CEE 102L ULO1c - Grade: B+ PDF

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UNIVERSITY OF MINDANAO College of Engineering Education Computer Engineering Program

Physically Distanced but Academically Engaged Self-Instructional Manual (SIM) for Self-Directed Learning (SDL) Course/Subject: CEE 102/L: PHYSICS 1 FOR ENGINEERS Name of Teacher: Engr. MARIANNE G. WATA Co-Authors: Engr. MA. ANGELA I. ESTELLA Engr. CRIJAMAICA L. OCEÑA Engr. ALFONSE IRENEO T. ESTIBAL Engr. JAMES MARK P. GALLAWAN Engr. JAY CARLO S. AGUILAR Engr. JAMES MARK GALLAWAN Engr. RONNEL ESPARAGOZA Engr. DANIELYN PLAZOS Engr. NORRODIN MELOG

THIS SIM/SDL MANUAL IS A DRAFT VERSION ONLY; NOT FOR REPRODUCTION AND DISTRIBUTION OUTSIDE OF ITS INTENDED USE. THIS IS INTENDED ONLY FOR THE USE OF THE STUDENTS WHO ARE OFFICIALLY ENROLLED IN THE COURSE/SUBJECT.EXPECT REVISIONS OF THE MANUAL.

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Table of Contents Big Picture in Focus: ULO-1c ............................................................................................................ 3 Metalanguage ......................................................................................................................................... 3 Essential Knowledge ............................................................................................................................... 3 Keywords ................................................................................................................................................ 3 Force ................................................................................................................................................... 3 Newton’s First Law.............................................................................................................................. 4 Mass .................................................................................................................................................... 4 Newton’s Second Law ......................................................................................................................... 4 Gravitational Force ............................................................................................................................. 7 Newton’s Third Law ............................................................................................................................ 8 Forces of Friction .............................................................................................................................. 11 Self-Help................................................................................................................................................ 13 Let’s Check ............................................................................................................................................ 14 Let’s Analyze ......................................................................................................................................... 15 In a Nutshell .......................................................................................................................................... 18

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Big Picture in Focus: ULO-1c. Recall and apply the concept of force and systems of forces

Metalanguage In the ULOb we discussed principles of rectilinear motion while ignoring the interactions affecting the motion. In this section, the motion of an object is described considering the influences and causes of its movement. You will be able to answer the following questions: 1. Why does the motion of an object change? 2. What might cause one object to remain at rest and another object to accelerate? 3. Why is it generally easier to move a small object than a large object? For you to demonstrate ULOc, you will need to have an operational understanding of the main factors to be considered in motion – that is, the forces acting on an object and the mass of the object. The following Newton’s Law of Forces are the key principle in the analysis of systems of forces and the achievement of this ULO-1c. 1. First law states that an object moves at constant velocity unless acted on by a force. 2. Second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The net force acting on an object equals the product of its mass and acceleration. 3. Third law states that in every applied force there is a always a reaction opposite to the exerted force. Please note that you will also be required to refer to the previous principles found in ULO-1b. Essential Knowledge This section discusses on Newton’s three laws of motion and his law of gravity, the concept of force on a more fundamental level. Dynamics is the branch of classical mechanics concerned with the study of forces and their effects on motion. Isaac Newton defined the fundamental physical laws which govern dynamics in physics. As long as the system under study doesn’t involve objects comparable in size to an atom or traveling close to the speed of light, classical mechanics provides an excellent description of nature. Keywords Force inertia

equilibrium Action force

Reaction force Frictional Force

Static friction Dynamic friction

Force The basic understanding of force refers to an interaction with an object that causes it to move. A physical contact between two objects referred to as contact forces that causes the object’s velocity to change. However, forces do not always cause motion. For example, 3

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when you are sitting, a gravitational force acts on your body and yet you remain stationary. You can push on a large boulder and not be able to move it. Another applied force called field forces doesn’t involve any direct physical contact such as gravitational force between to masses, electric force between two charges, and magnetic force. These forces act through empty space. Fig. 12 shows examples of forces applied to various objects.

