ΣF = 0 and Στ = 0 In 2-d: ΣF X = 0 and ΣF Y = 0 Goal: Write expression for Στ and ΣF

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1 Thur Sept 24 Assign 5 Friday Exam Mon Oct 5 Morton 235 7:15-9:15 PM if conflict Today: Rotation and Torques Static Equilibrium Sign convention for torques: (-) CW torque (+) CCW torque Equilibrium ΣF = 0 and Στ = 0 In 2-d: ΣF X = 0 and ΣF Y = 0 Goal: Write expression for Στ and ΣF

2 A person of weight 800 N is sitting on a chair of weight 10 N. The chair is supported by a rope over a pulley. The person pulls down on the rope with a force of F to support themself. What force, F, is required to hold themselves stationary? (1) 400N (2) 405N (3) 800N (4) 810N (5) 1600N (6) 1620N F F ΣF = ma 2F F G = 0 2F = (800N+10N) Consider seat as part of person. Two ropes pulling up on person and seat, each with same force F G

3 A person of weight 800 N is sitting on a chair of weight 10 N. The chair is supported by a rope over a pulley. The person pulls down on the rope with a force of F to support themself. What force is required to accelerate upward at 0.5 m/s 2? (1) 362N (2) 384N F g = mg (3) 400N m = F g /g (4) 405N (5) 426N m = 810N/9.8m/s 2 (6) 483N m = 82.6 kg F F ΣF = ma 2F F G = ma 2F = (800N+10N) + ((82.6kg)0.5m/s 2 ) F G

4 Block B of mass 1.5 kg is accelerating downward at a rate of 3.0 m/s 2. Block A is connected by a massless string. There is no friction between Block A and the table. What is the tension in the string? +y +x F N T ΣF y = ma y A F g T B F g +y F g T = ma y T = F g - ma y T = 1.5* *3 T = 10.2 N

5 Block B of mass 1.5 kg is accelerating downward at a rate of 3.0 m/s 2. Block A is connected by a massless string. There is no friction between Block A and the table. What is the mass of Block A? +y +x F N T ΣF x = ma x A T B +y T = ma x 10.2N = m * 3m/s 2 m = 3.4kg F g F g

6 A 200 N box is hanging from a rope. Two ropes attach the box to the ceiling at the angles given. What is the tension in ropes 2 and 3? T 1 T 2 θ 1 =30 θ 2 = 60 ΣF x = ma x -T 1 cosθ 1 + T 2 cosθ 2 = 0 T 1 = T 2 (cos θ 2 /cos θ 1 ) T 1 = T 2 (0.577) T 3 200N ΣF y = ma y T 1 sinθ 1 + T 2 sinθ 2 200N= 0 T 1 (0.5) + T 2 (0.866) = 200 N T 2 (0.577) (0.5) + T 2 (0.866) = 200 N Solve: T 2 = 173 N T 1 = T 2 (0.577) = 100. N

7 A 200 N box is hanging from a rope. Two ropes attach the box to the ceiling at the angles given. What is the tension in rope 3? (1) 50 N (2) 86 N (3) 100 (4) 136 N (5) 173 N (6) 200 N T 1 T 2 θ 1 =30 θ 2 = 60 Point where T 3 attaches to box, a=0 ΣF = ma T 3 200N = 0 T 3 = 200 N T 3 200N

8 Three boxes are accelerating to the right at a rate of 2.0 m/s 2. All 3 have non-zero mass. How do T 1, T 2, and T 3 relate? (1) T 1 = T 2 = T 3 (2) T 1 < T 2 < T 3 (3) T 3 < T 2 < T 1 (4) T 1 < T 2 = T 3 (5) T 3 = T 2 < T 1 (6) T 1 = T 2 < T 3 all 3 boxes same acceleration ΣF X = ma T1 = m A a T2 = (m A +m B )a T3 = (m A +m B +m C )a

9 Example: Three Boxes Three boxes are accelerating to the right at a rate of 2.0 m/s 2. The tension in rope 2 (T2) is 6 N. The mass of block B is 1 kg. The mass of block C is 2 kg. Find the tension in rope 1 (T1). Find T1: T1 B T2 ΣF X = ma T2 - T1 = ma T1 = T2 - ma T1 = 6N (1kg)*(2.0m/s 2 ) T1 = 4N

