Rotational Mechanics Part II Torque. Pre AP Physics

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Kathryn Banks
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1 Rotational Mechanics Part II Torque Pre AP Physics

2 We have so far discussed rotational kinematics the description of rotational motion in terms of angle, angular velocity and angular acceleration. Now we will discuss what causes this rotational motion. When one or more forces act on an object at a location other than the center of mass of an object the object has a tendency to rotate. This rotating effect is TORQUE. DEFINITION OF TORQUE Torque is a vector quantity that measures the tendency of a force to rotate an object about an axis. Torque = (Magnitude of force) (Lever arm)

3 The axis about which the rotation occurs is called the axis of rotation and the point about which the rotation occurs is referred to as a PIVOT POINT. = F l Cross Product = F lsin Torque Magnitude F is the force magnitude and l is called the lever arm (moment arm) and it is the perpendicular distance (d ) from the pivot point to the line of action of the force.

4 Force vector Pivot point Lever arm

5 Line of action of Force vector Lever arm Force vector Pivot point Length

6 Force vector Pivot point length Lever arm Line of action of Force vector

7 Some more important points about torque Engineers will often use the term "moment" to describe what physicists call a "torque". We will adopt a convention that defines torques that tend to cause clockwise rotation as negative and torques that tend to cause counter-clockwise rotation as positive. Torques are always defined relative to a point. It is incorrect to simply say the "torque of F". Instead you must say the "torque of F relative to point X". More general definition for the torque is given by the vector (or cross product). When a force acting at a point which has position vector r relative to an origin O the torque exerted by the force about the origin is defined as r F

9 Note that the metric unit of torque is the Newton meter. The English unit of torque is the Foot- pound. (Yes, the units are backward written!) Also remember that another name for a N m is Joule an energy unit. But torque is NOT an energy (scalar) quantity. Torque is a vector quantity. The cross product of two vectors always yields another vector as an answer.

10 Example #1 What is the torque produced by a 10-N force on a rod if the perpendicular distance (torque arm) from the force to the axis of rotation point (pivot point) is 4.5-m? Torque = F d = (10-N) (4.5-m) Torque = 45.0 N-m

11 Example #2 What is the torque produced by the forces F 1, F 2, and F 3 about the pivot point of the beam in the figure below? 30 o F 2 = 100-N Pivot Point 2.5-m F 3 = 250-N 2.0-m 1.0-m F 1 = 50-N (1) T 1 = F 1 d T 1 = (50-N) (1-m) T 1 = 50 N-m

12 30 o F 2 = 100-N Pivot Point 2.5-m F 3 = 250-N (2) 2.0-m T 2 = F 2 d 1.0-m F 1 = 50-N T 2 = (100-N) (? ) 30 o 2-m 100-N 2m Sin 30 o = 1.0-m T 2 = (100-N) (1-m) T 2 = 100 N-m

13 30 o F 2 = 100-N Pivot Point 2.5-m F 3 = 250-N 2.0-m 1.0-m F 1 = 50-N (3) T 3 = F 3 d T 3 = (250-N) (2.5-m) T 3 = 625 N-m

14 Again, torque, like displacement is a vector quantity. However, in our text, torque is treated as a scalar quantity for the purpose of making the concept a little easier to study at first. Having stated that, we still need to look at this rotating force in terms of clockwise or counterclockwise motion For this class let s adopt the following conventions: (1) Clockwise Torque s are considered negative. (2) Counterclockwise Torque s are considered positive. +

15 To determine the sign of the torque, imagine that it is the only torque acting on the object and that the object is free to rotate. If two or more forces are acting on an object then there may be more than one torque acting on the object. In that case you normally find the net torque using the formula: Net = So if we looked at example #2 again our net torque acting on the beam would be:

16 30 o F 2 = 100-N Pivot Point 2.5-m F 3 = 250-N 2.0-m 1.0-m F 1 = 50-N Net = ( 50 N-m) + ( 100 N-m) + (+625 N-m) = +475 N-m Net Or the net torque acting on that beam is 475 N-m counterclockwise.

17 Maintaining Rotational Equilibrium In solving problems that have net forces acting on an object it s important that we look at Newton s second law (F = ma), but we also need to look at the rotational effects (if any) of those forces. In effect if the object is to be in equilibrium (still or moving at constant speed) then the following must exist F x = 0; F y = 0 and T = 0. Now if the net force on an object is zero ( F x = 0; F y = 0), then the object is said to be in translational equilibrium. If the net torque on an object is zero ( T = 0), then the object is in rotational equilibrium. Sometimes this rotational equilibrium statement is called the second condition for equilibrium.

18 Note that in solving problems with the second condition of equilibrium the axis about which the torque is calculated can be arbitrary. Also note that if a force acts along a line passing through the axis of rotation it will not produce any torque. Again please remember the following conventions in the problem solving involving rotational motion (1) Forces upward and to the right are considered positive. (2) Forces downward and to the left are considered negative. (3) Clockwise Torque s are considered negative. (4) Counterclockwise Torque s are considered positive.