Fig. 12 Newton’s First Law Imagine a heavy slab lying on the floor, if you do nothing the slab remains in its position unmoved. But if you try to push it, you may move the slab because of the applied force. Now consider, a smooth and waxed floor, moving the slab would be easier and requires less force. Newton’s Law describes these phenomena into three Laws of Motion. Newton’s first law of motion sometimes called the law of inertia states that an object moves with a velocity that is constant in magnitude and direction unless a non-zero net force acts on it. This law explains what happens to an object that has no net force acting on it. The net force on an object is defined as the vector sum of all external forces exerted on the object. In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity. In other words, when no force acts on an object, the acceleration of the object is zero. The tendency of an object to resist any attempt to change its velocity is called inertia. Mass Inertia is the tendency of an object to continue its motion in the absence of a force. On the other hand, mass is a measure of the object’s resistance to changes in its motion due to a force. The greater the mass of a body, the less it accelerates under the action of a given applied force. From that definition, we can say that the acceleration is inversely proportional to the mass under a given force,

Newton’s Second Law

𝒎=

𝑭 𝒂

The second law answers the question of what happens to an object that does have a net force acting on it. Simply stated, Newton’s Second Law states that the acceleration of an 4

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object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, we can describe the second law as 󰇍 ∑𝑭 󰇍𝒂 = 𝒎 󰇍 is the vector sum of all where 󰇍𝒂 is the acceleration of the object, 𝑚 is its mass, and ∑ 𝑭 forces acting on it. Force is a vector quantity, in terms of its components we write as ∑ 𝑭𝒙 = 𝒎𝒂𝒙 and ∑ 𝑭𝒚 = 𝒎𝒂𝒚 When there is no net force on an object, its acceleration is zero, which means the velocity is constant (Newton’s First Law). By dimension analysis, the SI standard unit of force is ∑𝐹 𝑎= 𝑚 𝒎 = 𝟏 𝑵𝒆𝒘𝒕𝒐𝒏 ∑ 𝐹 = 𝑚𝑎 = 𝒌𝒈 ∙ 𝒔𝟐 In the U.S. customary system, the unit of force is the pound (lb) which is 𝑠𝑙𝑢𝑔 ∙ 𝑓𝑡/𝑠 2 The conversion from newtons to pounds is given by 1 N = 0.225 lb. Example 1 An airboat with mass 3.50 x102 kg, including the passenger, has an engine that produces a net horizontal force of 7.70 x102 N, after accounting for forces of resistance. a. Find the acceleration of the airboat. Apply Newton’s second law and solve for the acceleration, 𝐹𝑛𝑒𝑡 7.70 𝑥 102 𝑁 = 𝐹𝑛𝑒𝑡 = 𝑚𝑎 → 𝑎 = = 𝟐. 𝟐𝟎 𝒎/𝒔𝟐 𝑚 3.50 𝑥 102 𝑁

b. Starting from rest, how long does it take the airboat to reach a speed of 12.0 m/s? Apply kinematics velocity equation from ULOb, 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 𝑚 𝑚 𝑚 12.0 = 0 + (2.20 2 ) 𝑡 𝑠 𝑠 𝑠 𝒕 = 𝟓. 𝟒𝟓 𝒔 c. After reaching that speed, the pilot turns off the engine and drifts to a stop over a distance of 50.0 m. Find the resistance force, assuming it’s constant. Let 12.0 m/s of (b) be the initial velocity. Also, know that the 50.0 m is the displacement ∆𝑥 , 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎(𝑥𝑓 − 𝑥𝑖 ) 𝑚 𝑚 2 0 = (12.0 ) + 2𝑎(50.0𝑚) 𝑠 𝑠 𝑎 = −1.44 𝑚/𝑠 2 Substitute the acceleration into Newton’s second law, finding the resistance force: 𝐹𝑟𝑒𝑠𝑖𝑠𝑡 = 𝑚𝑎 = (3.50𝑥 102 𝑘𝑔)(−1.44𝑚/𝑠 2 ) = −𝟓𝟎𝟒 𝑵 5

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Example 2 Two horses are pulling a barge with mass 2.00 x103 kg along a canal, as shown in Fig. 13. The cable connected to the first horse makes an angle of 𝜃 = 30.0° with respect to the direction of the canal, while the cable connected to the second horse makes an angle of 𝜃 = − 45.0°. Find the initial acceleration of the barge, starting at rest, if each horse exerts a force of magnitude 6.00x102 N on the barge. Ignore forces of resistance on the barge.

