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Applications of Force ► Anyone who has never made a mistake has never tried anything new. -Albert Einstein Albert EinsteinAlbert Einstein ► Overview Vector Components Force Lab Review of Common Forces (Dynamics) Free Body Diagrams Incline Planes Multi-mass Problems Static Equilibrium and Torque Atwood Machine Lab Momentum and Collisions
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Vector Components ► Motion is rarely in a straight line. We need to analyze forces and motion that follows curved lines or motion that changes direction abruptly. ► Motion Equations: The Big Four Uniform motion (constant velocity) ► V = Δd Δt (velocity and displacement) Δt (velocity and displacement)
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► Uniformly accelerated motion A = v 1 – v 2 Δt Δt 1.V = v i + at 2.d = (v 1 + v 2 )t The Big Four 2 3.d = v i t + ½ at 4.v f 2 = v i 2 + 2ad
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Vector Addition and Subtraction ► In Physics 11 we solved vector problems graphically ► Now we will use trig and right angles in a more precise method vectors at right angles (SOHCAHTOA) rarely vectors are at right angles so we will resolve the vectors into their will resolve the vectors into their components components
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Vector Addition ► Step 1: resolve all vectors into their x and y components (be careful to indicate if the components are negative or positive – what quadrant they are in) ► Step 2: calculate the total x and y components by addition ► Step 3: draw a summary right angle triangle and use the pythagorean theorem to calculate the hypotenuse (resultant) and angle (direction)
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Vector Components and Dynamics ► Newton’s 1 st and 2 nd laws of motion are: An object at rest or in uniform motion will remain at rest or in uniform motion unless acted upon by an external force F(net) = ma net force is the vector sum of all the forces acting on a body
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Equilibrium - Stationary ► When the net forces add to zero the body is said to be at equilibrium. ► The force that will counter the net force to make the body stationary is called an equilibriant. ► Problem: calculate the equilibriant for the forces: F 1 – 425 N [E 63° N] F 1 – 425 N [E 63° N] F 2 – 385 N [W 15° N] F 2 – 385 N [W 15° N]
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