Torque and Static Equilibrium

Transcription

1 Torque and Static Equilibrium Ch. 7.3, 8.1 (KJF 3 rd ed.) Phys 114 Eyres Torque 1

2 Causing rotation: Torque Three factors affect the how a force can affect the rotation: The magnitude of the force The distance, r, out from pivot where the Force is applied The angle in which the force is applied Hinge is the pivot Only the force component perpendicular to the bar contributes to the torque. r is distance from pivot to where force is applied Pearson Education, Inc. Torque Torque is due to the component of the force perpendicular to the radial line. The SI unit of torque is the Newton-meter (Nm). 2

3 Sign of Torque (direction) The force causes the pencil to turn counterclockwise (a positive torque) The force causes the pencil to turn clockwise (a negative torque) about the axis of rotation? Sign of Torque does NOT always match the sign of the Force 2014 Pearson Education, Inc. Torque A steel band exerts a horizontal force of 80.0 N on a tooth at point B. What is the torque on the root of the tooth about point A? θ 80 N 3

4 Torque (42 ) Tooth will rotate CCW which is positive rotation. Thus the sign of the torque is positive. A steel band exerts a horizontal force of 80.0 N on a tooth at point B. What is the torque on the root of the tooth about point A? 48⁰ θ 80 N A Second Interpretation of Torque F 4

5 Torque Which force would be most effective in opening the door? Static Equilibrium 5

6 Where is the gravitational force exerted on a rigid body? The balance point is the location of the center of mass We can assume that the gravitational force exerted on an object is exerted at the location of its center of mass. That is why the object's center of mass is sometimes called the object's center of gravity. Notice that the center of mass is where the hammer would balance if supported at that point Pearson Education, Inc. Conditions of Equilibrium Translational Static Equilibrium An object is at rest with respect to the observer and the components of the sum of the forces exerted on it in the x- and y- directions are zero. Rotational Static Equilibrium A rigid body is at rest with respect to the observer and the sum of the torques about any axis of rotation is zero. Σ 0 Σ 0 Σ0 6

7 Static Equilibrium? Force x Force y r Torque 0-6 N 0.2 m +1.2 Nm 0 +9 N 0 m 0 Nm 0-2 N 0.5 m -1.0 Nm 0-1 N 0.2 m -0.2 Nm Σ 0 Σ0 Yes: In static Equilibrium $ Pearson Education, Inc. 6!"# 1 2 Force x Force y r Torque Example 8.1 F barbell m Nm Weight 0-0N m 0 Nm F bicep 0 +F b 0.04 m +Fb(0.04m) Nm F joint 0 -Fj 0 0 Nm ' ')718!" +,-? 0,*3!"+,- 45& 6 & & ' (!)"*!" +,-? ' /0,/122!" +,- 450 Σ 0 Σ / / ; 0 / ; 0 ; Pearson Education, Inc. 7

8 Lifting Problem A person bending forward to lift a load with his back rather than with his knees can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. Consider the model of a person bending forward to lift a 200-N object. The spine and upper body are represented as a uniform horizontal rod of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is Find the tension in the back muscle and the compressional force in the spine. Στ 0 ΣF 0 Lifting Consider a person bending forward to lift a 200-N object. The spine and upper body are represented as a uniform horizontal rod of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, is angled at 12.0 from the spine. The force labelled as R is really a normal force by the joint on the upper torso. It is drawn as two components in this diagram. This is partly due to not knowing its direction ahead of time. Στ 0 ΣF 0 For this example, we will assume that the upper torso is 1 meter long. Since the center of mass was described as half way up the torso, it would be 0.5 meters from the pivot joint. The erector spinalis muscle is 2/3 of the way up the spine or 2/3 meters from the pivot joint. 8

9 2/21/2017 T Lifting N W F Force Caused BY ON X Y r Torque (τ) W F N T Write equations for a static equilibrium condition T Force table n W F Force Caused BY ON X Y r Torque (τ) W Earth Upper Torso N 0.5 m 0.5(350) CW rotation -175 Nm F Barbell /Arm Upper Torso N 1.0 m 1.0(200) CW rotation -200 Nm N Joint Upper Torso 0 0 T Erector Spinalis Muscle Upper Torso Write equations for a static equilibrium condition.?? $ m 0 +? ? 0 $ (? ) CCW rotation + $ (? ) 0 175Nm 200Nm+ $? 0 9

10 T Force table n W F 175Nm 200Nm+ $? 0 Solve for Ty Solve for T and Tx with the angle and Ty known. 12⁰ ? 0 +? 0 Solve for n y Solve for n with n x and n y known. Also solve for the direction of n. Solve for n x Follow the same steps for your prelab. Be careful, the upper torso length in the prelab is NOT 1 m. 10