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Forces in Equilibrium & Motion along an Incline Chapter 7.1.

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Presentation on theme: "Forces in Equilibrium & Motion along an Incline Chapter 7.1."— Presentation transcript:

1 Forces in Equilibrium & Motion along an Incline Chapter 7.1

2 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. 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 Ex. 1: Equilibrium A 100 N sign is hung by two wires as seen below. What is the tension in the wires? Physics is Fun F g = 100 N FAFA FBFB    = 15 °

4 Diagram the Problem Physics is Fun F g = 100 N FAFA FBFB  y x System   = 15 ° y x F Bx F By F Ay F Ax F g = 100 N FAFA FBFB

5 State the Known & Unknown What is known? F g = 100N θ = 15 ° What is not known? F A F B

6 Perform Calculations Isolate the x and y components separately. Since the sign is not moving, F net = ma = 0 in both the x and y directions. x – direction: -F Ax + F Bx = 0 -F A cosθ + F B cosθ = 0 F A cosθ = F B cosθ y – direction: F A sinθ + F B sinθ – F g = 0 2F A sinθ = F g F A = (100N)/(sin(15 ° )(2) F A = 193 N

7 Motion on an Incline When objects are not on a flat level surface: A portion of the gravitational force is directed along the surface. The normal force is not equal to the weight of the object. Choose a coordinate system such that the x-axis is directed parallel to the slope. FNFN FgFg F gx F gy x y

8 Ex. 2: Motion on an Incline Determine the rate of acceleration of a 25 kg wooden crate as it slides down a wooden ramp with a coefficient of friction,  = 0.2. The angle the ramp makes with the horizontal is 30 . What is known? m = 25 kg  = 0.2 θ = 30  v i = 0 m/s What is not known? a = ?  System FNFN FgFg F gx F gy

9 Diagram and Solve the Problem y-direction: F net,y = F N – F gy Since F net,y = 0 F N = F gy = mgcosθ x-direction: F net,x = F gx – F f ma = mgsinθ – μF N ma = mgsinθ - μmgcosθ a = g(sinθ – μcosθ) a = 9.8 m/s 2 (sin 30 ° - μcos 30 °) a = 3.2 m/s 2   F g(y) FfFf FNFN FgFg F g(x) y-axis x-axis

10 Key Ideas Equilibrium: When an object is at rest, or when an object is in motion at a constant speed in a straight line.


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