Presentation is loading. Please wait.

Presentation is loading. Please wait.

Physics 121.

Similar presentations


Presentation on theme: "Physics 121."— Presentation transcript:

1 Physics 121

2 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

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

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

5 (2)(10)(L) + (1)(10)(L/2) = (T sin 30)(L)
Solution The String   = 0 (2)(10)(L) + (1)(10)(L/2) = (T sin 30)(L) T = 50 N

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

7 All the forces on the beam are shown
Solution 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

8  Fx = 0 So H = T cos 300 or H = 50 * 0.87 or H = 44 N
Solution 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 = or V = 5 N 20 N 10 N H V T sin 30 T cos 30

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

10 Example 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 N / m2 ]

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

12 That’s all folks!


Download ppt "Physics 121."

Similar presentations


Ads by Google