Fig. 13 Compute the total x- and y- component forces exerted by the horses, 𝐹1𝑥 = 𝐹1 𝑐𝑜𝑠𝜃 = (6.00𝑥 102 𝑁)(cos 30.0°) = 5.20𝑥 102 𝑁 𝐹2𝑥 = 𝐹2 𝑐𝑜𝑠𝜃 = (6.00𝑥 102 𝑁)(cos −45.0°) = 4.24𝑥 102 𝑁 ∑ 𝐹 = 𝐹1𝑥 + 𝐹2𝑥 = 9.44 𝑥 102 𝑁 𝑥

𝐹1𝑦 = 𝐹1 𝑠𝑖𝑛𝜃 = (6.00𝑥 102 𝑁)(sin 30.0°) = 3.00𝑥 102 𝑁 𝐹2𝑦 = 𝐹2 𝑠𝑖𝑛𝜃 = (6.00𝑥 102 𝑁)(sin −45.0°) = −4.24𝑥 102 𝑁 ∑ 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 = −1.24 𝑥 102 𝑁

Obtain the components of the acceleration, ∑ 𝐹𝑥 9.44𝑥 102 𝑁 𝑚 𝑎𝑥 = = = 0.472 2 3 𝑚 2.00 𝑥10 𝑘𝑔 𝑠 ∑ 𝐹𝑦 −1.24 𝑥102 𝑁 𝑚 = = −0.0620 2 𝑎𝑦 = 3 𝑚 2.00 𝑥10 𝑘𝑔 𝑠 Calculate the magnitude of the acceleration, 𝑎 = √𝑎𝑥2 + 𝑎2𝑦 = √(0.472 Calculate the direction,

𝜃 = tan−1

𝑚 2 𝑚 2 ) = 𝟎. 𝟒𝟕𝟔 𝒎/𝒔𝟐 ) + (−0.0620 𝑠2 𝑠2

𝑎𝑦

−0.0620𝑚/𝑠 2 = −𝟕. 𝟒𝟔 ° = 0.472 𝑚/𝑠 2 𝑎𝑥 6

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Example 3 A hockey puck having a mass of 0.30 kg slides on the frictionless, horizontal surface of an ice rink. Two hockey sticks strike the puck simultaneously, exerting the forces on the puck shown in Fig. 14. The force 󰇍󰇍󰇍 𝐹1 has a magnitude of 5.0 N, and is directed at 𝜃 = 20° below the x-axis. The force 󰇍󰇍󰇍 𝐹2 has a magnitude of 8.0 N and its direction is 𝜙 = 60° above the xaxis. Determine both the magnitude and the direction of the puck’s acceleration. Compute the total x- and y- component forces exerted by the sticks, 𝐹1𝑥 = 𝐹1 𝑐𝑜𝑠𝜃 = (5.0𝑁)(cos −20°) = 4.7𝑁 𝐹2𝑥 = 𝐹2 𝑐𝑜𝑠𝜃 = (8.0𝑁)(cos 60°) = 4.0𝑁 ∑ 𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 = 8.7𝑁

𝐹1𝑦 = 𝐹1 𝑠𝑖𝑛𝜃 = (5.0𝑁)(sin −20°) = −1.7𝑁 𝐹2𝑦 = 𝐹2 𝑠𝑖𝑛𝜃 = (8.0𝑁)(sin 60°) = 6.9𝑁 ∑ 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 = 5.2𝑁

