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An Unbalanced Force FAFA FfFf FgFg FNFN

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What is an unbalanced force? According to Newton’s Second Law of Motion: –an unbalanced force is one that causes an object to accelerate. Positively or negatively (increasing or decreasing velocity) or Change direction. (a force that causes an object to move in a circular path)

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Strategies for solving force related problems. 1.Draw a free-body diagram that summarizes all the forces that act on the object in the vertical (y) and horizontal (x) directions. 2.Write a mathematical expression that summarizes these forces in each direction. –F net(x) = F A – F f Since the length of the vector for F A is longer than the one for F f, the block should be accelerating on the positive x-direction. –F net(y) = F N – F g Since the length of the vectors for F N and F g are the same length, the net force in the vertical direction should be zero as we would expect. This also means that F N = F g. F A = applied force F f = frictional force F N = normal force F g = force due to gravity or weight

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Strategies for solving force related problems. 3.For horizontal surfaces: F N = F g = mg. 4.For inclined surfaces, the force due to gravity (weight) must be broken into two vectors; one that is perpendicular to the incline, and the other that is parallel to the incline. –F N = F g( ) = F g cos –F g(||) = F g sin

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Is there an applied force? Yes No Does the applied force cause the object to move? Yes Does the object accelerate? No v = 0 v <> 0 v = constant Yes Is the object in motion? No Yes F net = 0 v = 0 m/s: F A = F f(static) v = constant: F A = F f(kinetic) F net = F A - F f F net = F f F net = 0 If no friction F net = F A a ≠ 0 Forces Acting on Objects in the Horizontal Direction a = 0 F net = ma

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No Forces in the Horizontal Direction FgFg FNFN If there is no horizontally applied force, then the object will be: stationary (v = 0 m/s) or in motion, sliding along a frictionless surface at a constant velocity (v = constant). Under both circumstances, F net = 0 N since there is no acceleration. Return to Flow Chart

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FfFf FgFg FNFN Under the circumstance where the object is in motion and being acted upon by a only a frictional force, the object will experience an unbalanced force due to friction (F net = F f ). For example, assume the block has a mass of 10 kg, and undergoes an acceleration of -5.0 m/s 2 as it slides from the left to the right. 1.Determine the frictional force. 2.Determine the normal force. 3.Determine the coefficient of friction. v Friction in the Absence of an Applied Force

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FfFf FgFg FNFN 1.As per our free-body diagram, the only force acting in the x-direction is friction. Therefore: F net = -F f F f = ma F f = (10kg)(-5m/s 2 ) = -50N 2.On a horizontal surface, the normal force equals the weight. F N = F g F N = mg F N = (10kg)(9.81m/s 2 ) = 98.1N 3.Since F f = F N,. v Return to Flow Chart

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FAFA FfFf FgFg FNFN When multiple forces act on an object, they need to be summed up to determine the net force (F net = F A - F f ). Assume that there is an applied force of 120N acting on a 10kg block to the right that causes it to go through an acceleration of 4m/s 2. 1.Determine the net force. 2.Determine the frictional force. 3.Determine the coefficient of friction. An Applied Force with Friction

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1.The net force is simply determined by multiplying the object’s mass by its acceleration. F net = ma F net = (10kg)(4m/s 2 ) = 40N 2.From the free-body diagram, the net force is the sum of the forces acting in the x- direction. We will then solve the relationship for the frictional force. F net = F A - F f F f = F A - F net F f = 120N – 40N = 80N 3.Since F f = F N,. FAFA FfFf FgFg FNFN Return to Flow Chart

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FAFA FgFg FNFN When the surfaces are considered frictionless, the applied force will equal the net force (F net = F A ). For example: Assume that there is an applied force of 120N acting on a 10kg block to the right. 1.Determine the acceleration of the block. Since F net = F A, a = F A /m An Applied Force with No Friction Return to Flow Chart

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FAFA FfFf FgFg FNFN When objects move at a constant velocity (same speed in a linear direction) then the net force will be zero. Similarly, if the object is not moving, then the net force must be zero as well. Why? In both cases, the acceleration is zero, hence (F net = ma = 0) And if the net force is zero, then the applied force must equal the frictional force (F A = F f ). Example: A wooden 10kg block is placed on a wooden floor. Case #1: the block is stationary. Case #2: the block is moving at a constant velocity. Stationary or Moving at a Constant Velocity

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FAFA FfFf FgFg FNFN Case #1: Determine the maximum force that can be applied to the block without causing it to move. Case #2: Determine the applied force required to cause the block to move at a constant velocity. In both cases, you will need to refer to your reference table to find the appropriate values for the coefficient of friction for wood on wood. Case #1: Stationary s = 0.42 F A = F f F A = F N F A = mg F A = (0.42)(10kg)(9.81m/s 2 ) = 41.2N Case #2: Constant Velocity k = 0.30 F A = F f F A = F N F A = mg F A = (0.30)(10kg)(9.81m/s 2 ) = 29.4N Return to Flow Chart Stationary or Moving at a Constant Velocity

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Special Case – The Elevator Question 1.When talking about elevators, they spend a short period of time at the beginning and end of their ascent accelerating and decelerating, respectively. 2.All motion and forces occur in the vertical direction. 3.Write a mathematical expression that summarizes these forces. –F net(y) = F N – F g Note: F N will be greater than F g if the acceleration is in the upward direction and vice-versa if the acceleration is in the downward direction.

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Note: You are weightless when in free-fall! Special Case – The Elevator Question

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Unbalanced Forces and Uniform Circular Motion Whenever an object moves in a circular path, it experiences an unbalanced force. The unbalanced force always acts perpendicular to the direction of motion and towards the center of the circular path. A centripetal force is an unbalanced force, which is also a net force except that the motion is circular instead of linear.

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Uniform Circular Motion – A Horizontal Path(FBD’s) Car on RoadThe RotorBall on String FfFf FgFg FNFN FfFf FfFf v v F c = F f F c = F N F c = F T

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Uniform Circular Motion – A Vertical Path(FBD’s) Roller CoasterFerris WheelBall on String v v F c = F T + F g F c = F T - F g F c = F N - F g F c = F N + F g Top Bottom F c = F N - F g

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Friction Ffriction = μFNormal.

Friction Ffriction = μFNormal.

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