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Atwood Machines

Atwood Machines. Method #1:. Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following. (a) The acceleration of the 4-kilogram block

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Atwood Machines

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  1. Atwood Machines

  2. Method #1: Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following. (a) The acceleration of the 4-kilogram block (b) The tension in the string supporting the 4-kilogram block (c) The tension in the string connected to the l-kilogram block

  3. Step 1: Determine Direction of Motion Identify which side has the greater mass—this will determine the direction of motion.

  4. Step 2: Draw Free Body Diagram If there are multiple objects on one side, then treat them as 1 unit Since both sides are connected by the same string, the value for T is the same -T1 +T1 - + - + 33 kg +M1g -M2 g

  5. Step 3: Write Net Force Formulas for Each Side SF = m2a SF = +T1 - m2g +T1 - m2g = m2a SF = m1a SF = -T1 + m1g -T1 + m1g = m1a -T1 - + +T1 - + 33 kg -m2 g +m1g + Value given to the variable pointing in the same direction as the motion - Value given to the variable pointing in the opposite direction of the motion

  6. Step 4: Add up the two formulas +T1 + -m2g = m2a -T1 + m1g = m1a + + m1g - m2g = m1a + m2a

  7. Step 5: Solve for acceleration m1g - m2g = m1a + m2a m1g - m2g = (m1+ m2)a m1g - m2g = a (m1+ m2) (4kg)(10m/s2) - (3kg)(10m/s2) = a (4kg+ 3kg) 10kg m/s2 = a 7kg 1.43 m/s2 = a

  8. Step 6: Solve for Tension in string 1 -T1 + m1g = m1a m1g - m1a = T1 m1 (g-a) = T1 4kg (10 m/s2 - 1.4 m/s2 ) = T1 34.4 N = T1

  9. Next Solve for Tension in String 2 +T1 SF = m3a SF = T1 - m3g – T2 T1 - m3g – T2 = m3a T1 - m3g – m3a = T2 34.4 N – (2kg)(10 m/s2) - (2kg)(1.4m/s2) 34.4 N – 20 N – 2.8 N = T2 11.6 N = T2 - + -m3 g -T2 • Substitute values for T1 and acceleration to solve for T2 • 1kg mass can simply be ignored

  10. Alternative Solution for Tension in String 2 SF = m4a SF = T2 - m4g T2 - m4g = m4a T2 = m4g + m4a T2 =(2kg)(10 m/s2 + 1.4m/s2) T2 = 11.4 N - + +T2 -m4 g • Ignore 2 kg mass completely

  11. Method #2: Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following. (a) The acceleration of the 4-kilogram block (b) The tension in the string supporting the 4-kilogram block (c) The tension in the string connected to the l-kilogram block

  12. Step 1: Determine Direction of Motion Identify which side has the greater mass—this will determine the direction of motion.

  13. Step 2: Draw a Free Body Diagram for Each Object T1 T1 T2 m2g T2 m1g m4g

  14. Step 3: Indicate the direction in which object is accelerating T1 T1 T2 a a a m2g T2 m1g m4g

  15. Step 4: Write the net force formulas for each object (+ values in same direction of movement) T1 T1 T2 a a a m2g T2 m1g m4g T2 – m1g = m1a m4g – T1 = m4a T1 - m2g – T2 = m2a

  16. Step 5: Add up the formulas m4g – T1 = m4a T1 - m2g – T2 = m2a T2 – m1g = m1a m4g - m2g –m1g = m4a + m2a + m1a

  17. Step 6: Solve for acceleration m4g - m2g –m1g = m4a + m2a + m1a m4g - m2g –m1g = (m4+ m2 + m1) a m4g - m2g –m1g = a (m4+ m2 + m1) 40N – 20N – 10N = a (7 kg) 1.4 m/s2 = a

  18. Step 7: Solve for Tension m4g – T1 = m4a T2 – m1g = m1a m4g – m4a = T1 T2 = m1a + m1g 40 N – 5.6 N = T1 T2 = 1.4 N + 10 N 34.4 N= T1 T2 = 11.4 N

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