10 Three Boxes Three boxes are accelerating to the right at a rate of 2.0 m/s 2. The tension in rope 2 (T2) is 6 N. The tension in rope 1 (T1) is 4 N. The mass of block B is 1 kg. The mass of block C is 2 kg. What is the mass of block A? kg kg kg kg kg kg A T1 ΣF X = ma T1 = ma 4N = m*(2 m/s 2 ) m = 2 kg

11 Example: Three Boxes Three boxes are accelerating to the right at a rate of 2.0 m/s 2. The tension in rope 2 (T2) is 6 N. The mass of block B is 1 kg. The mass of block C is 2 kg. Find the tension in rope 3 (T3). ΣF X = ma Find T3: T3 T2 = ma T2 C T3 T3 = ma + T2 T3 = (2kg)*(2.0m/s 2 ) +6N T3 = 10N Can also use all 3 blocks: T3 = (2kg+1kg+2kg) * 2m/s 2

12 Example: Three Boxes Three boxes are accelerating to the right at a rate of 2.0 m/s 2. The mass of Block A is 2 kg. The mass of block B is 1 kg. The mass of block C is 2 kg. Suppose friction were involved. µ S = 0.45 and µ K =0.35. What is the magnitude of the tension T1? ΣF X = ma +y In this case: x F N ΣF y = ma T1 F FRICT = ma +x y x F T1 = F FRICT + ma x F N F g = 0 f T1 = 6.9N + 2kg*2m/s 2 A F N =F g =19.6N T1 = 10.9 N F FRICT = µ K F N F g F FRICT = µ K F g F FRICT = 0.35*2kg*9.8m/s 2 T1

13 A board is placed in a support. F1 and F2 push down on the board. If the board stays stationary, how do the forces compare? (1) F1 > F2 (2) F1 = F2 (3) F1 < F2 F1 F2 Support (fulcrum)

14 Motion Translational linear Rotational All motion can be broken into these parts Now dealing with Extended Objects Chapter 9: Torques, Static Equilibrium, Center of Mass Equilibrium acceleration zero Static Equilibrium acceleration AND velocity zero Torque Tendency to rotate, a twist A force may or may not apply a non-zero torque Object in equilibrium if forces are balanced AND if torques are balanced.

15 Two forces are exerted on a wheel as shown. If the wheel is stationary, how do the two forces compare? (1) F1 > F2 (2) F1 = F2 (3) F1 < F2 F1 F2 While we use beams or boards a lot, this works for any shaped object.

16 Equilibrium Equilibrium Accel = 0 Static Equilibrium Accel = 0 AND velocity=0 Stationary What happens? What happens? If Forces applied at different points, can get a twist

17 'Twists' Hold and push down or up at different places What happens if let go of 'hold'? Hold Hold 'Twist' depends on force and position force applied If applying torque and object in equilibrium, must be another torque somewhere

18 Torque Simple Situation r F Axis If the force is perpendicular to the object to which the force is applied: τ = F r where r is the distance from the 'axis of rotation So need to choose an axis of rotation before calculating torque τ (greek lowercase tau) is the torque Units: N*m (enter them this way into computer) For static problems, axis of rotation can be anywhere

19 Which of the following forces provides the greatest torque about the given axis? (the black dot represents the axis of rotation) 2F 2F 1.5r r F A B 2.5r C (1) A (2) B (3) C (4) All same τ A = 2F*1.5r = 3F*r τ B = 2F*r = 2F*r τ C = F*2.5r = 2.5F*r τ A : CW (-) τ B : CCW (+) τ C : CCW (+) Sign Convention: (+) for CCW rotation; (-) for CW

20 What force, F, is required to balance the meter stick? (1) 0.5N (2) 1.0N (3) 2.0N 25cm 1.0N 50cm Support (fulcrum) F Στ = 0 +(1.0N)(25cm) - F(50cm) = 0 F = 0.5N Note opposing directions of torques Convention: + CCW, -CW

21 A force of 0.5N is applied downward 30cm to the right of the fulcrum. Where, relative to the fulcrum, would one exert a force of 0.7N downward to balance the the stick? (1) 21cm to the left (2) 21 to the right (3) 42 cm to the left (4) 42 cm to the right Στ = 0 21cm 0.7N 30cm Support (fulcrum) -(0.5N)(30cm) + (0.7N)L = 0 L = 21.4cm 0.5N

22 A force of 0.5N is applied downward 30cm to the right of the fulcrum and a force of 0.7N is exerted 21cm left of the fulcrum. What is the force exerted by the fulcrum on the board? (1) 0.2N Up (2) 0.2N Down (3) 0.6N Up (4) 0.6N Down (5) 1.2N Up (6) 1.2N Down ΣF Y = 0 0.7N 21cm 30cm Support (fulcrum) 0.5N -(0.5N) - (0.7N) + F FULCRUM = 0 F FULCRUM = 1.2N

23 Solving Static Equilibrium Problems Draw FBD Draw axes Select axis of rotation Write expressions for ΣF X = 0, ΣF Y = 0 and Στ = 0 Solve Where should you choose axis? If static equilibrium and you don t know a force and you don t care about that force, choose axis so that force gives zero torque!

