# Newton’s Laws of Motion. HFinks '072 6/2/2015 Basic Concepts  Force – push or pull on an object - Vector quantity  Mass – amount of matter in a body.

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Newton’s Laws of Motion

HFinks '072 6/2/2015 Basic Concepts  Force – push or pull on an object - Vector quantity  Mass – amount of matter in a body. - Measured with a balance - Scalar quantity

HFinks '073 6/2/2015 First Law  Also known as the law of inertia  An object will remain at rest or continue to move at a constant speed unless acted upon by a net force.  Net force – vector sum of all forces acting on an object.

HFinks '074 6/2/2015 Second Law  When a net force acts on a body, the acceleration is directly proportional to the force and indirectly proportional to the mass.  Equation: ∑F = ma or F net = ma Net force on an object = mass x acceleration Net = sum of the forces  Units of measurement for force metric system - Newton (N) metric system - Dyne (dyn) english system - Pound (lb)

HFinks '075 6/2/2015 Third Law For every action, there is an equal and opposite reaction. Forces occur in pairs. Net force and acceleration will always be in the same direction.

HFinks '076 6/2/2015 Examples and Labeling of Forces

HFinks '077 6/2/2015 Applied Force (F a )  Pushing, pulling or lifting an object.  Applied force is a vector quantity.

HFinks '078 6/2/2015 Applied Force (F a ) Example 1: Lifting an object FaFa

HFinks '079 6/2/2015 Applied Force (F a ) Example 2: Pulling an Object FaFa

HFinks '0710 6/2/2015 Applied Force (F a ) Example 3: Pushing an Object FaFa

HFinks '0711 6/2/2015 Applied Force (F a ) Example 4: Pulling an object up an inclined plane FaFa

HFinks '0712 6/2/2015 Normal Force (F N )  Normal force, F N, is a component of the force the surface exerts on an object.  Normal force is always perpendicular to the surface.  Normal force is a vector quantity.

HFinks '0713 6/2/2015 Normal Force (F N ) Example 1: Block on floor FNFN

HFinks '0714 6/2/2015 Normal Force (F N ) Example 2: Block on wall FNFN

HFinks '0715 6/2/2015 Normal Force (F N ) Example 3: Block on inclined plane FNFN

HFinks '0716 6/2/2015 Weight (W) Weight – depends on the acceleration due to gravity. - Measured with a spring scale. - Vector quantity - Is a force - Equation: W = mg - Units of measurement metric system - Newton metric system - Dyne english system - Pound

HFinks '0717 6/2/2015 Comparing Units F = ma W = mg = kg m/s 2 = kg m/s 2 = N = N N = Newton

HFinks '0718 6/2/2015 Weight (W) and Mass (m)  Weight and mass are not the same.  Weight changes with a change in acceleration due to gravity.  Mass does not change with a change in acceleration due to gravity. Example:  Your weight will change if you go from the earth to the moon.  Your mass will remain the same.

HFinks '0719 6/2/2015 Weight (W) Example 1: Block on floor W

HFinks '0720 6/2/2015 Weight (W) Example 2: Block on wall W

HFinks '0721 6/2/2015 Weight (W)  Example 3: Block on inclined plane W

HFinks '0722 6/2/2015 Friction  Occurs when objects are in contact  Acts opposite to the direction of motion  Two kinds of frictional forces Static (object at rest) = f s Kinetic (object is moving) = f k  Equations Static f s max =µ s F N Kinetic f k = µ k F N

HFinks '0723 6/2/2015 Coefficient of Friction (µ)  Ratio of the frictional force to the normal force. That is, you are comparing one force to another. Static Kinetic (sliding) µ s = f s µ k = f k F N F N  No unit of measurement because you are dividing a force by a force.

