1443-501 Spring 2002 Lecture #17 Dr. Jaehoon Yu 1. Conditions for Equilibrium 2. Center of Gravity 3. Elastic Properties of Solids • Young’s Modulus • Shear Modulus • Bulk Modulus Today’s Homework Assignment is the Homework #8!!! 2nd term exam on Wednesday, Apr. 10. Will cover chapters 10 -13. Conditions for Equilibrium What do you think does the term “An object is at its equilibrium” mean? The object is either at rest (Static Equilibrium) or its center of mass is moving with a constant velocity (Dynamic Equilibrium). When do you think an object is at its equilibrium? Translational Equilibrium: Equilibrium in linear motion Is this it? The above condition is sufficient for a point-like particle to be at its static equilibrium. However for object with size this is not sufficient. One more condition is needed. What is it? Let’s consider two forces equal magnitude but opposite direction acting on a rigid object as shown in the figure. What do you think will happen? F d d F 0 CM -F Apr. 1, 2002 The object will rotate about the CM. The net torque 0 acting on the object about any axis must be 0. For an object to be at its static equilibrium, the object should not have linear or angular speed. v 0 0 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 CM 2 More on Conditions for Equilibrium To simplify the problems, we will only deal with forces acting on x-y plane, giving torque only along z-axis. What do you think the conditions for equilibrium be in this case? The six possible equations from the two vector equations turns to three equations. F 0 F F x 0 y 0 0 0 z What happens if there are many forces exerting on the object? If an object is at its translational static equilibrium, and if the net torque acting on the object is 0 about one axis, the net torque must be 0 about any arbitrary axis. r’ r5 O O’ Net Force exerting on the object F F 1 F 2 F 3 0 Net torque about O O r 1 F 1 r 2 F 2 r 3 F 3 r i F i 0 Position of force Fi about O’ Net torque about O’ Apr. 1, 2002 O' ' i r r i r' r '1F 1 r '2 F 2 r 1 r ' F 1 r 2 r ' F 2 r i F i r ' F i 1443-501 rSpring F 2002 r '0 Dr. J. Yu, Lecture #17 O' i i O 0 3 Center of Gravity Revisited When is the center of gravity of a rigid body the same as the center of mass? Under the uniform gravitational field throughout the body of the object. Let’s consider an arbitrary shaped object The center of mass of this object is CM CoG m x m x M m my my m M xCM i i i i i yCM i i i i i Let’s now examine the case with gravitational acceleration on each point is gi Since the CoG is the point as if all the gravitational force is exerted on, the torque due to this force becomes m1 g1 m2 g 2 xCoG m1 g1 x1 m2 g 2 x2 If g is uniform throughout the body Apr. 1, 2002 Generalized expression for different g throughout the body m1 m2 gxCoG m1 x1 m2 x2 g m x 1443-501 Spring 2002 m Dr. J. Yu, Lecture #17 xCoG i i i xCM 4 Example 12.1 A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N, respectively. If the support (or fulcrum) is under the center of gravity of the board and the father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board by the support? 1m F MFg x n MBg D MFg Since there is no linear motion, this system is in its translational equilibrium F F Therefore the magnitude of the normal force x 0 y MBg MF g MDg n 0 n 40.0 800 350 1190N Determine where the child should sit to balance the system. The net torque about the fulcrum M g 0 M g 1.00 M g x 0 B F D by the three forces are Therefore to balance the system x M F g 1.00m 800 1.00m 2.29m the daughter must sit MDg 350 Apr. 1, 2002 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 5 Example 12.1 Continued Determine the position of the child to balance the system for different position of axis of rotation. Rotational axis 1m F MFg x n x/2 MBg D MFg The net torque about the axis of rotation by all the forces are M B g x / 2 M F g 1.00 x / 2 n x / 2 M D g x / 2 0 Since the normal force is n MBg MF g MDg The net torque can M B g x / 2 M F g 1.00 x / 2 be rewritten M B g M F g M D g x / 2 M D g x / 2 M F g 1.00 M D g x 0 Therefore Apr. 1, 2002 x MFg 800 1.00m 1.00m 2.29m MDg 350 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 What do we learn? No matter where the rotation axis is, net effect of the torque is identical. 