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Focus on Free Body Diagrams

Focus on Free Body Diagrams. AP Physics B Kernion. FBD Analysis. Keys Massless ropes have the same tension throughout Objects in rest or in uniform motion have Σ F = 0 Objects which are accelerating have Σ F = ma Establish a F.O.R. and stick with it

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Focus on Free Body Diagrams

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  1. Focus on Free Body Diagrams AP Physics B Kernion

  2. FBD Analysis • Keys • Massless ropes have the same tension throughout • Objects in rest or in uniform motion have ΣF = 0 • Objects which are accelerating have ΣF = ma • Establish a F.O.R. and stick with it • Normal force = apparent wt. = support force applied by a surface • 4 step approach • Draw a picture • make a FBD • Sum the forces • Solve for the unknown

  3. Example 1 • The elevator is accelerating downward. • Find the apparent weight of the person • ΣF = ma • N-mg = m(-a) • N = mg – ma • N = m (g - a) • Result: a feeling of “lightness” +y N mg

  4. Example 2 • Find the tension in the massless rope connecting the two masses F 1 kg 2 kg +y Frictionless surface N N +x T F T mg mg • ΣFx = max • T = (1 kg)a • ΣFx = max • F - T = (2 kg)a • a = (F – T) / 2 • T = (F – T) / 2 • 2 T + T = F • T = F/3

  5. Friction Static Friction: Present when the two surfaces are not moving relative to each other if the surfaces would move without friction being present ff≤ μs N Kinetic Friction: Present when the two surfaces are moving relative to each other ff = μk N

  6. Example 3 • Find the force pulling the two blocks to the right +y motion F m1 m2 +x Coefficient of Friction = μk • ΣFy = m2ay • N – m2g = 0 • N2 = m2g • ΣFy = m1ay • N – m1g = 0 • N1 = m1g N N ff T ff F T • ΣFx = m2ax • F - T – ff = m2a • T = F – ff - m2a • T = F – μN2 - m2a • T = F – μm2g - m2a mg • ΣFx = m1ax • T – ff = m1a • T – μN1 = m1a • T = m1a + μN1 • T = m1a + μm1g mg m1a + μm1g = F – μm2g - m2a F = m1a + μm1g + m2a + μm2g F = a (m1+ m2) + μg (m1 + m2)

  7. Example 4 Objects on an inclined plane Subscripting the forces m2 m1 θ N1,s block 1 ff2,1 N2,1 ff1,s m1gcosθ ff1,2 m1gsinθ m2gsinθ m2gcosθ N1,2

  8. Example 5 +y motion m1 frictionless surface +x m2 • ΣFy = m2ay • T – m2g = m2(-a) • T= m2g – m2a • ΣFx = m1ax • T = m1a • m2g – m2a = m1a • a(m1+m2) = m2g • a = m2g / (m1 + m2) T =m1m2g / (m1 + m2) The weight of block 2 which equals T if the system is held at rest

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