HFinks '0724 6/2/2015 Kinetic Frictional Force (f k )  The frictional force always act opposite to the applied force. Think of these two forces as being x- components. One is positive and the other is negative. FaFa fkfk Direction of motion

HFinks '0725 6/2/2015 Kinetic Frictional Force (f k) Example 1: Box pulled across a floor fkfk FaFa

HFinks '0726 6/2/2015 Kinetic Frictional Force (f k ) Example 2: Object pulled up an inclined plane fkfk FaFa

HFinks '0727 6/2/2015 Kinetic Frictional Force (f k )  Example 3: Object sliding down an inclined plane  No applied force fkfk

HFinks '0728 6/2/2015 Static Frictional Force (f s ) - Friction between atoms or molecules - No applied force

HFinks '0729 6/2/2015 Frictional Force (f s ) Example 1: Box at rest on a floor fsfs

HFinks '0730 6/2/2015 Frictional Force (f s )  Example 2: Object at rest on an inclined plane fsfs

HFinks '0731 6/2/2015 F net = ma Applications of this equation in the x direction Applied Force (F a ) and Frictional Force (f k )

HFinks '0732 6/2/2015 F net = ma Applications of this equation in the y direction Normal Force (F N) and Weight (W) Applied Force (F N ) and Weight (W)

HFinks '0733 6/2/2015 Applied Force (F a ) and Weight (W) Example 1: Lifting FaFa W F a – W = ma

HFinks '0734 6/2/2015 Normal Force (F N ) and Weight (W) Example 2: Block on floor FNFN W F N - W = ma

HFinks '0735 6/2/2015 Solving Force Problems 1. Label all forces 2. Write an expression for the forces acting in the x direction. 3. Write an expression for the forces acting in the y direction. 4. Substitute the known values into the expressions. 5. Solve for the unknown.

HFinks '0736 6/2/2015 Example 1 What is the magnitude and direction of F N ? Mass = 20.0 kg FNFN W = mg = 196 N F N - W = ma a.Label forces (No forces in x direction) b. Write expression c. F N - W = ma F N = W + ma =196 N + (20.0 kg)(0 m/s 2 ) F N = 196 N, upward a y = 0 m/s 2 At rest

HFinks '0737 6/2/2015 Example 2 A 12.0 kg object is pulled upward by a massless rope with an acceleration of 3.00 m/s 2. What is the tension (T) in the rope? F a = T W = (12.0 kg)(9.80 m/s 2 ) = 118 N T – W = ma T = 118 N + (12.0 kg)(3.00 m/s 2 ) T = W + ma T = 154 N m = 12.0 kg a = 3.00 m/s 2

HFinks '0738 6/2/2015 Example 3 A 12.0 kg object is accelerating downward at 3.00 m/s 2. What is the tension (T) in the rope? F a = T W = (12.0 kg)(9.80 m/s 2 ) = 118 N T – W = ma T = 118 N + (12.0 kg)(-3.00 m/s 2 ) T = W + ma T = 82.0 N m = 12.0 kg a = -3.00 m/s 2

HFinks '0739 6/2/2015 Example 4 A 100. N crate is pulled across the floor. The tension in the rope is 75.0 N. Calculate the maximum frictional force. (µ k = 0.5) fkfk F a = T FNFN W = 100 n a. Label forces b. Write expressions a F k = µ k F N = (0.5)(100 N) F k = 50.0 N

HFinks '0740 6/2/2015 Example 4 Calculate the resultant force (F net) in Example 4. F k =50.0 N F a = T = 75.0 N F N =100. N W = 100 n a. Label forces b. Write expressions a F y = F N – W F x = T – f k F N = W = 75.0 N – 50.0 N F y = 100. N – 100. N F x = 25.0 N F y = 0 N

HFinks '0741 6/2/2015 Example 4 Calculate the acceleration of the object in Example 4. f k = 50.0 N F a = T= 75.0 N FNFN W = 100. N m = w = 100. N = 10.2 kg g 9.80 m/s 2 a. Label forces b. Write expressions for direction of motion only. T – f k = ma a = 75.0 N – 50.0 N a = T – f k 10.2 kg m a = 2.45 m/s 2 a

