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Torque Torque is defined as the tendency to produce a change in rotational motion.

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Presentation on theme: "Torque Torque is defined as the tendency to produce a change in rotational motion."— Presentation transcript:

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2 Torque Torque is defined as the tendency to produce a change in rotational motion.

3 Torque is a twist or turn that tends to produce rotation. * * * Applications are found in many common tools around the home or industry where it is necessary to turn, tighten or loosen devices.

4 What makes something rotate in the first place? TORQUE AXIS How do I apply a force to make the rod rotate about the axis? Not just anywhere!

5 TORQUE To make an object rotate, a force must be applied in the right place.To make an object rotate, a force must be applied in the right place. the combination of force and the distance from the axis to the point of application (L) is called TORQUE.the combination of force and the distance from the axis to the point of application (L) is called TORQUE. Force, F lever arm, L Axis

6 Torque = force times lever arm Torque = F  L

7 Torque example F L What is the torque on a bolt applied with a wrench that has a lever arm of 30 cm with a force of 30 N? Torque = F  L = 30 N  0.30 m = 9 N m For the same force, you get more torque with a bigger wrench  the job is easier!

8 Units for Torque Torque is proportional to the magnitude of F and to the distance L from the axis. Thus, a tentative formula might be:  = FL Units: N  m or lb  ft 60 cm 40 N  = (40 N)(0.60 m) = 24.0 N  m,  = 24.0 N  m,

9 Direction of Torque Torque is a vector quantity that has direction as well as magnitude. Turning the handle of a screwdriver clockwise (negative) and then counterclockwise (positive) will advance the screw first inward and then outward.

10 Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. Positive torque: Counter-clockwise, out of page cw ccw Negative torque: clockwise, into page

11 Line of Action of a Force The line of action of a force is an imaginary line of indefinite length drawn along the direction of the force. F1F1 F2F2 F3F3 Line of action

12 The Moment Arm The lever arm of a force is the perpendicular distance from the line of action of a force to the axis of rotation. F2F2 F1F1 F3F3 L L L

13 Calculating Torque Read problem and draw a rough figure.Read problem and draw a rough figure. Extend line of action of the force.Extend line of action of the force. Draw and label Lever arm.Draw and label Lever arm. Calculate the Lever arm if necessary.Calculate the Lever arm if necessary. Apply definition of torque:Apply definition of torque: Read problem and draw a rough figure.Read problem and draw a rough figure. Extend line of action of the force.Extend line of action of the force. Draw and label Lever arm.Draw and label Lever arm. Calculate the Lever arm if necessary.Calculate the Lever arm if necessary. Apply definition of torque:Apply definition of torque:  = FL Torque = force x Lever arm

14 Torque If we know the angle  between F and r, we can calculate torque!If we know the angle  between F and r, we can calculate torque! t = F Lt = F L –r is the total length – F is force L = r sin   = r sin  F –  is angle between F and r The SI unit of torque is the Nm.The SI unit of torque is the Nm. F  Hinge (rotates) Direction of rotation Extend the line of action r L

15 Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate L  = (80 N)(.12m sin 60 o ) = 8.31 N m L

16 Net Torque An object is in “Equilibrium” when: 1.There is no net force acting on the object 2.There is no net Torque In other words, the object is NOT experiencing linear acceleration or rotational acceleration.

17 What mass is needed for the levers to be in equilibrium? Weights are attached to 8 meter long levers at rest. Determine the unknown weights below 20 N ?? 20 N ?? 20 N ??

18 What mass is needed for the levers to be in equilibrium? F 1 = 20 N r1 = 4 mr2 = 4 m F 2 =?? 0 = -F 2 r 2 + F 1 r 1 0 = -(F 2 )(4)+ (20)(4) F 2 = 20 N … same as F 1 Upward force from the fulcrum produces no torque (since r = 0)

19 What mass is needed for the levers to be in equilibrium? 0 = -F 2 r 2 + F 1 r 1 0 = -(F 2 )(2) + (20)(4) F 2 = 40 N F 2 =?? F 1 = 20 N r1 = 4 mr2 = 2 m (force at the fulcrum is not shown)

20 More interesting problems (the pivot is not at the center of mass) Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg. Determine the unknown weight below. 20 N?? CM

21 Trick: gravity applies a torque “equivalent to” (the weight of the lever)(R cm )  cm =(mg)(r cm ) = (100 N)(2 m) = 200 Nm More interesting problems (the pivot is not at the center of mass) 20 N ?? CM Weight of lever Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg.

22 Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg. Determine the unknown weight below. F1 = 20 N CM R1 = 6 m R2 = 2 m F2 = ?? Rcm = 2 m Fcm = 100 N 0 = -F 2 r 2 + F 1 r 1 + F cm R cm 0 = -(F 2 )(2) + (20)(6)+(100)(2) F 2 = 160 N

23 Diving board bolt A 8 meter long diving board with a mass of 40 kg. a. Determine the downward force of the bolt. The Pivot point is not at the center of mass  cm = (392 N) 2 m = 784 Nm F 1 r 1 = F 2 r 2 +  cm

24 Diving board A 8 meter long diving board with a mass of 40 kg. a.Determine the downward force of the bolt. (Balance Torques) bolt R cm = 2 R 1 = 2 F cm = 392 N F bolt = ? N F 1 r 1 =  cm = 784 Nm F1 = 784Nm = 392 N 2m

25 Diving board A 4 meter long diving board with a mass of 40 kg. b. Determine the upward force applied by the fulcrum. (Balance Forces) bolt F cm = 392 NF bolt = 392 N F = 784 N The total upward force has to equal the total downward for the object to be stable.

26 Remember: An object is in “Equilibrium” when: a.There is no net Torque b. There is no net force acting on the object

27 Torque with two supports 1.Label all the forces 2.Choose a pivot point 3.Write the equation for net torques FL Fr Fcm

28 Torque with two supports T net = 0 -F1 L1 + Fcm L cm + F2 (0)=0 -F1 L1 + Fcm L cm + F2 (0)=0 F1 F2 Fcm Pivot point L1 L4 cm Lcm

29 0 = F 1 (0) -755 (1.40) – 167 (2.75) + F 2 (5.50M) F 2 = 276 N F 2 + F 1 = N F 1 = 646 N F1F1 F2F2 Pick a pivot pt. What is the force exerted at each end?


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