KINS 382 LAB 18

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KINS 382 - LAB 18

• Torque, Levers, C of E, Momentum, and Impulse
  Momentum

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Question 1

• Draw a 1st class lever.
• The applied forces moment arm is 1.5m long.
• The resistance is 6 Newtons (N) and has a moment
  arm of 2m.
• If there is no movement in the system, how large
  is the applied force?

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Answer 1

• 1st class lever
  
• Fulcrum
• Propulsive Forces are on opposite sides of the
  fulcrum. A seesaw is a 1st class lever.

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Answer 1 (Cont)

• T(resistive) T(propulsive)
• Tr Fr(lr) or Fp Tp/lp
• Tr 6N(2m) 12Nm
• Fp 12Nm / 1.5m 8N

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Question 2

• A lever 2.5m long has an applied force of 10 N.
• a. How large is the resistance if its lever arm
  is 0.4m?
• b. What class lever is described? (Assume the
  system is in equilibrium)

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Answer 2

• Tp Tr
• Tp Fp(lp)
• Tp 10N(2.5m)
• Tp 25Nm
• Fr 25Nm / 0.4m
• Fr 62.5N
• A first class lever

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Question 3

• The amount of torque which must be generated by
  the biceps brachii at the elbow in order to
  maintain the arm in an isometric contraction is
  50 Nm.
• If the biceps attaches to the radius at a
  perpendicular distance of 3 cm from the joint
  center, the force produced by the biceps must be
  _____?
• Convert your answer to pounds.

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Answer 3

• Tp 50Nm, lp 3cm (.03m), fp ?
• fp Tp/lp
• fp 50Nm / .03m
• fp 1666.67 N
• fp 1666.67 N x 1lb / 4.45 N 374.53lb

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Question 4

• If the attachment to the radius is increased 2 cm
  from the joint center in problem 3, how much
  more or less force do the biceps generate?

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Answer 4

• Tp 50Nm, lp 5cm (.05m), fp ?
• fp 50Nm / .05m
• fp 1000N
• fp 1000N x 1lb/4.45N 224.72lb
• Less force with a longer lever arm and vice
  versa. Because less is necessary to exist at
  equilibrium.

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Question 5

• Draw all three types of levers.
• Write a definition and provide an example of each
  lever.

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Answer 5

• 1st class The applied force and resistance are
  on opposite sides of the fulcrum or axis. Look
  at answer 1 for a picture.
• 2nd class The applied force and resistance are
  on the same side of the axis with the resistance
  closer to the axis or fulcrum.
• F(propulsive)
• F(resistive)
• Fulcrum

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Answer 5 (Cont)

• 3rd class Force and resistance are on the same
  side of the axis, but the applied force is closer
  to the axis.
• Fr
• Fp
• Fulcrum

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Question 6

• Consider the implications that the insertion of a
  muscle has on its sport.
• What type of athlete would prefer a longer
  propulsive lever arm?
• A longer resistive lever arm?

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Answer 6

• Speed athletes prefer a longer resistive lever
  arm
• Strength athletes prefer a longer propulsive
  lever arm

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Question 7

• Marks bicep inserts 4 cm further out on his
  forearm than Scotts.
• Each performs a bicep curl with a 200 lb barbell.
  
• All things being equal (they have the same amount
  of torque), who will generate more force?
• Who will be stronger?
• Who will be faster?

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Answer 7

• The individual with the insertion point 4cm
  further from the axis or rotation will
• Will not need to generate as much force because
  the longer lever arm will increase torque
• Will be stronger
• Will be slower

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Question 8

• Eric weighs 190 lbs and is running at 5 m/s.
• If Nate weighs 280 pounds and wants to have the
  same linear momentum as Eric how fast will he run?

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Answer 8

• Linear Momentum (P), P m(v)
• Eric(m1), m1 190lb or 86.36kg, v1 5m/s
• Nate(m2), m2 280lb or 127.27kg, v2 ?
• Eric P1 86.36kg(5m/s) 431.8kg(m/s)
• Nate wants the same P as Eric
• Nate v2 P/m
• v2 431.8kg(m/s) / 127.27kg 3.39m/s

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Question 9

• Ali drops a superball from a height of 2 meters
  onto the gym floor.
• If the coefficient of restitution between the
  ball and the floor is 0.96, how high will the
  ball bounce?

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Answer 9

• ht(drop) 2m, e .96, ht(rebound) ?
• e
• .96
• .96² ht(rebound) / 2m
• .92 ht(rebound) / 2m
• 2m(.92) ht(rebound)
• 1.84m ht(rebound)

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Question 10

• Daniel and Walter push Walters car to get it
  moving as quickly as possible before they climb
  in.
• If they apply an average force of 100 N in the
  direction of motion to the 150 kg vehicle for 7
  seconds before jumping in what is the cars speed
  at that point?

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Answer 10

• F 100N, m 150kg, t 7s, V ?
• Impulse Ft mv (Final) - mv (Original)
• 100N(7s) 150kg(v) - 150kg(0 m/s)
• 700Ns 150kg (v)
• 700kg(m/s) / 150kg v
• 4.67 m/s v

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Question 11

• Jo (60kg) is running down the hall with a
  velocity of 6 m/s. She collides head on with
  Jennifer (65kg) moving at 5 m/s.
• If the two ladys entangle and continue traveling
  together as a unit following a collision, what is
  their combined velocity?

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Answer 11

• Jo m1 60kg, v1 6m/s
  Jennifer m2 65kg, v2 5m/s
• Find the post collision velocity (state
  direction)
• m1v1 m2v2 (m1 m2) v
• 60kg(6m/s) 65kg(-5m/s) 60kg 65 (v)
• 360kg(m/s) -325kg(m/s) 125kg (v)
• 35kg(m/s) 125kg (v)
• 35kg(m/s) / 125kg v
• .28m/s v (In the positive direction, toward
  Jen).

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Question 12

• Tara (58.5kg) sits 1.5m from the axis of rotation
  of a seesaw. What torque does she cause?
• How far away on the other side of the axis of
  rotation must Kari (63kg) sit in order to balance
  the seesaw.

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Answer 12

• m1 58.5kg, l1 1.5m, m2 63kg, l2 ?, T
  ?
• T F x d (d the lever arm, l1 and l2)
• Convert mass to Newtons m1 58.5kg x 9.81N/1kg
  573.88N
• T 573.88N x 1.5m 860.83Nm (Tara)
• Given Tp Tr
• F 63kg x 9.81N/1kg 618.03N
• 860.83Nm 618.03N (l2)
• 860.83Nm / 618.03N l2 1.39m

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Question 13

• Shannon and Kristine sit on opposite sides of a
  play ground seesaw. If
• Shannon weighing 200 N, is 1.5m from the seesaws
  axis of rotation.
• Kristine weighing 190 N, is 1.6m from the
  seesaws axis of rotation.
• Whos end will the seesaw fall toward.

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Answer 13

• F1 200N, l1 1.5m, F2 190N, l2 1.63m
• T1 200N (1.5m)
• T1 300Nm
• T2 190N (1.63m)
• T2 304Nm
• The seesaw will drop toward Kristine.

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Conclusion