An Unbalanced Force

1. FN An Unbalanced Force FA Ff Fg

2. 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)

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

4. Strategies for solving force related problems. • For horizontal surfaces: • FN = Fg = mg. • 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. • FN = Fg() = Fgcos • Fg(||) = Fgsin

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

6. No Forces in the Horizontal Direction FN Fg • 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, Fnet = 0 N since there is no acceleration. Return to Flow Chart

7. Friction in the Absence of an Applied Force FN v Ff Fg • 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 (Fnet = Ff). • For example, assume the block has a mass of 10 kg, and undergoes an acceleration of -5.0 m/s2 as it slides from the left to the right. • Determine the frictional force. • Determine the normal force. • Determine the coefficient of friction.

8. Friction in the Absence of an Applied Force FN v Ff Fg • As per our free-body diagram, the only force acting in the x-direction is friction. Therefore: • Fnet = -Ff • Ff = ma • Ff = (10kg)(-5m/s2) = -50N • On a horizontal surface, the normal force equals the weight. • FN = Fg • FN = mg • FN = (10kg)(9.81m/s2) = 98.1N • Since Ff = FN, . Return to Flow Chart

9. An Applied Force with Friction FN FA Ff Fg • When multiple forces act on an object, they need to be summed up to determine the net force (Fnet = FA - Ff). • 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/s2. • Determine the net force. • Determine the frictional force. • Determine the coefficient of friction.

10. An Applied Force with Friction FN FA Ff Fg • The net force is simply determined by multiplying the object’s mass by its acceleration. • Fnet = ma • Fnet = (10kg)(4m/s2) = 40N • 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. • Fnet = FA - Ff • Ff = FA - Fnet • Ff = 120N – 40N = 80N • Since Ff = FN, . Return to Flow Chart

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

12. Stationary or Moving at a Constant Velocity FN FA Ff Fg • 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 (Fnet = ma = 0) • And if the net force is zero, then the applied force must equal the frictional force (FA = Ff). • 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.

13. Stationary or Moving at a Constant Velocity FN FA Ff Return to Flow Chart Fg • 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 FA = Ff FA = FN FA = mg FA = (0.42)(10kg)(9.81m/s2) = 41.2N Case #2: Constant Velocity k = 0.30 FA = Ff FA = FN FA = mg FA = (0.30)(10kg)(9.81m/s2) = 29.4N

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

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

16. 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.

17. Uniform Circular Motion – A Horizontal Path(FBD’s) Car on Road Ball on String The Rotor Ff Ff v v FN Ff Fg Fc = Ff Fc = FT Fc = FN

18. Uniform Circular Motion – A Vertical Path(FBD’s) Roller Coaster Ball on String Ferris Wheel v Top Fc = FN + Fg Fc = FT + Fg Fc = FN - Fg Bottom v Fc = FN - Fg Fc = FN - Fg Fc = FT - Fg