Static Equilibrium; Torque

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Branden Park
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1 Static Equilibrium; Torque

2 The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium.

3 The first condition for equilibrium is that the net force on the object is zero. This is known as translational equilibrium.the conditions can be stated as follows: Generally: F Net = 0 Using components: ΣF x = 0 ΣF y = 0

4 Example: A 10.0 kg chandelier is suspended by wires as shown. What are the tensions in all the wires (Θ 1 = 30 0 and Θ 2 =60 0 )?! 2 wire (2)! 1 wire (1) wire (3)

5 Solution: F T3 According to the FBD [fig (1) ], ΣF y = 0 mg=f T3 =98 N +y +x Fig (1) m mg All the wires are connected at a common point. At this point all the tensions are pulling as shown (on right [fig (2)] and the tensions must add to zero. F T2 Fig (2)! 2 mg F T1! 1 If solving with components one must break the tensions of wires (1) and (2) into x an y components [fig (3)]. Write components in terms of the magnitudes of the tensions F T2 sin! 2 F T1 sin! 1 ΣF x = 0 F T1 cosθ 1 =F T2 cosθ 2 F T1 cos30=f T2 cos60 F T2 cos! 2 Fig (3) F T1 cos! 1 ΣF y = 0 F T2 sinθ 2 +F T1 sinθ 1 =mg F T2 sin60 +F T1 sin30=(10)(9.8) mg One has two equations and two unknowns. By using substitution one can find that: F T1 = 49 N F T2 = 84.9 N You should also be able to solve this by constructing a 2-d vector diagram (head to tail method).

6 Activity: Static 2-d forces Connect masses m 1 and m 2 to mass m 3 over the pulleys as shown and a static situation is achieved. Make a note of the value of the masses and measure the angles. Assume the pulleys are massless and frictionless (and also the strings). Derive the following relationships from FBDs of the masses m 1,m 2, and m 3 : string 1 m 1 θ 1 θ 2 m 3 m2 string 2 m 1 gcosθ 1 = m 2 gcosθ 2 m 3 g = m 1 gsinθ 1 + m 2 gsinθ 2 Using the known masses and measured angles, verify the above equations (in box). The LHS of each equation in theory should equal the RHS of each equation. Perhaps the assumptions made are not entirely accurate and of course experimental error cannot be avoided.

7 Example: A heavy physics book (m= 4.50 kg) is being held by a force (F) at an angle of 60 0 (Θ) against a vertical wall with friction (μ s = 0.35) as shown. a) What is the minimum force F required to keep it from sliding down the wall? b) What is the maximum force F that can be applied without it sliding up the incline?! m " s F

8 Solution: (see mr. martin for full solution) a) F= 42.4 N b) F= 63.8 N

9 Torque The second condition of equilibrium is that there be no torque around any axis (the net torque is zero); the choice of axis is arbitrary. This is called rotational equilibrium. Lets examine what torque actually is.

10 Torque is an angular rotation of an object caused by a force. The location where the force acts on the object is important when calculating torques. Torque is a vector. There can be clockwise (cw), or counterclockwise (ccw) torques. Using a rotating wheel as an example, if it were rotating with a non-zero net torque, it would have an angular acceleration. A Wheel rotating with a constant speed, or that is stationary has a net torque of zero about its axis of rotation.

11 To make an object start rotating, a force is needed; the position and direction of the force matter as well. The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.

12 A longer lever arm is very helpful in rotating objects. A longer lever arm will provide a larger torque.

13 Here, the lever arm for F A is the distance from the knob to the hinge; the lever arm for F D is zero since the line of action of FD passes through the hinge; and the lever arm for F C is as shown(r C ).

14 Units of torque: Nm The torque is defined as: Where r is the perpendicular distance, measured form some location, to the line of action of the force. Or another way of stating it is: τ p = rf sinθ Where r is measured to the line of action of the force F, and the angle ϴ is the angle between the directions of r and F. The quantities r and F are vectors and have a direction. The subscript p represents the location about which the torque is calculated. It is often calculated about a pivot point (such as the hinges shown in the diagrams)but it does not have to be.

15 Torque and line of action: In this situation if you calculated the torques either measuring along l 1 or l 2 to the line of action of the force they would be equal. In most cases if it is a beam such as this it is easier to measure along the beam (l 1 ). l is equivalent to r. F τ p = l 1 F sinθ 1 = l 2 F sinθ 2 l 1 θ 1 P l 2 θ 2 Line of action

16 Example: Calculate the torque about point p due to the individual force and state whether it is a clockwise or counterclockwise torque. solution: T= (3)(20)(sin 40)= 39 Nm counter clockwise. What angle would cause the greatest torque about p?

17 If an object were stationary it is obviously not moving or rotating. The condition for rotational equilibrium is that the net torque is zero ( τ P Net = 0 ) as was stated earlier in the power point. This condition is often stated in the following way: Σ τ P CW = Σ τ P CCW This just means that the sum of all clockwise torques must equal the sum of all counter clockwise torques about location p. The location about which torques are taken is completely arbitrary in equilibrium. However there are more convenient locations commonly used.

18 Example: If there are only two forces acting on the object below as shown: a) What is the torque about P caused by F 1? b) What is the torque about P caused by F 2? c) Would the beam be in rotational equilibrium if these were the only forces acting on it?

19 Solution: a) 37.5 Nm cw b) 96.4 Nm ccw c) No. There is a non-zero net torque ( 58.9 Nm counterclockwise)

20 Center of Mass When drawing FBD s when using torque is important to label precisely where the forces act. The gravitational force acting on the object is labeled at its centre of mass, which is usually the object's geometric centre. This assumes that the object has a uniform density. Further investigation: What is the difference between the centre of mass and centre of gravity (CG)? For the situations we are considering there is no difference.

