force equations. Σ F2 = T – m2g = m1a Σ F1 = - T + m1g = m1a -T +m1g m1 m1 m2

  9. Problem #68 and 58 is just like Atwood’s machine. Use the same analysis.For #57, use the block & pully analysis on the “Forces Advanced B” presentation on my website

  10. Problem #69: Dumbwaiter: a trickier Atwood’s machine • First analyze the forces on the man and do F=ma. N + T –mg =ma • Then analyze the forces on the elevator and do F= ma. 2T – (M+m)g = (M +m)a • This will generate two equations and two unknowns. Solve simultaneously by solving the top equation for T and plugging into the bottom equation. This will yield a = { 2N + (M-m)g} / ( m - M). Once you know a, find T.

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