Levers F IN Class 1 Lever Fulcrum between input effort and output load F OUT Small force with long lever arm can exert large force with short level arm

24 Levers F IN Class 1 Lever Fulcrum between input effort and output load F OUT Small force with long lever arm can exert large force with short level arm Disadvantage Load doesn't move as far Mechanical Advantage: MA = F OUT F IN Pliers, scissors Catapult Oars Crowbar Seesaw

25 At which point and at which direction would the least amount of force be required to hold the lever stationary? Longest lever arm - perpendicular force

26 Torque More Formal Pick axis. Can calculate torque due to any force. Torque (τ) = (mag of Force) x (perpendicular lever arm) Axis of Rotation τ = F r sinθ τ = F r τ = 0 Perpendicular Lever arm line drawn through hinge and perpendicular to line of action.

27 Torque Perpendicular Lever Arm axis May have to calculate using trig Be careful to make sure you use right trig function Perpendicular Lever arm line drawn through hinge and perpendicular to line of action.

28 Two Different Approaches Find lever arm through trig Look at Components of force r θ F θ F θ F d d, r τ = F r τ = F (d cos θ) τ = F τ = F d τ = (F cosθ) d

29 A cook holds a 2.00kg carton of milk at arm's length. What force F b must be exerted by the bicep muscles? (ignore the weight of the forearm for now). Rubber Band Hold

30 A cook holds a 2.00kg carton of milk at arm's length. The bicep muscle exerts a force F b on the forearm. What is the expression for the torque of the bicep muscle on the forearm? Use the elbow as the axis of rotation. (1) (F b cosθ) 8cm (2) (F b sinθ) 8cm (3) F b 8cm (4) -(F b cosθ) 8cm (5) -(F b sinθ) 8cm (6) -F b 8cm (7) None of the above

31 A cook holds a 2.00kg carton of milk at arm's length. The bicep muscle exerts a force F b on the forearm. What direction is the force of the elbow on the forearm? (Choose 1,3,5,7 only if exactly on axis. Choose zero for no force) To balance torque, need to push down. To balance forces horizontally, need to push to left.

32 A cook holds a 2.00kg carton of milk at arm's length. What force F b must be exerted by the bicep muscles? (ignore the weight of the forearm for now). "Write the expression for the sum of the torques." Write out each torque in terms of variables or quantities given. Choose axis at elbow since don't know those forces yet. Στ = 0 F g (25cm+8cm) F b cos(75º)(8cm) + F ELBOW (0cm) = 0 F g = 2.00kg*9.8 = 19.6N F b = 312N

33 What is the magnitude of the horizontal component of the force that the elbow exerts on the forearm? (1) 61.2N (2) 292N (3) 301N (4) 307N (5) 312N (6) 332N ΣF x = 0 F b sin(75) F ELBOW,X = 0 F ELBOW,X = 312N*sin(75) = 301N

34 What are the magnitude and direction of the force exerted on the forearm by the elbow? If we can find the components of the force, we can reconstruct the full force vector. F ELBOW,X = 301N 301N θ 61N ΣF y = 0 -F g + F b cos(75) F ELBOW,Y = 0 F ELBOW,Y = 312N*cos(75) 19.6N F ELBOW,Y = 61N F! = θ = 301N N 61N tan 1 & # $! = % 301N " 2 = 307N! 11.4

Equilibrium. Today: Center of Gravity Static Equilibrium. Sign convention for torques: (-) CW torque (+) CCW torque

Equilibrium. Today: Center of Gravity Static Equilibrium. Sign convention for torques: (-) CW torque (+) CCW torque Tue Sept 29 Assign 5 Friday Exam Mon Oct 5 Morton 235 7:15-9:15 PM Email by Wednesday afternoon at the latest if conflict Q/A Sunday 1-3,4-6,7-9 Today: Center of Gravity Static Equilibrium Sign convention