21 Solving Statics Problems 1. Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act. 2. Choose a coordinate system and resolve forces into components. 3. Write equilibrium equations for the forces. 4. Choose any axis (location p) perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously. 5. Solve. *Some problems may only require forces, or only torque, while others require both.

22 Example: A 5.00 kg mass is placed on top of a horizontal 4.00 m long beam, 1.50 m from the right end of the beam. In turn the beam (mass 10.0 kg ) is supported by two supports at it s ends kg A 1.50 m B a) Draw the FBD of the beam. b) Determine the forces exerted from the supports (A and B) on the beam.

23 Solution: a) F A P m 2 g F B A 1.50 m B m 1 g b) Translational equilibrium: ΣF y = 0 F A +F B = m 1 g+m 2 g= 147 Rotational equilibrium: Στ P CW = Στ P CCW 2(10)(9.8)sin (2.5)(5)(9.8)sin90 0 =4F B F B = 79.6 N Sub back into to force equation to find F A F A = 67.4 N

24 Example: A metre stick (m 1 = 25 grams) is resting horizontally on a support, and the meter stick supports a suspended 15 gram mass 10.0 cm from its right end. Where is the support positioned relative to the left end of the metre stick? Start by drawing the FBD of the metre stick. X=? m 2

25 Solution: F N p 50cm 40-L L 10 cm m 1 g One can take torques about point p, and will define L the distance from p to the force m 2 g. Its not always necessary to convert units to the mks system. I will leave the length units in cm. The condition: Στ P CW = Στ P CCW m 2 g And recalling that: τ p = rf sinθ L(15)g=(40-L)25g, L= 25 cm, and x= 65 cm *Another option could be to solve for F N using forces, then take torque about the left end of the meter stick to find x.

26 Example: A 5.00 m long beam of mass 10.0 kg is supported by a horizontal wire with a tension 1500 N connected 2.00 m (x) from its end. The beam is also supported by the frictionless hinge at the bottom of the beam. The beam supports an unknown suspended mass M and its end. The angle ϴ is a) Draw the FBD b) What is the value of mass M? c) What force does the hinge exert on the beam? State magnitude and direction. x m M ϴ hinge

27 Solution: a) It is convenient to label the reaction force R with two unknown component vectors. Once the components are found the magnitude and direction of R can be determined N Ry 30 0 Mg b) To find M one can take torque about position p and measure along the length of the beam. The force exerted by the hinge would not appear in the relationship. Στ P CW = Στ P CCW p 60 0 Rx mg (2.5)(10)(9.8)sin30 + (5)M(9.8)sin30 = 3(1500)sin60 M = 154 Kg c) Using forces one can determine Rx and Ry. You need to define the +y and +x direction for forces. +y +x

28 c) continued ΣF x = 0 Rx= 1500 N ΣF y = 0 Ry = mg + Mg = N Adding the components of R head to tail one finds: R= 2.2x10 3 N, ϴ= 47 0 ϴ R Rx Ry

29 Example: A 3.00 m long ladder (m= 20.0 kg) leans against the wall as shown(θ=40 0 ). If one assumes no friction between the wall and the ladder what is the minimum coefficient of friction required between the ground and the ladder to stop it from sliding? Start by drawing the FBD. ϴ

30 Solution: ΣF x = F N1 F N1 = f, recall f = µf N 2 ΣF y = 0 mg = F N 2 = 20(9.8)= 196 N Στ P CW = Στ P CCW 50 0 mg f 40 0 F N2 p (3)F N1 sin40=1.5(20)(9.8)sin50, F N1 = N F N1 = f = µf N 2 µ = F N1 F N 2 = /196= 0.596

31 The previous technique may not fully solve all statics problems, but it is a good starting point.

32 If a force in your solution comes out negative (as F A will here), it just means that it s in the opposite direction from the one you chose. This is trivial to fix, so don t worry about getting all the signs of the forces right before you start solving.

33 If there is a cable or cord in the problem, it can support forces only along its length. Forces perpendicular to that would cause it to bend.

34 Example: A 6.00 m long beam (m 1 = 10.0 kg) is supported by a wire ( 1.00 m from end) and a hinge. In turn the beam supports a hanging 5.00 kg (m 2 ) mass 2.00 m from the hinge. a) Draw the FBD of the beam. b) Determine the tension (F T ) in the wire, and the force (F) that the hinge exerts on the beam. hinge 2 m wire kg m

35 Solution: a) Fy Fx F T 2 m m m 2 g Taking torque about the hinge: m 1 g 40 0 Στ P CW = Στ P CCW 2[(5)(9.8)]sin50 + 3[(10)(9.8)]sin50 =5F T sin30 F T = 120 N ΣF x = 0 Fx = F T cos70 = N ΣF y = 0 Fy + F T sin70 = 5(9.8) + 10(9.8) Fy= N F Fy! Fx F = 53.4 N, Θ =

36 These same principles can be used to understand forces within the body.

37 The angle at which this man s back is bent places an enormous force on the disks at the base of his spine, as the lever arm for F M is so small.

38 Stability and Balance If the forces on an object are such that they tend to return it to its equilibrium position, it is said to be in stable equilibrium.

39 If, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium.

40 An object in stable equilibrium may become unstable if it is tipped so that its center of gravity is outside the pivot point. Of course, it will be stable again once it lands!

41 People carrying heavy loads automatically adjust their posture so their center of mass is over their feet. This can lead to injury if the contortion is too great.

HATZIC SECONDARY SCHOOL

HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT STATIC EQUILIBRIUM MULTIPLE CHOICE / 33 OPEN ENDED / 80 TOTAL / 113 NAME: 1. State the condition for translational equilibrium. A. ΣF = 0 B. ΣF