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Equilibrium & Newton’s Laws of Motion

Equilibrium & Newton’s Laws of Motion. Tensional Forces. Equilibrium. Newton’s 1 st Law of Motion When the forces on an object are balance, it is said to be in equilibrium. When an object is in equilibrium, it is not accelerating.

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Equilibrium & Newton’s Laws of Motion

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  1. Equilibrium & Newton’s Laws of Motion Tensional Forces

  2. Equilibrium • Newton’s 1st Law of Motion • When the forces on an object are balance, it is said to be in equilibrium. • When an object is in equilibrium, it is not accelerating. • An object that is not accelerating is stationary or moving at constant speed in a straight line. You balanced forces during the force table lab.

  3. Equilibrium in 2-Dimensions • When objects are in equilibrium: • ax and ay = 0 • It then follows that: • ΣFx and ΣFy = 0 • Because Fnet = ma and a = 0.

  4. Ex. 1: Equilibrium • A 100 N sign is hung by two wires as seen below. What is the tension in the wires? FB FA  Physics is Fun  = 15° Fg = 100 N

  5. FB FA y y  x x Physics is Fun System Fg = 100 N   FBx FAx FA FB FAy FBy   = 15° Fg = 100 N Diagram the Problem

  6. State the Known & Unknown • What is known? • Fg = 100N • θ = 15° • What is not known? • FA • FB

  7. y x  FBx FAx FA FB FAy FBy   = 15° Fg = 100 N Perform Calculations • Isolate the x and y components separately. • Since the sign is not moving, Fnet = ma = 0 in both the x and y directions. • x – direction: • -FAx+ FBx = 0 • -FA cosθ + FB cosθ = 0 • FA cosθ = FB cosθ

  8. y x  FBx FAx FA FB FAy FBy   = 15° Fg = 100 N Perform Calculations • y – direction: • FAy + FBy – Fg = 0 • FA sinθ + FB sinθ – Fg = 0 • 2FA sinθ = Fg(due to symmetry, FAy = FBy) • FA = (100N)/((sin15°)(2)) • FA = 193 N

  9. Ex. 2: Equilibrium • An engine has a weight of 3150 N. The engine is positioned above the engine compartment using rope, a pulley and a ring as shown. Find the tension in each of the sections of rope

  10. State the Known & Unknown • What is known? • W = 3150 N • θ1 = 10° • θ2 = 80° • What is not known? • T1 • T2

  11. Perform Calculations • Isolate the x and y components separately. • Since the engine is not moving, Fnet = ma = 0 in both the x and y directions. • x – direction: • -T1sin θ1 + T2sin θ2 = 0 • -T1sin 10° + T2sin 80° = 0 (1) • y – direction: • T1cos θ1 - T2cos θ2 – W = 0 • T1cos 10° - T2cos 80° - W = 0 (2) • Solve (1) for T1 and substitute into (2).

  12. Perform Calculations (cont.)

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