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Learn about static equilibrium conditions, Free Body Diagrams, Young’s modulus, Shear modulus, Bulk modulus. Prepare for Exam 3 discussing torque, force expressions, and axis of rotation choice. Solve various statics problems involving planks, seesaws, ladders, and more with detailed examples. Understand center of mass, torque conditions for balance, and deformation concepts in elastic objects. Explore practical applications and real-world scenarios in statics engineering.
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Lecture 21 Goals: • Discussion conditions for static equilibrium • Use Free Body Diagrams prior to problem solving • Introduce Young’s, Shear and Bulk modulus Exam 3: Wednesday, April, 18th 7:15-8:45 PM
Statics Equilibrium is established when In 3D this implies SIX expressions (x, y & z)
Free Body Diagram Rules • For forces zero acceleration reflects the motion of the center of mass • For torques the position of the force vector matters • We are free to choose the axis of rotation (usually one that simplifies the algebra)
Statics: Using Torque T1 T2 M x CM L/2 L/4 y Mg x • Now consider a plank of mass M suspended by two strings as shown. • We want to find the tension in each string:
Statics: Using Torque T1 T2 M x CM L/2 L/4 y Mg x • Now consider a plank of mass M suspended by two strings as shown. • We want to find the tension in each string:
Statics: Using Torque T1 T2 M x CM L/2 L/4 y Mg x • Now consider a plank of mass M suspended by two strings as shown. • Using a different torque position
Another Example: See-saw 60 kg 30 kg 30 kg • Two children (60 kg and 30 kg) sit on a 3.0 m long uniform horizontal teeter-totter of mass 30 kg. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. • Position of the small boy?
N Example: Soln. 60 kg 30 kg 0.5 m 1 m 300 N 300 N 600 N 30 kg • Draw a Free Body diagram (assume g = 10 m/s2) • 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5 m from pivot point d
Exercise: Statics • A 1.0 kg ball is hung at the end of a uniform rod exactly 1.0 m long. The system balances at a point on the rod 0.25mfrom the end holding the mass. • What is the mass of the rod? (a) 0.5 kg (b) 1 kg (c) 2 kg 1m 1kg
The classic ladder problem m=0 N’ N mg f • A uniform ladder of length L and mass m is propped up against a frictionless wall. There is friction with the ground with static coefficient m. The angle the ladder makes with the ground is q. • What is the minimum value of m to keep the latter from slipping?
The classic ladder problem m=0 N’ N mg f • A uniform ladder of length L and mass m is propped up against a frictionless wall. There is friction with the ground with static coefficient m. If the angle the ladder makes with the ground is q. • What is the minimum value of m that keeps the ladder from slipping?
Example:Statics 3 1 2 • A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case. • In which case(s) does the box tip over ? (a)all (b)2 & 3(c) 3 only
A stable equilibrium 1 • Potential energy with tipping (left and right) • x-axis corresponds to the x-position of the center-of mass
Balancing Acts • So, where is the center of mass for this construction? • Between the arrows!
Example problem • The board in the picture has an average length of 1.4 m, mass 2.0 kg and width 0.14 m. The angle of the board is 45°. The bottle is 1.5 m long, a center of mass 2/5 of the way from the bottom and is mounted 1/5 of way from the end. If the sketch represents the physics of the problem. • What are the minimum and maximum masses of the bottle that will allow this system to remain in balance? -0.1 m -0.8 m 0.0 m -0.5 m 0.1 m 0.4 m
Example problem -0.1 m -0.8 m 0.0 m -0.5 m 0.1 m 0.4 m • What are the minimum and maximum masses of the bottle that will allow this system to remain in balance? • X center of mass, x, -0.1 m < x < 0.1 m 2.0 kg
Example problem -0.1 m -0.8 m 0.0 m -0.5 m 0.1 m 0.4 m • What are the minimum and maximum masses of the bottle that will allow this system to remain in balance? • Check torque condition (g =10 m/s2) • x, -0.1 m < x < 0.1 m 2.0 kg
Real Physical Objects • Representations of matter • Particles: No size, no shape can only translate along a path • Extended objects are collections of point-like particles They have size and shape so, to better reflect their motion, we introduce the center of mass Rigid objects: Translation + Rotation Deformable objects: Regular solids: Shape/size changes under stress (applied forces) Reversible and irreversible deformation Liquids: They do not have fixed shape Size (aka volume) will change under stress
For Thursday • All of chapter 12
Some definitions F L0 L V0 F V0 - V • Elastic properties of solids : • Young’s modulus: measures the resistance of a solid to a change in its length. • Bulk modulus: measures the resistance of solids or liquids to changes in their volume. elasticity in length volume elasticity