Equilibrium

1. Equilibrium

2. The First Condition of Equilibrium In a situation involving equilibrium, there is no acceleration (no change in velocity). Thus the net force (sum of all the forces) is zero.

3. First Condition of Equilibrium : The Net Force is Zero In equilibrium, the sum of all force vertical components is zero and the sum of all force horizontal components is zero.

4. The Balancing Force The force that balances given forces is the force opposite and equal to the sum of the given forces. This balancing force produces a sum that is a point, zero force. In the top diagram, the red force is the sum of the black and blue forces. In the bottom diagram, the red force is the force balancing the black and blue forces.

5. An Equilibrium Problem Given the weight of w is 100.0 N, find the tension forces in the ropes A and B.

6. A 100.0 N B An Equilibrium Problem : Method 1 (Easiest) Given the weight of w is 100.0 N, find the tension forces in the ropes A and B. Solution using the Sine Law: A = 100.0 N and B = 100.0 N sin 45 sin 75 sin 60 sin 75 A = 100.0Nsin45 = 73.20508…= 73.20 N sin 75 B = 100.0Nsin60 = 89.6575… = 89.66 N Free Body Diagram sin 75

7. An Equilibrium Problem : Method 2 Given the weight of w is 100.0 N, find the tension forces in the ropes A and B. Bx – Ax = 0 Bcos30 – Acos30 = 0 .707107B - .866025A = 0 .707107B = .866025A B = .866025A = 1.224745A .707107 Ay[up] + By[up] – 100.0[up] = 0 Asin30 + Bsin30 = 100.0 .5A + .707107B = 100.0{2 equations with 2 unknowns} .5A + .707107(1.224745A) = 100.0 {substitution from above} 1.366025A = 100.0 A = 73.20 N B = 1.224745(73.205) = 89.66 N

8. T1 T1y T2 T1x W A Second Equilibrium Problem An object is suspended as shown. Given T1 is 50.0 N, find the object’s weight and T2. Free Body Diagram Since the weight is to be determined, T2 and T1x can be ignored. T1[up] + W (weight down) = 0 T1[up] = - W[down] = W[up] 50.0sin30 = W {cancelling the [up] directions} W = 25.0 N [down] T2 = 50.0cos30 = 43.3 N [left]

9. The Centre of Gravity The point at which an object would balance on a pivot point is referred to as the centre of gravity. For a uniform object, the centre of gravity is at its mid point. For equilibrium problems, all the mass and all the weight of the object can be considered to be concentrated at the centre of gravity.

10. The Centre of Gravity for Irregular-Shaped Objects The centre of gravity for irregularly-shaped objects is determined by suspending an object from 2 different points with plumb lines. Where the plumb lines intersect is the centre of gravity. Many times the centre of gravity is physically outside of the irregularly-shaped object. Objects rotate about their centre of gravity.

11. Centre of Gravity and Balance Objects balance at their centres of gravity. Objects suspended on a pivot will move so that their centre of gravity lines up beneath the pivot.

12. Torque Torque measures the turning ability of an object about a pivot point. Torque is measured by the product of the rotating force and its perpendicular distance from the pivot. Torque ( ) = f x d . The unit of torque is a Nm is the Greek letter tau (small letter)

13. θ r F θ A General Formula for Torque For forces not perpendicular to the torque arm (left below), = rFsinθ = Frsinθ = FLcosa (right below).

14. Force causing + torque Force causing - torque Positive and Negative Torques A torque causing a counter-clockwise rotation is given a positive value while a torque causing a clockwise rotation is given a negative value.

15. 30 N Force 2 m from fulcrum causing + torque 20 N Force 3 m from fulcrum causing - torque 60 Nm of + torque 60 Nm of - torque The Second Condition of Equilibrium : CT and CCT = 0 In a given plane for a pivoting object, there can be either a clockwise torque (positive: +) or a counter-clockwise torque (negative: -). In an equilibrium situation, the clockwise torque equals the counter-clockwise torque. CT = CCT.

16. Centre of Gravity at Fulcrum causes no torque Centre of Gravity to left of fulcrum (non-uniform object) causes + torque Centre of Gravity treated as a force causing torque The Centre of Gravity and Torque In problems with a uniform beam, the centre of gravity is in the centre of the object and causes no rotation (torque) if the pivot is also around the centre. If the centre of gravity is not at the pivot point, it causes a torque like any force

17. B A Arbitrary Position of the Point of Rotation For many torque problems, the point of rotation is placed at a hinge or fulcrum. But in some problems the point of rotation is arbitrarily assigned. In the example below, either A or B could be assigned as fulcrums or pivots.

18. Levers A lever is a bar that pivots about a point called the fulcrum. In a lever clockwise torque equals counterclockwise torque and the sum of the forces down equals the sum of the forces up (the net forces add to zero).