HFinks '0742 6/2/2015 Inclined Plane

HFinks '0743 6/2/2015 Example 5 A 20.0 kg object is at rest on an inclined plane (µ s = 0.10) What is the value of f s ? Ө = 10.0º f s = µ s F N fsfs FNFN W Ө F N ≠ W Reason: They aren’t opposite each other “W” is a vector quantity and is acting at an angle. Calculate x- and y-components, Next

HFinks '0744 6/2/2015 Example 5 W = mg = 196 N fsfs FNFN WxWx Ө W y = W cos Ө W x = W sin Ө Note: Ө is the same for both the large and small right triangles. W x will always act “down” the incline.. ( -W x ) Next WyWy W Ө

HFinks '0745 6/2/2015 Example 5 F N = W y fsfs FNFN WxWx Ө F s = µ s F N = µ s W y = µs ( W cos Ө) = (0.10)(196 N)(cos 10.0º) f s = 19.3 N WyWy W Ө

HFinks '0746 6/2/2015 Example 6 A 20.0 kg object is sliding down an incline at a constant velocity under the influence of gravity. Find the value of f K. fkfk

HFinks '0747 6/2/2015 Example 6 Label Forces and write expressions fkfk FNFN WxWx Ө Object isn’t moving in the y direction. So…..don’t use the y expression: F N - W y = ma Object is moving along the incline. Use…. ∑F x f k - W x = ma a = 0. Object is moving at a constant velocity. So….. F k = W x Next WyWy W Ө

HFinks '0748 6/2/2015 Example 6 fkfk FNFN WxWx Ө f k = W x = W sin Ө =196N sin 10.0º f k = 34.0 N Next WyWy W Ө

HFinks '0749 6/2/2015 Example 7 A 20.0 kg object is pulled up a 10.0º incline and µ k = 0.05. The object is accelerated at a rate of 1.25 m/s 2. What is the value of the applied force? W = 196 N fkfk FNFN WxWx Ө Next WyWy W Ө FaFa (i) F a - f k - W x = ma F a = f k + W x + ma *’a” is positive. The object is moving up the incline. f k = µ k F N = µ k W y f k = µ k W cos Ө

HFinks '0750 6/2/2015 Example 7 W = 196 N fkfk FNFN WxWx Ө WyWy W Ө FaFa (i) F a - f k - W x = ma F a = f k + W x + ma (ii)F a = ( µ k W cos Ө) +W sin Ө + ma = (0.05)(196 N)(cos 10.0º) + (196 N)(sin 10.0º) + (20.0 kg)(1.25 m/s 2 ) = 9.65 N + 34.0 N + 25.0 N F a = 68.7 N

HFinks '0751 6/2/2015 Reviewing Concepts 1. A push or pull is a(n) ______ 2. Quantities that require magnitude, direction, and point of origin for their description are ____ quantities. 3. A physical quantity that can affect the motion of an object is a(n) ____. 4. When no net force acts on a body, either no force acts on the body or the ___ of all force acting on the body is ____.

HFinks '0752 6/2/2015 Answers 1. Force 2. Vector 3. Force 4. Vector sum, zero

HFinks '0753 6/2/2015 Reviewing Concepts 5. If there is no net force acting on a body, it will continue in its state of __ or will continue moving along a(n) __line with ___speed. 6. The property of matter that is the concern of the first law of motion is__. 7. A common unbalanced force that makes it difficult to prove Newton’s first law of motion experimentally is ___

HFinks '0754 6/2/2015 Answers 5. rest, straight, constant 6. Inertia 7. friction

HFinks '0755 6/2/2015 Apparent Weight Person on Scale in Elevator

HFinks '0756 6/2/2015 Email me if you see typing or calculator errors. Thank you. (hfphysics@aol.com)

HFinks '0757 6/2/2015 Things to Remember  Weight will not change when you are near the surface of the earth.  The normal force, F N, is changing. It is the scale reading.  Forces occur in pairs. The person exerts a force on the scale and the scale exerts a force on the person.  The inertia of the person wants to remain at rest. The elevator floor and scale will either push up on the person causing an increase in scale reading or drop out from underneath the person causing a decrease in scale reading.

