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Transcript
Unit Two: Dynamics Section 4: Forces and acceleration Newton’s Second Law Newton’s first law states that an object does not accelerate unless a net force is applied to the object. This relates back to the first law (an object will continue with the same velocity unless a force acts upon it). Newton’s Second Law But how much will an object accelerate when there is a net force? The larger the force the larger the acceleration. Therefore acceleration is directly proportional to mass. Acceleration also depends on mass. The larger the mass, the smaller the acceleration. Therefore acceleration is inversely proportional to mass. We say that a massive body has more INERTIA than a less massive body. Newton’s Second Law: Summary A net force is needed to accelerate an object. A larger force means more acceleration. A larger mass means less acceleration as it is harder to move. Newton’s Second Law - Newton’s Law of Motion (Summary) Force = mass x acceleration Fnet = ma ***The acceleration is in the same direction as the net force. Newton’s Second Law Examples Ex. 1: What net force is required to accelerate a 1500. kg race car at +3.00m/s2? Draw a FBD to show the net force. Practice Problems Page 163, Questions 1, 2, 3 Putting it All Together Now that we have considered Newton’s Second Law, you can use that to analyze kinematics problems with less information than we have used previously We can either use dynamics information to then apply to a kinematic situation or vice versa Newton’s Second Law Examples Ex. 2: An artillery shell has a mass of 55 kg. The shell is fired from a gun leaving the barrel with a velocity of +770 m/s. The gun barrel is 1.5m long. Assume that the force, and the acceleration, of the shell is constant while the shell is in the gun barrel. What is the force on the shell while it is in the gun barrel? Example 2.5 A 25kg crate is slid from rest across a floor with an applied force 72N applied force. If the coefficient of static friction is 0.27, determine: The free body diagram. Include as many of the forces (including numbers) as possible. The acceleration of the crate. The time it would take to slide the crate 5.0m across the floor. FBD FN=250N Fa=72N Ff=? Fg=-250N Use the frictional force equation to determine the magnitude of the frictional force F f FN F f (.27)( 250 N ) F f 66 N F f 66 N The net force is the sum of the forces (acting parallel or anti-parallel) Fnet Fnet Fnet Fnet Fi F f Fa 66 N 72 N 5.8 N Use Newton’s Second Law to solve for the acceleration Fnet ma 5.8 N (25kg)a 2 a 0.23m / s Use kinematics to solve for the time taken to cross the floor 2 at d (t ) v0t d 0 2 2 2 0.23m / s t 5 .0 m 2 2(5.0m) t 2 0.23m / s t 6 .6 s Practice Problems Page 168, questions 4 to 8 Example 3 A baby carriage with a mass of 50. kg is being pushed along a rough sidewalk with an applied horizontal force of 200. N and it has a constant velocity of 3.0 m/s. A) What other horizontal force is acting on the carriage and what is the magnitude and direction of that force? B) What value of applied horizontal force would be required to accelerate the carriage from rest to 7.0 m/s in 2.0 s? Example 3 A) Force of friction must be equal in magnitude (so 200.N) in the opposite direction to the force applied (of the baby carriage moving forward). B) First find a using a = (vf – vi)/t = 2.0m/s2 Now find Fnet = ma = 50x2 = 100N Now use Fnet = Fapp – Ff Fapp = 300N = 3.0 x 102 N Example 4 A horizontal force of 50. N is required to pull an 8.0 kg block of aluminum at a uniform velocity across a horizontal wooden desk. What is the coefficient of kinetic friction? Example 4 You know that Force of friction is equal and opposite to Force applied. Therefore, Ff = 50.N. You know that Force of friction = coefficient of friction x normal force. Normal Force = - Force of gravity = -mg Fn = 78.48N Sub in and rearrange to find that coefficient = 0.64 (no units) Example 5 A 75 kg man stands in an elevator. Draw a free body diagram and determine what the force the elevator exerts on him will be when A) the elevator is at rest B) the elevator is moving upward with a uniform acceleration of 2.0 m/s2 C) the elevator is moving downward with a uniform velocity of 2.0 m/s D) the elevator is moving downward with a uniform acceleration of 2.0 m/s2 Example 5 A) B) C) D) Fnet = 0N, Fapp = 0N Fnet = 150N [up] Fnet = 0N Fnet = 150N [down]