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Horizontal Pulleys

Horizontal Pulleys. m 1 = 70 kg. m 2 = 30 kg. What are the forces acting on this system? Think of two separate FBD. FBD-N. F N. F T. No Friction. F T. w 70. = mg = 70 kg (10 m/s 2 ) = 700 N. F T. F N. w 30. = mg = 30 kg (10 m/s 2 ) = 300 N. F T. w 30. w 70.

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Horizontal Pulleys

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  1. Horizontal Pulleys

  2. m1 = 70 kg m2 = 30 kg What are the forces acting on this system? Think of two separate FBD. FBD-N FN FT No Friction FT w70 = mg = 70 kg (10 m/s2) = 700 N FT FN w30 = mg = 30 kg (10 m/s2) = 300 N FT w30 w70 Treat this as two different systems and join them later.

  3. m1 = 70 kg m2 = 30 kg FN FT w70 = mg = 70 kg (10 m/s2) = 700 N FN FT ΣF70y = ? ΣF70x = ? w70 ΣF70y = 0 ΣF70x = m70a ΣF70y = FN – w70 ΣF70y = FT 0 = FN – w70 m70a = FT FN = w70 a = ? FN = 700 N FT = ?

  4. m1 = 70 kg m2 = 30 kg FT FT w30 = mg = 30 kg (10 m/s2) = 300 N w30 ΣF30 = ? ΣF30 = - m30a ΣF30= FT – w30 -m30a = FT – m30g a = ? FT = ?

  5. So what do we know? FN = 700 N FT = m70a -m30a = FT – m30g Can’t I plug this into here? Here’s where they join. -m30a = m70a– m30g -m70a -m70a -m30a - m70a = - m30g a (m30 + m70) = m30g (m30 + m70) (m30 + m70) a = m30g (m30 + m70) a = 300 N 100 kg = 3 m/s2

  6. m1 = 70 kg m2 = 30 kg The FT for both systems is the same. The tension in the string pulling sideways is the same the tension in the string hanging downward. FN Now we can find FT. FT FT w70 w30 Let’s check to make sure they are the same for both systems: ΣF70x = ? ΣF30 = - m30a ΣF70x = m70a FT30 = ? ΣF30= FT – w30 FT70 = ? ΣF70y = FT FT30 = -m30a + m30g -m30a = FT – m30g FT70= m70a m70a = FT FT30 = -30 kg (3 m/s2) + 30 kg (10 m/s2) FT70= 70 kg (3 m/s2) FT30 = -90 N + 300 N FT70= 210 N same FT30 = 210 N

  7. Find vf =? Find d =? vi = 0 d = vit + ½ at2 a = 3 m/s2 d = ½ at2 d = 3 m/s2 (10 s )2 2 t = 10 s vf = vi + at d = 150 m vf = at vf = 3 m/s2 (10 s) Find ΣF30 = ? vf = 30 m/s ΣF30 = - m30a OR ΣF30= FT – w30

  8. Cheating…how to figure this out without all that work! (Too bad you can’t do this on the test) FT FN FN = 700 N combine FT w30 w70 w70 +w30 = 1000 N Fnet = 300 N Fnet = ma a = Fnet m a = 300 N 100 kg a = 3 m/s2

  9. m1 = 30 kg m2 = 70 kg DO FOR HOMEWORK FBD-N No Friction

  10. m1 = 70 kg m2 = 30 kg FBD-P FN Ff FT Friction FT w70 = mg = 70 kg (10 m/s2) = 700 N FT FN Ff w30 = mg = 30 kg (10 m/s2) = 300 N FT w30 w70 What should we expect to happen to the acceleration?

  11. m1 = 70 kg m2 = 30 kg FN Ff FT w70 = mg = 70 kg (10 m/s2) = 700 N FN Ff FT Ff =µFN ΣF70x = ? ΣF70y = ? w70 Ff = (.1) (700 N) ΣF70x = m70a ΣF70y = 0 ΣF70y = FT - Ff ΣF70y = FN – w70 Ff = 70 N 0 = FN – w70 m70a = FT - Ff FN = w70 a = ? FT = ? FN = 700 N

  12. m1 = 70 kg m2 = 30 kg FT FT w30 = mg = 30 kg (10 m/s2) = 300 N w30 ΣF30 = ? ΣF30 = - m30a ΣF30= FT – w30 -m30a = FT – m30g a = ? FT = ?

  13. Here’s what we know: FN = 700 N FT = m70a + Ff -m30a = FT – m30g Here’s where they join. -m30a = m70a + Ff– m30g -m70a -m70a -m30a - m70a = Ff - m30g a (m30 + m70) = Ff - m30g (m30 + m70) (m30 + m70) a = Ff -m30g (m30 + m70) a = 70 N - 300 N 100 kg = 2.3 m/s2

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