HFinks '0758 6/2/2015 Think About Your ride on an Elevator The elevator door closes and  You are moving upward  Slowing down as it comes to a stop  You are moving downward  Slowing down as it comes to a stop Do you feel heavier, lighter or nothing?

HFinks '0759 6/2/2015 Elevator at Rest Person standing on scale in elevator. F N – W = ma a = 0 m/s 2 F N = W W FNFN

HFinks '0760 6/2/2015 Example 8. Elevator at Rest A 70.0 kg person stands on a scale in an elevator that is at rest. What is the scale reading? F N = W = mg = (70.0 kg)(9.80 m/s 2 ) F N = 686 N W FNFN 3, please

HFinks '0761 6/2/2015 Elevator Moving Upward and Speeding Up F N – W = ma F N = W + ma W FNFN a Any brakes on this tub?

HFinks '0762 6/2/2015 Example 9. Elevator Moving Upward and Speeding Up Mass of person = 70.0 kg Weight = 686 N The elevator is accelerating upward at 1.10 m/s 2. What is the scale reading? F N – W = ma F N = W + ma = 686 N + (70.0 kg)(1.10 m/s 2 ) F N = 763 N W FNFN a

HFinks '0763 6/2/2015 Elevator Moving Upward and Slowing Down F N – W = m(-a) F N = W - ma W FNFN a Whoa. I’m Getting dizzy.

HFinks '0764 6/2/2015 Example 10. Elevator Moving Upward and Slowing Down Mass of person = 70.0 kg Weight = 686 N The elevator is accelerating downward at 1.10 m/s 2. What is the apparent weight? F N – W = m(-a) F N = W – ma = 686 N – (70.0 kg)(1.10 m/s 2 ) F N = 609 N W FNFN a

HFinks '0765 6/2/2015 Elevator Moving Downward and Speeding Up F N – W = m(-a) F N = W - ma W FNFN a My goodness. Please stop.

HFinks '0766 6/2/2015 Example 11. Elevator Moving Downward and Speeding Up Mass of person = 70.0 kg Weight = 686 N A = 1.10 m/s 2 F N – W = m(-a) F N = W – ma = 686 N – (70.0 kg)(1.10 m/s 2 ) F N = 609 N W FNFN a

HFinks '0767 6/2/2015 Elevator Moving Downward and Slowing Down F N – W = ma F N = W + ma W FNFN a Yea… fresh air soon

HFinks '0768 6/2/2015 Example 12. Elevator Moving Downward and Slowing Down Mass of person = 70.0 kg Weight = 686 N F N – W = ma F N = W + ma = 686 N + (70.0 kg)(1.10 m/s 2 ) F N = 763 N W FNFN a

HFinks '0769 6/2/2015 Elevator in Free Fall F N – W = ma F N = 0 N - W = ma -mg = ma -g = a W FNFN a Help!!!!!

HFinks '0770 6/2/2015 Example 13. Elevator in Free Fall F N – W = ma F N = 0 N (No contact) - W = ma -mg = ma -g = a -9.80 m/s 2 = a W FNFN a

HFinks '0771 6/2/2015 Example 14. Applied Force (T) on Elevator A woman stands on a scale in a moving elevator. Her mass is 70.0 kg and the combined mass of the elevator and scale is an additional 790. kg. Starting from rest, the elevator accelerates upward. During the acceleration, there is a tension (T) of 8500. N in the hoisting cable. What is the reading on the scale during the acceleration? T –W e - W w = m t a a = 8500. N – 7742 N - 686 N (70.0 + 790.)kg a = 0.0837m/s 2 F N = W + ma = mg + ma = m(g + a) = 70.0 kg(9.80 + 0.0837)m/s 2 F N = 692 N T FNFN W w =(70.0kg)(9.80 m/s 2 ) =686 N W e = (790kg)(9.80 m/s 2 )=7742 N a scale

