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Chapter 4 Forces and the Laws of Motion. Changes in Motion When we think of Force, we typically imagine a push or pull exerted on an object. When we think.

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Presentation on theme: "Chapter 4 Forces and the Laws of Motion. Changes in Motion When we think of Force, we typically imagine a push or pull exerted on an object. When we think."— Presentation transcript:

1 Chapter 4 Forces and the Laws of Motion

2 Changes in Motion When we think of Force, we typically imagine a push or pull exerted on an object. When we think of Force, we typically imagine a push or pull exerted on an object. Force can be defined as causing a change in the motion of an object. Force can be defined as causing a change in the motion of an object.

3 Force… Can make an object accelerate Can make an object accelerate Can make an object decelerate Can make an object decelerate Can make an object change direction Can make an object change direction

4 Newton Force is measured in newtons. Force is measured in newtons. A newton (N) is defined as the amount of force that will accelerate a 1kg object by 1m/s 2 A newton (N) is defined as the amount of force that will accelerate a 1kg object by 1m/s 2 N = kgm/s 2 N = kgm/s 2 An objects weight, in lbs. Is the measure of the force of gravity on that object. An objects weight, in lbs. Is the measure of the force of gravity on that object. 1lb = 4.4N 1N =.23lbs 1lb = 4.4N 1N =.23lbs

5 Types of Forces Contact Forces result from physical contact between two objects Contact Forces result from physical contact between two objects Throwing a ball Throwing a ball Braking in your car Braking in your car Field forces do not require contact Field forces do not require contact Gravity Gravity Magnetism Magnetism All fundamental forces. All fundamental forces.

6 Force (cont) Force is a vector. Its effect depends on both its magnitude and direction Force is a vector. Its effect depends on both its magnitude and direction We can use force diagrams, which represent force vectors with arrows, to help us understand all the forces acting on an object. We can use force diagrams, which represent force vectors with arrows, to help us understand all the forces acting on an object. We will assume all forces act on the center of an object. We will assume all forces act on the center of an object.

7 Force Diagrams A force diagram isolates one object and illustrates all the forces acting on that object. A force diagram isolates one object and illustrates all the forces acting on that object. Force diagrams should always include the force of gravity Force diagrams should always include the force of gravity If the object is on the ground the Normal force will negate the force of gravity. To keep the object from accelerating through the ground. (or lifting off of it) If the object is on the ground the Normal force will negate the force of gravity. To keep the object from accelerating through the ground. (or lifting off of it) The force of friction will always oppose motion. The force of friction will always oppose motion.

8 Normal Force The normal force the force exerted on on object by the floor. The normal force the force exerted on on object by the floor. It keeps the forces in the y direction equal to zero. It keeps the forces in the y direction equal to zero. If there is no vertical forces other than gravity, then the normal force will equal gravity but in the opposite direction. If there is no vertical forces other than gravity, then the normal force will equal gravity but in the opposite direction. If there are vertical forces, then the normal force will simply be enough to cancel the other vertical forces. If there are vertical forces, then the normal force will simply be enough to cancel the other vertical forces.

9 Friction The force of friction depends on the surfaces in contact. The force of friction depends on the surfaces in contact. There are two types of Friction There are two types of Friction Static Friction - opposes initial motion of an object (F s,max ) Static Friction - opposes initial motion of an object (F s,max ) Kinetic Friction - opposing force on a moving object (F k ) Kinetic Friction - opposing force on a moving object (F k ) Kinetic Friction is always less than static friction Kinetic Friction is always less than static friction

10 Coefficient of Friction Forces of both static friction and kinetic friction are dependant on the normal force acting on an object multiplied by the coefficient of friction for the contact pairs in question. Forces of both static friction and kinetic friction are dependant on the normal force acting on an object multiplied by the coefficient of friction for the contact pairs in question. F s,max =  s (F n ) F s,max =  s (F n ) F k =  k (F n ) F k =  k (F n ) Table 4-2 lists both  s and  k for several surface pairs Table 4-2 lists both  s and  k for several surface pairs Coefficient of Friction is a unit-less value Coefficient of Friction is a unit-less value

11 Example How much force would be needed to start moving a 2kg aluminum weight on a steel surface? (F s,max ) How much force would be needed to start moving a 2kg aluminum weight on a steel surface? (F s,max ) F n = F g = m w a g = 2kg (9.8m/s 2 ) = 19.6 N F n = F g = m w a g = 2kg (9.8m/s 2 ) = 19.6 N F s,max =  s (F n ) = (0.61)(19.6N) = 12 N F s,max =  s (F n ) = (0.61)(19.6N) = 12 N

