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CHAPTER-12 Equilibrium and Elasticity. Ch 12-2 Equilibrium A body is said to be in equilibrium if: i) Translational equilibrium F net =dp/dt= 0; p=mv=constant;

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Presentation on theme: "CHAPTER-12 Equilibrium and Elasticity. Ch 12-2 Equilibrium A body is said to be in equilibrium if: i) Translational equilibrium F net =dp/dt= 0; p=mv=constant;"— Presentation transcript:

1 CHAPTER-12 Equilibrium and Elasticity

2 Ch 12-2 Equilibrium A body is said to be in equilibrium if: i) Translational equilibrium F net =dp/dt= 0; p=mv=constant; v=constant ii) Rotational equilibrium  net = dl/dt =0; l=I  =0;  = constant Zero and Non-Zero constant values of v and  1) Static equilibrium v=  =0; F net =  net =0 2) Dynamical equilibrium v=  0; F net =  net =0

3 Ch 12-3 Requirement of Static Equilibrium For static equilibrium of a rigid body: i)  F ext =0 ii)  ext =0 iii) P=  p i =0

4 Ch 12 Checkpoint 1 The figure gives six overhead views of a uniform rod on which two or more forces acts perpendicular to the rod. If the magnitudes of the forces are adjusted properly ( but kept nonzero), in which situation can be rod in static equilibrium? Answer: c, e, f

5 Ch 12-2 Center of Gravity Center of gravity (cog) of a body is a point where gravitational force Fg effectively acts on a body. For a constant value of g for all elements of body com and cog coincides

6 Example: Sample Problem 12-1 A uniform beam and a block is resting on two scales. What do the scales read? Draw FBD Solve for two conditions of static equilibriums 1)  F=0 and  =0  F=F l +F r -Mg-mg=0 2) Taking the moment about the point of action of F l :  =F r xL-mgx(L/2)-Mgx(L/4)=0 F r -mg/2-Mg/4=0; F r =mg/2+Mg/4 F l +F r -Mg-mg=0 F l = Mg+mg- F r

7 Ch 12 Checkpoint 3 The figure gives an overhead view of a uniform rod in static equilibrium. (a) Can you find the magnitude of the unknown forces F1 and F2 by balancing the forces. (b) If you wish to find the magnitude of force F2 by using single equation, where should you place a rotational axis? (C) The magnitude of F2 turns out to be 65 N, what then is the magnitude of F1 3. (a) no; (b) at site of F1, perpendicular to plane of figure; (c)10x4d+F 1 X2d+30xd=20x8d 2F 1 d=160d-70d=90d F 1 = 45N

8 Ch 12 Checkpoint 4  In the figure a stationary 5 kg rod AC is held against a wall by a rope and friction between rod and wall. The uniform rod is 1m long, and angle  =30°.  (a) If you are to find the magnitude of the force T on the rod from the rope with a single equation, at what labeled point should a rotational axis be placed? With that choice of axis and counterclockwise torques positive, what is the sign of  (b) the torque  W due to rod weight and  (c) the torque due to  r due to pull on the rod by the rope?  (d) Is the magnitude of  r greater than,less than, or equal to magnitude of  W  (a) at C (to eliminate forces there from a  torque equation);  (b) plus;  (c) minus;  (d) equal

9 Ch 12-7 Elasticity  A rigid body is said to be elastic if we can change its dimension by pulling, pushing, twisting, or compressing them  Stress: deforming force per unit area  Strain: fractional change in dimension  Tensile stress associated with stretching  Shearing Stress associated with a force || surface area  Hydraulic stress from a fluid on an immersed object to shrink its volume by an amount  V

10 Ch 12-7 Elasticity  Over linear (elastic range) Stress=modulus x strain  For further increase in stress beyond yield strength Sy of the specimen, the specimen becomes permanently deformed  For additional increase in stress beyond yield strenth Sy, the specimen eventually rupture at ultimate strengthSu

11 Ch 12-7 Elasticity  Tension and Compression: For simple tension or compression the stress is force F/area A (  to the direction of application of force). The strain is fractional change  L/L in the length of specimen. The modulus of for tensile and compressive stresses is called the Young modulus E= (F/A)/ (  L/L)  Shearing, the shearing stress = force F/area A (||to the direction of application of force). The strain is fractional change  x/L in the length of specimen.Shear modulus G= (F/A)/ (  x/L)  Hydraulic Stress For Hydraulic Stress the stress is fluid pressure p (force F/area A ) The strain is fractional change in volume  V/V. The modulus of for hydraulic stresses the Bulk modulus B= p/(  V/V)


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