Physics 121

9. Equilibrium and Elasticity 9.1 Statics: Forces in Equilibrium 9.2 Conditions for Equilibrium 9.3 Solving Statics Problems 9.4 Measurement and Uncertainty 9.5 Stability and Balance 9.6 Elasticity: Stress and Strain

What is the tension in the string? Example 9.1 . . . The String What is the tension in the string? 300 2 kg W = 1 kg

The Conditions for Equilibrium  Fx = 0  Fy = 0   = 0 300 2 kg W = 1 kg

(2)(10)(L) + (1)(10)(L/2) = (T sin 30)(L) Solution 9.1 . . . The String   = 0 (2)(10)(L) + (1)(10)(L/2) = (T sin 30)(L) T = 50 N

Are there any (reaction) forces at the hinge? Example 9.2 . . . The Hinge Are there any (reaction) forces at the hinge? 300 2 kg W = 1 kg

All the forces on the beam are shown Solution 9.2 . . . The Hinge All the forces on the beam are shown Solution continues on the next slide 20 N 10 N H V T sin 30 T cos 30

 Fx = 0 So H = T cos 300 or H = 50 * 0.87 or H = 44 N Solution 9.2 . . . The Hinge  Fx = 0 So H = T cos 300 or H = 50 * 0.87 or H = 44 N  Fy = 0 So V + 50 sin 300 = 10 + 20 or V = 5 N 20 N 10 N H V T sin 30 T cos 30

Young’s Modulus = Stress / Strain Elasticity Stress = Force / Area Strain =  L / L Young’s Modulus = Stress / Strain

Example 9.3 . . . The Piano A 1.6 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.30 cm when tightened. [ Young’s Modulus for steel is E = 2.0 x 10 11 N / m2 ]

Solution 9.3 . . . The Piano strain =  L / L = 0.003/1.6 = 1.875 x 10 - 3 m stress = F / A = F/  r2 = F /3 .1 x 10 - 6 m2 E = stress / strain 2.0 x 10 11 = F /(3 .1 x 10 - 6 )(1.875 x 10 - 3 ) F = 1200 N

That’s all folks!