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

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Presentation on theme: "Tension is equally distributed in a rope and can be bidirectional 30N Draw FBD block Draw FBD hand Tension does not always equal weight:"— Presentation transcript:

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

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

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: Up is positive down is negative

4 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.

5 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” )

6 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.

7 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

8 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

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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