19 Example #3 A horizontal rod AB, is 10-m long. It weighs 500-N and its center of gravity, C, is located 3-m to the right of A. At A a force of 1000-N acts straight down. At B a force of 750-N acts straight down. At D, 2-m to the left of B, a force of 400-N acts straight up. At E, 1.0-m to the right of A, a force of 750-N acts straight up. (A) What is the magnitude and direction of the force that must be used to produce equilibrium and (B) what is the location of this force, with respect to point A? A B

20 750-N 400-N 1.0-m E 10-m D 2.0-m A B C? 3.0-m 1000-N 500-N F 750-N (A) In order for the beam to be translational equilibrium, F x = 0; F y = 0. Now there are no horizontal (x-direction) forces, so we need only consider the forces in the y-direction. In order for the forces in the y-direction to be in equilibrium the forces up minus the forces down must equal zero.

21 Or: F + (+750-N) + (+400-N) + ( 1000-N) + ( 500-N) + ( 750-N) = 0 F = N. Notice that the answer is positive. This tells us that the direction we drew the vector-arrow for the unknown force was in the correct direction (B) To find the location of the point of application of this force: (1) Choose a pivot point to calculate the sum of the torques. Since the problem asks for the location with respect to point A choose point A as the location of the pivot point.

22 750-N 400-N 1.0-m E 10-m D 2.0-m A 3.0-m C? 1000-N 500-N F 750-N (2) Calculate the sum of the torques about that chosen pivot point and set equal to zero. Counterclockwise Torque T = 0 (+750 N)(1 m) + ( 500 N)(3 m) + (+1100 N)(?) + (+400 N)(8m) + ( 750 N)(10 m) = 0 Clockwise Torque B

23 Note that for the positive torque (counterclockwise) producing forces you might use a positive sign for the forces and a negative sign for counterclockwise torque producing forces. (3) Solve the resulting equation for the unknown distance.? = 750 N-m 1500 N m 3200N m 7500N m 1100 N? = 4.59-m Note that if you get a negative distance, you just chose the location on the wrong side of the pivot point.

24 Example #4 Andy and Dana were cleaning windows on the outside of the 35 th floor of one of the skyscraper banks in downtown Fort Worth to earn enough spending money for the prom so that they could hear their senior song together one more time. The uniform fifteenmeter long plank they are standing on weighs 700 Newtons and each person weighs exactly 500 Newtons. How much tension, in Newtons, is exerted on each rope if Dana is standing five meters from Andy and Andy is standing seven meters from the center of the plank? (Assume that there is only one rope at each end of the plank.) 5 m 15 m

25 F B F A 7.5 m 15 m 5 m 0.5 m 700 N 500 N 500 N F y = 0 F B 700 N 500 N 500 N + F A = 0 F B = 1700 N F A

26 Choosing point B as the pivot point gives the equation: F B F A 7.5 m 15 m 5 m 0.5 m T = N 500 N 500 N ( 700)(7.5) + ( 500)(9.5) + ( 500)(14.5) + (+F A )(15) = 0 F A = 1,150 N F B = 1,700 N 1,150 N = 550 N

27 Example #5 Josh pushed a 10 cm x 30 cm wooden plank that is 12 meters long out over the ledge of a building that is 50 m tall so that he could sit on the edge of the plank to drop stones and test certain laws of physics. If the plank is uniform and weighs 450 N, how far from the edge of the building, in meters, could Josh sit if he weighs 500 N? 12 m 6? B? 450 N 500 N

28 12 m 6? B? 450 N 500 N F y = 0 F B = 0 F B = 950 N (This is the force that the corner of the building is pushing upward with.) Choosing point B (building corner) as the pivot point gives the equation: T = 0 ( 500)(?) + (+450)(6?) = 0? = 2.84 m

29 Example #6 One day Big Al, who weighs 900 N, was sitting at one end of a seesaw, while Liz, who weighs 550 N, was sitting on the other end up in the air. Along came Mighty Mutt, who weighs 400 N, and seeing their predicament jumps on the middle (fulcrum where the seesaw is balanced) and walks toward Liz until the seesaw is parallel to the ground. If the seesaw is 8 meters long, how far away from Liz is Mighty Mutt, in meters, when all concerned are in equilibrium?

30 ? F Fulcrum 550 N 400 N 900 N F y = 0 F Fulcrum = 0 F Fulcrum = 1850 N Choosing Liz as the pivot point gives the equation: T = 0 ( 400)(?) + (+1850)(4) + ( 900)(8) = 0? = m Note, that by choosing Liz as the pivot point the distance found is the distance needed.

31 Example #7 When Trevor drives his truck unto the scales his front wheels exert a force of 10,000 N. When he drives his truck forward so that the back wheels are now on the scales the scales read 8600 N. If the distance between the front and back wheels is 2.75 m: (A) what is the weight of the truck in Newtons? (B) Where is the center of gravity of the truck with respect to the front wheels?

32 2.75 m CG 8600 N? 10,000 N (A) F y = 0 10,000 N N F CG = 0 F CG = 18,600 N Choosing the front tire as the pivot point: (B) T = 0 (+18,600)(?) + ( 8600)(2.75) = 0? ? = 1.27 m behind the front wheel

33 ? F Fulcrum 550 N 400 N 900 N F y = 0 F Fulcrum = 0 F Fulcrum = 1850 N Choosing Fulcrum as the pivot point gives the equation: T = 0 (+550)(4) + (400)(4?) + ( 900)(4) = (?) 3600 = (?) = 0? = ( 200) ( 400)? = m

Chapter 12 Static Equilibrium

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