Physics 141Mechanics Lecture 17 Equilibrium and Elasticity Yongli Gao A rigid body has six degrees of freedom: three for linear motion and three for rotational motion. It can rotate about any of the three axes. If the sum of all external forces is zero, the momentum of the rigid body is conserved. If the sum of all the external torques is zero about any axis, the angular momentum of the rigid body is conserved. The balance of torque is with respect to any axis. Usually we choose one through which most of the action lines of the unknown forces pass. The gravitational force is treated as acting at CM.

Example: Balance The principle of a balance can be understood by looking at the equilibrium conditions. About the support point About the point m 2 is attached Clearly about any point the total torque must be zero. m1gm1g m2gm2g F l1l1 l2l2

Example: Support on the Wall m is support by a string and a massless rod of length l. Find the force at the other end O. Solution: Choose the torque about O   O BxBx ByBy T mgmg l m O

Example: Ladder on Smooth Wall Suppose a ladder of mass m and length l is leaning on a smooth wall with an angle  to the vertical, what is the force by the ground and by the wall? Solution: How much is the downward force F applied at the wall contact when the ladder start to slip if the static friction coefficient to the ground is  ? mgmg  f1f1 N1N1 N2N2

How far along the length of the ladder a person of mass M can go before the ladder start to slip if the static friction coefficient to the ground is  s ? Solution: mgmg f1f1 N1N1 N2N2 MgMg

Example: Vertical Block An rectangular block of mass m, length l and width w is hanging at one corner. What is the horizontal force F one must apply at the bottom so that l is vertical? Solution: We can also find  and T l w mgmg T  F

Stress and Strain A solid is not exactly rigid. It always deforms when we apply a force on it. If the force is not too strong to cause permanent damage, the solid object can recover its original shape once the force is removed. The phenomenon is called elasticity. Stress is the strength of the force causing the deformation of an object, typically measured by the force on unit area, and therefore the unit of stress is N/m 2, or Pascal (Pa). Strain is the relative deformation of the object caused by the stress, and is therefore typically unitless. Elastic modulus is the ratio of stress and stain elastic_modulus=stress/strain

Youngs Modulus Tensile stress =F/A Tensile strain=  L/L Young's modulus (material dependent) Compare to Hooke's law (material and size dependent) Y steel = 2.0x10 11 Pa L LL F cross section area A

Shear Modulus Shear stress =F/A Shear strain=x/h=tan  ≈  shear modulus (material dependent) S steel = 0.84x10 11 Pa x F cross section area A h

Bulk Modulus Bulk stress =F/A=P (hydraulic stress) Bulk strain =  V/V bulk modulus (material dependent) B steel = 1.70x10 11 Pa Yield strength S y is the stress above which permanent deformation takes place. Ultimate strength S u is the stress at which the object ruptures. pressure P volume change  V