19. Archimedes and the Lever The Greek mathematician, Archimedes, discovered the law of the lever. Many of the machines he designed used levers. To make the point about the power of a lever for lifting he said, “Give me a place to stand on, and I will move the earth.”

20. Classes of Levers: Determined by Fulcrum Position First class levers have the fulcrum between the effort and load forces. Second class levers have the fulcrum to one side of both the the effort and load forces with the effort force farther from the fulcrum than the load force. Third class levers have the fulcrum to one side of both the the effort and load forces with the load force farther from the fulcrum than the effort force.

21. Advantages of Levers First class levers can produce force or motion advantages. Second class levers produce force advantages. Third class levers produce motion advantages (force disadvantage).

22. Example 1 A torque of 24.0 Nm is needed to tighten a nut. If a person applies a force of 100 N at a right angle to the wrench, what is the minimum length of wrench required? From τ = rFsinθ, r = τ/Fsinθ so r = 24.0 Nm/100 N = 0.240 m = 24.0 cm

23. T = ? L 30.0 o W = 100 N Example 2 A 25.0 N uniform beam is attached to a wall by means of a hinge. Attached to the other end of this beam is a 100 N weight. A rope also helps to support the beam as shown. What is the tension in the rope?

24. T H V 100 N Wb Example 2 Continued The tension in the rope, T, Is obtained by considering torque Using the free body diagram (right), The sum of all torques is zero so LTsin30 - .5L25sin90 – L100sin90=0 Tsin30 - 12.5sin90 – 100sin90 = 0/L = 0 .5T – 12.5 - 100 = 0 .5T = 112.5 T = 225 N

25. 30o 225sin30 = 112.5 N T H 225 N 225cos30 = 194.86 N V 100 N Wb Example 2 Continued The forces of the wall on the beam can be calculated from the relation that the sum of the forces up and down equals zero. The tension force,T, Can be resolved into its horizontal and vertical components. Since T has a horizontal component right of 195 N, the wall must exert a force of 195 N left. The downward forces add to 125N so the upward forces must add to 125 N. Given that T has an upward component of 112.5 N, the wall must also exert an upward force of 12.5 N to balance the downward force of 125N. 112.5[up] + x[wall up] = 100N[down] + 25N[down]

26. T2 T1 1.0 m 30o 5.0 m Another Equilibrium Problem A uniform beam 5.0 m long has a weight of 200 N on it and is suspended by three ropes, as shown. If an 800 N object is placed as shown in the diagram, what is the tension in each of the ropes?

27. 2.5 m 520 N 520sin30 = 260 N 1.0 m T2 T1 520cos30 = 450 N 30o Arbitrary Rotation Point 200 N Beam Weight 800 N Equilibrium Problem Continued The sum of torques is zero so 800(1) + 200(2.5) – 5Tsin30 = 0 T = 1300/5sin30 = 520 N Resolving T into its vertical and horizontal components, vert. = 260 N, horiz. = 450 N. The wall’s force T3 to the left is 450 N. T3

28. 2.5 m 520 N 520sin30 = 260 N 1.0 m T2 T1 520cos30 = 450 N 30o Arbitrary Rotation Point 200 N Beam Weight 800 N Equilibrium Problem Continued Since the sum of all forces up and down is zero, 800 + 200 down = T2 + 260 up. 1000 – 260 = T2 up 740 N = T2

29. 1.2 kg 2.2 kg 1.8 m Centre of Gravity Problem Two steel balls are joined by a “massless” bar. Find the centre of gravity from the 1.2 kg mass. Consider the centre of gravity as a pivot and let x be the distance from the 1.2 kg mass to this pivot. Then 1.8 – x is the distance from the 2.2 kg mass to the pivot. Since the torques are equal: 1.2(x) = 2.2(1.8 – x) 1.2x = 3.96 – 2.2x 3.4x = 3.96 X = 3.96/3.4 = 1.165 = 1.2 m from the 1.2 kg mass

30. 4.1 m 4.1 - x 2.2 kg 2.9 m 2.8 kg x y 2.6 kg 2.9 - y Centre of Gravity Problem Find the centre of gravity for these three ball joined by “massless” bars. (Measure from the centre of the 2.8 kg mass.) Consider the centre of mass between the 2.8 kg and other masses. Then: 2.8x = 4.8(4.1 – x) 2.8x = 19.68 – 4.8x 7.6 x = 19.68 and x = 2.6 Consider the centre of mass between the 2.6 kg and other masses. Then: 5y = 2.6(2.9 – y) 5y = 7.54 -2.6y 7.6y = 7.54 and y = 1.0 Treating the centre of the 2.8 kg mass as the origin, the centre of mass is located at (2.6, -1.0)

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