HFinks '0772 6/2/2015 Reviewing Concepts 8. If an object is stationary, its inertia tends to keep it ___; if an object is in motion, its inertia tends to keep it ___. 9. The acceleration of a body is __ proportional to the mass of the body. 10. The acceleration of a body is in the __ direction as the applied force.

HFinks '0773 6/2/2015 Answers 8. Stationary, in motion 9. Inversely or indirectly 10. Same

HFinks '0774 6/2/2015 Reviewing Concepts 11. The acceleration of a body is ___ proportional to the force exerted. 12. If a body is moving in a straight line and a force is applied in the direction of its motion, the body will __ in speed as long as the force continues. 13. If a body is moving in a straight line and a force is applied in the direction __ to its motion, the velocity is changing, the body may stop, and the body may move in the opposite direction.

HFinks '0775 6/2/2015 Answers 11. Directly 12. Increase 13. opposite

Applied Force Acting At An Angle with the Horizontal F N ≠ W Always calculate and use the x- and y- components of a vector acting at an angle

HFinks '0777 6/2/2015 Email me if you see typing or calculator errors. Thank you. (hfphysics@aol.com)

HFinks '0778 6/2/2015 Vector Acting at an Angle F a(x) = F a cos Ө = 20.0 N cos 20.0 º F a(x) = 18.8 N F a(y) = F a sin Ө = 20.0 N sin 20.0 º F a(y) = 6.84 N Ө = 20.0 º F a = 20.0 N

HFinks '0779 6/2/2015 ∑F y Object Moving to the right  F N ≠ W  F N = W – F a(y)  Why? ∑F y = F N + F a(y) – W  F N + F a(y) – W = ma a = 0 m/s 2 in the y direction So … F N + F a(y) – W = 0 Or F N = W – F a(y)  F a(y) = F sin Ө FNFN FaFa W fkfk Ө

HFinks '0780 6/2/2015 ∑F x Object Moving to the right F a(x) = F cos Ө ∑F x = F a(x) - f k So ….F a(x) - f k = ma f k = µ k F N = µ k (W – F a(y) ) or = µ k (mg - F a sin Ө) FNFN FaFa W fkfk Ө

HFinks '0781 6/2/2015 Example 16 A box weighing 450. N is pulled along a level floor at constant speed by a rope that makes an angle of 30.0 º with the floor. If the force on the rope is 300. N, what is the coefficient of sliding friction? constant speed --- a = 0 m/s 2 F cos Ө - µ k (W – F a(y) ) = ma F cos Ө - µ k (W – F a(y) ) = 0 F cos Ө - µ k (W – F a(y) ) = - F cos Ө - µ k (W – F a(y) ) = - F cos Ө - - µ k = - F cos Ө - (W – F a(y) ) - = 300. N cos 30.0 º - 450. N - 300. sin 30.0 º - µ k = 0.866 FNFN FaFa W fkfk Ө

HFinks '0782 6/2/2015 Example 17 A 20.0 kg box is pulled along a horizontal surface by a force of 40.0 N, which is supplied at a 30.0 º angle. The coefficient of kinetic friction is 0.30. Calculate the acceleration. F a(x) - f k = ma a = F a(x) - f k a = F cos Ө - µ k (W – F a(y) ) m = (40.0N cos 30.0 º ) – (0.30) [196 N – (40.0 sin 30.0 º )] 20.0 kg 34.6 N – 52.8 N = 20.0 kg a = -.907 m/s 2 FNFN FaFa W fkfk Ө W = (20.0 kg)(9.80 m/s 2 ) = 196 N

HFinks '0783 6/2/2015 Reviewing Concepts 14. The maximum frictional force between stationary objects is known as ___ friction, and the frictional force between objects that are sliding with respect to one another is known as __ friction. 15. Starting friction my be (less than, greater than) sliding friction.

