Ch 13: Rotational EquilibriumSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
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Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
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Ch 17: Fluid Mechanics
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Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
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Ch 25: Resistors & DC Circuits
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Ch 32: Wave Optics
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Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Solution: A solid bar of length L has a mass m 1. The bar is fastened by a pivot at one end to a wall which is at an angle θ with respect to the horizontal. The bar is held horizontally by a vertical cord that is fastened to the bar at a distance x from the wall. A mass m2 is suspended from the free end of the bar. Find the tension in the cord. 1. T = (m1 + 1/2 m2) (L/x) gcosθ 2. T = (m1 + m2) gsinθ 3. T = (m1 + m2) gcosθ 4. T = (m1 + m2) (L/x) (g/2) 5. T = 0 6. T = (m1 + 1/2 m2) (L/x) g 7. T = (m1 + 1/2 m2) (L/x) gsinθ 8. T = (1/2 m1 + m2) (L/x) gsinθ 9. T = (1/2 m1 + m2) (L/x) g 10. T = (1/2 m1 + m2) (L/x) gcosθ

Problem

A solid bar of length L has a mass m 1. The bar is fastened by a pivot at one end to a wall which is at an angle θ with respect to the horizontal. The bar is held horizontally by a vertical cord that is fastened to the bar at a distance x from the wall. A mass m2 is suspended from the free end of the bar. Find the tension in the cord.

1. T = (m1 + 1/2 m2) (L/x) gcosθ

2. T = (m1 + m2) gsinθ

3. T = (m1 + m2) gcosθ

4. T = (m1 + m2) (L/x) (g/2)

5. T = 0

6. T = (m1 + 1/2 m2) (L/x) g

7. T = (m1 + 1/2 m2) (L/x) gsinθ

8. T = (1/2 m1 + m2) (L/x) gsinθ

9. T = (1/2 m1 + m2) (L/x) g

10. T = (1/2 m1 + m2) (L/x) gcosθ