6 Example 12.2 A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm. FB Since the system is in equilibrium, from the translational equilibrium condition F 0 O F F F mg 0 l mg F From the rotational equilibrium condition F 0 F d mg l 0 d x y B U U U Thus, the force exerted by the biceps muscle is B FB d mg l mg l 50.0 35.0 FB 583N d 3.00 Force exerted by the upper arm is FU FB mg 583 50.0 533N Apr. 1, 2002 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 7 Example 12.3 A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the horizontal. If 600N person stands 2.00m from the wall, find the tension in the cable, as well as the magnitude and direction of the force exerted by the wall on the beam. R 53.0o 53.0o FBD Rsin Rcos 8m From the rotational equilibrium Using the translational equilibrium Apr. 1, 2002 First the translational equilibrium, using components 200N 600N 2m T F F Tsin53 Tcos53 R cos T cos 53.0 0 y R sin T sin 53.0 600 N 200 N 0 T sin 53.0 8.00 600 N 2.00 200 N 4.00m 0 T 313N R cos T cos 53.0 And the magnitude of R is R sin T sin 53.0 600 N 200 N 800 313 sin 53.0 tan 71 . 7 313 cos 53.0 1 x T cos 53.0 313 cos 53.0 R 582 N cos cos 71.1 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 8 Example 12.4 A uniform ladder of length l and weight mg=50 N rests against a smooth, vertical wall. If the coefficient of static friction between the ladder and the ground is ms=0.40, find the minimum angle min at which the ladder does not slip. First the translational equilibrium, using components P l FBD mg n O F F f Thus, the normal force is x f P 0 y mg n 0 n mg 50 N The maximum static friction force just before slipping is, therefore, f smax m s n 0.4 50 N 20 N P From the rotational equilibrium min Apr. 1, 2002 l mg cos min Pl sin min 0 2 mg 1 50 N tan 1 tan 51 2P 40 N O 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 9 Elastic Properties of Solids We have been assuming that the objects do not change their shapes when external forces are exerting on it. It this realistic? No. In reality, the objects get deformed as external forces act on it, though the internal forces resist the deformation as it takes place. Deformation of solids can be understood in terms of Stress and Strain Stress: A quantity proportional to the force causing deformation. Strain: Measure of degree of deformation It is empirically known that for small stresses, strain is proportional to stress The constants of proportionality are called Elastic Modulus Elastic Modulus Three types of Elastic Modulus Apr. 1, 2002 1. 2. 3. stress strain Young’s modulus: Measure of the resistance in length Shear modulus: Measure of the resistance in plane Bulk modulus: Measure of the resistance in volume 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 10 Young’s Modulus Let’s consider a long bar with cross sectional area A and initial length Li. Li Fex After the stretch F Tensile Stress ex A Young’s Modulus is defined as Fex Fex=Fin A:cross sectional area Tensile stress Lf=Li+DL Tensile strain Tensile Strain Fex Tensile Stress A Y DL Tensile Strain Li DL Li Used to characterize a rod or wire stressed under tension or compression What is the unit of Young’s Modulus? Force per unit area 1. For fixed external force, the change in length is Experimental proportional to the original length Observations 2. The necessary force to produce a given strain is proportional to the cross sectional area Elastic limit: Maximum stress that can be applied to the substance before it becomes permanently 1443-501deformed Spring 2002 Apr. 1, 2002 Dr. J. Yu, Lecture #17 11 Shear Modulus Another type of deformation occurs when an object is under a force tangential to one of its surfaces while the opposite face is held fixed by another force. A Dx h After the stress Fixed face Shear stress Shear strain F=fs Shear Stress Tangential Force F Surface Area the force applies A Shear Strain Dx h F Shear Stress A S Shear Modulus is defined as Dx Shear Strain h Apr. 1, 2002 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 12 Bulk Modulus F Bulk Modulus characterizes the response of a substance to uniform squeezing or reduction of pressure. V After the pressure change F V’ F F Normal Force F Volume stress Pressure Surface Area the force applies A =pressure If the pressure on an object changes by DP=DF/A, the object will undergo a volume change DV. Bulk Modulus is defined as Because the change of volume is reverse to change of pressure. Apr. 1, 2002 DF Volume Stress DP A B D V DV Volume Strain Vi Vi Compressibility is the reciprocal of Bulk Modulus 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 13 Example 12.7 A solid brass sphere is initially under normal atmospheric pressure of 1.0x105N/m2. The sphere is lowered into the ocean to a depth at which the pressures is 2.0x107N/m2. The volume of the sphere in air is 0.5m3. By how much its volume change once the sphere is submerged? Since bulk modulus is DP B DV Vi The amount of volume change is DV DPVi B From table 12.1, bulk modulus of brass is 6.1x1010 N/m2 The pressure change DP is DP Pf Pi 2.0 107 1.0 105 2.0 107 Therefore the resulting 2.0 107 0.5 4 3 D V V V 1 . 6 10 m f i volume change DV is 6.11010 The volume has decreased. Apr. 1, 2002 1443-501 Spring 2002 Dr. J. Yu, Lecture #17 14