HFinks '0784 6/2/2015 Answers 14. Static (starting), kinetic (sliding) 15. Greater than

HFinks '0785 6/2/2015 Reviewing Concepts 16. The ratio of the force of sliding friction to the normal force pressing the surfaces together is called ___. 17. If an object moves along a horizontal surface, under the influence of a horizontal force, the normal force pressing the surfaces together is the __ of the object.

HFinks '0786 6/2/2015 Answers 16. Coefficient of kinetic friction 17. weight

Forces in Equilibrium Sum of components in x-direction = 0 Sum of components in y-direction = 0

HFinks '0788 6/2/2015 Example 18. Solving for Tension (T) I ∑F x = -T cos Ө + B = 0 T = B cos Ө ∑F y = T sin Ө - W = 0 T = W sin Ө T Ө W B

HFinks '0789 6/2/2015 Example 18. Solving for Tension (T) I Calculate the magnitudes of “T” and “B”. ∑F y = T sin Ө - W = 0 T = W sin Ө T = (0.500 kg)(9.80 m/s 2 ) sin 60 º T = 5.7 N ∑F x = -T cos Ө + B = 0 B = T cos Ө = 5.7 N cos 60 º B = 2.9 N T 60º m = 500 g B

HFinks '0790 6/2/2015 Example 19. Solving For Tension (T) II ∑F x = 0 -T 1 cos Ө 1 + T 2 cos Ө = 0 Solve for T 2 and substitute your answer in the statement below. ∑F y = 0 T 1 sin Ө 1 + T 2 sin Ө 2 - W = 0 T1T1 T2T2 Ө1Ө1 Ө2Ө2 W

HFinks '0791 6/2/2015 Example 19. Solving For Tension (T) II ∑F x = 0 -T 1 cos Ө 1 + T 2 cos Ө = 0 -T 1 cos 10 º + T 2 cos 5 º = 0 T 2 = T 1 cos 10 º cos 5 º T 2 = 0.989T 1 T1T1 T2T2 Ө 1 10ºӨ 2 = 5º W = 90 N Next

HFinks '0792 6/2/2015 Example 19. Solving For Tension (T) II ∑F y = 0 T 1 sin Ө 1 + T 2 sin Ө 2 - W = 0 T 1 sin 10 º + T 2 sin 5 º - 90 N = 0 0.174T 1 +(0.989T 1 )(0.087) = 90 N 0.174T 1 + 0.086T 1 = 90 N T 1 = 346 N T 2 = 0.989T 1 T 2 = 342 N T1T1 T2T2 Ө 1 10ºӨ 2 = 5º W = 90 N

HFinks '0793 6/2/2015 Reviewing Concepts Select the correct answer(s) 18. The newton is (a) the unit of force in the metric system (b) is the force required to accelerate 1 kg of mass at the rate of 1 m/s 2 (c) is a derived unit, (d) is 1 kg m/s 2

HFinks '0794 6/2/2015 Answer a, b, c, & d

HFinks '0795 6/2/2015 Reviewing Concepts Select the correct answer(s) 19. The force required to accelerate an object of known weight is a. Directly proportional to the weight of the object. b. directly proportional to the acceleration desired. c. Directly proportional to the acceleration due to gravitation d. Not related in any way to the acceleration due to gravitation.

HFinks '0796 6/2/2015 Answer a & b

HFinks '0797 6/2/2015 Reviewing Concepts Select the correct answer(s) 20. When an automobile is accelerated forward, a. The tires exert a forward action force on the road. b. The road exerts a forward reaction force on the tires. c. The tires exert a rearward action force on the road. d. The road exerts a rearward reaction force on the tires.

HFinks '0798 6/2/2015 Answer b & c

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