12 Example An object has a weight of 14,700 N. An object has a weight of 14,700 N. It is being pulled backward at an angle of 10º from the horizontal, with a force of 5,800 N. It is being pulled backward at an angle of 10º from the horizontal, with a force of 5,800 N. The object is experiencing 775N of friction. The object is experiencing 775N of friction. Draw that diagram Draw that diagram -14,700 N = F g 5,800 N = F p 10º 13,390 N = F n 755 N = F f

13 Resolving Force Diagrams Each force can be resolved into its x and y components using the sine and cosine functions. Each force can be resolved into its x and y components using the sine and cosine functions. The Resultant force can be found using the Pythagorean Theorem. The Resultant force can be found using the Pythagorean Theorem. F R = √(sum x forces) 2 + (sum y forces) 2 F R = √(sum x forces) 2 + (sum y forces) 2

14 Example Solve the example force diagram for Net Force (F net ) Solve the example force diagram for Net Force (F net ) Resolve x and y components of any “slanted” forces. (F p ) Resolve x and y components of any “slanted” forces. (F p ) F x = F cos  F y = F sin  F x = F cos  F y = F sin  F x = (5800N)cos10F y = (5800N)sin10 F x = (5800N)cos10F y = (5800N)sin10 F x = -5710 NF y = 1010 N F x = -5710 NF y = 1010 N

15 Sum x and y Forces X Forces X Forces F f = 755 N F f = 755 N F p,x = -5710 N F p,x = -5710 N F X = -4955 N F X = -4955 N Y Forces F g = -14,700 N F n =13,690 N F p,y = 1010 N F Y = 0 N

16 Example 2 A tug boat has a propeller on each of its sides. A tug boat has a propeller on each of its sides. Each motor is capable of producing 25,000N. (assume the boat is aligned with the compass) Each motor is capable of producing 25,000N. (assume the boat is aligned with the compass) If both the north and south motors are pointed south, the east motor is pointed 25º south of east, and the west motor is pointed 40º south of west, what is the result force? If both the north and south motors are pointed south, the east motor is pointed 25º south of east, and the west motor is pointed 40º south of west, what is the result force?

17 Newton’s 1st Law An object will continue to maintain its state of rest or of uniform motion unless it experiences a net external force. (Inertia) An object will continue to maintain its state of rest or of uniform motion unless it experiences a net external force. (Inertia) Examples: Examples: An car moving at a constant velocity experiences several forces but no net force An car moving at a constant velocity experiences several forces but no net force If you increase the force forward, the car will accelerate in that direction If you increase the force forward, the car will accelerate in that direction If you add a new force, brakes, in the opposite direction the car will decelerate If you add a new force, brakes, in the opposite direction the car will decelerate

18 Newton’s 2nd Law The acceleration of an object is directly proportional to the net external force and inversely proportional to the mass of the object. The acceleration of an object is directly proportional to the net external force and inversely proportional to the mass of the object. Force = mass x acceleration (F = ma) Force = mass x acceleration (F = ma)

19 Example If a 2000kg car experiences an acceleration of 10 m/s 2, what is the net external force acting on that object? If a 2000kg car experiences an acceleration of 10 m/s 2, what is the net external force acting on that object? F net = ma F net = ma F net = (2000kg)(10 m/s 2 ) = 20,000 N F net = (2000kg)(10 m/s 2 ) = 20,000 N

20 Example 2 If a 45N force is exerted on an object with a mass of 0.75kg, what is its acceleration? If a 45N force is exerted on an object with a mass of 0.75kg, what is its acceleration? F = ma F = ma a = F/m a = F/m a = (45N)/(0.75kg) = 60m/s 2 a = (45N)/(0.75kg) = 60m/s 2

21 Newton’s 3rd Law If two bodies interact, the magnitude of the force exerted on object 1 onto object 2 is equal to the magnitude of the force exerted by object 2 onto object 1, and these two forces will be in opposite direction If two bodies interact, the magnitude of the force exerted on object 1 onto object 2 is equal to the magnitude of the force exerted by object 2 onto object 1, and these two forces will be in opposite direction Every action has an equal and opposite reaction. Every action has an equal and opposite reaction. F 1 = F 2 F 1 = F 2 m 1 a 1 = m 2 a 2 m 1 a 1 = m 2 a 2

22 Example If a 3.10 kg rifle were hung from a ceiling and fires an 11.0g bullet which accelerates at 1340m/s 2, what force is experienced by the bullet and what acceleration is experienced by the gun? If a 3.10 kg rifle were hung from a ceiling and fires an 11.0g bullet which accelerates at 1340m/s 2, what force is experienced by the bullet and what acceleration is experienced by the gun? F b = (0.011kg)(1340m/s 2 ) = 14.7 N F b = (0.011kg)(1340m/s 2 ) = 14.7 N F b = F g = m g a g F b = F g = m g a g a g = F b /m g = (14.7N)/(3.10kg) = 4.74m/s 2 a g = F b /m g = (14.7N)/(3.10kg) = 4.74m/s 2


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