Obtain the components of the acceleration, ∑ 𝐹𝑥 8.7𝑁 𝑚 𝑎𝑥 = = = 29 2 𝑚 0.30𝑘𝑔 𝑠 ∑ 𝐹𝑦 6.9𝑁 𝑚 𝑎𝑦 = = = 23 2 𝑠 𝑚 0.30𝑘𝑔 Calculate the magnitude of the acceleration, 𝑎 = √𝑎2𝑥 + 𝑎𝑦2 = √(29

Calculate the direction,

𝜃 = tan−1

𝑎𝑦 𝑎𝑥

𝑚 2 𝑚 2 ) = 𝟑𝟒𝒎/𝒔𝟐 ) + (23 𝑠2 𝑠2

=

23𝑚/𝑠 2 = 𝟑𝟏 ° 29𝑚/𝑠 2

Gravitational Force The gravitational force is the mutual force of attraction between any two objects in the Universe, as shown in Fig. 15. Newton’s law of universal gravitation states that every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. That is, 𝑚1 𝑚2 𝐹𝑔 = 𝐺 𝑟2 Where G = 6.67x10-11 N m2/kg2 is the universal gravitation constant.

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Fig. 15 The magnitude of the gravitational force acting on an object of mass 𝑚 is called the weight 𝑤 of the object, given by 𝑊 = 𝑚𝑔 Where 𝑔 is the acceleration due to gravity. The SI standard unit for weight is Newton 𝑁. Newton’s Third Law Newton recognized, however, that a single isolated force couldn’t exist. Instead, forces in nature always exist in pairs. He stated on his third law that, if two objects interact, the force 󰇍󰇍󰇍󰇍󰇍 𝐹12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the 󰇍󰇍󰇍󰇍21 󰇍 exerted by object 2 on object 1: force 𝐹 󰇍󰇍󰇍󰇍 󰇍 = −𝐹󰇍󰇍󰇍21 󰇍󰇍󰇍 𝐹12 The force that object 1 exerts on object 2 is popularly called the action force, and the force of object 2 on object 1 is called the reaction force. The negative sign indicates that the reaction force is acted in the opposite direction of equal magnitude. There are many kinds of reaction forces such as friction force and normal force. Frictional Force refers to the force generated by two surfaces that contacts and slide against each other. Example of which is a slab on a rough surface, a book sliding on a smooth surface, and such. The normal force is the support force exerted upon an object that is in contact with another stable object or surface, it is acted perpendicular to the action force. It is the upward force that opposes the weight of an object. Example 1 A man of mass 75.0 kg and woman of mass 55.0 kg stand facing each other on an ice rink, both wearing ice skates. The woman pushes the man with a horizontal force of 85.0 N in the positive x-direction. Assume the ice is frictionless. 85.0𝑁 𝐹 = = 𝟏. 𝟏𝟑 𝒎/𝒔𝟐 𝑚 75.0 𝑘𝑔 b. What is the reaction force acting on the woman? Apply Newton’s third law of motion, finding that the reaction force R acting on the woman: 𝑅 = −𝐹 = −𝟖𝟓. 𝟎𝑵

a. What is man’s acceleration?

𝐹 = 𝑚𝑎 → 𝑎 =

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c. Calculate the woman’s acceleration.

𝐹 = 𝑚𝑎 → 𝑎 =

Example 2

𝐹

𝟐 −85.0𝑁 = 55.0 𝑘𝑔 = −1.55 𝒎/𝒔 𝑚

A traffic light weighing 1.00 x102 N hangs from a vertical cable tied to two other cables that are fastened to a support, as in Fig. 16a. The upper cables make angles of 37.0° and 53.0° with the horizontal. Find the tension in each of the three cables. Tension is described as the pulling force transmitted through a rope, string or wire. Because we are interested only at the body and the acting forces, a force diagram called free-body diagram (FBD) would be helpful in the analysis. Identify all the action and reaction forces, in this case, we have the weight (Fg) of the traffic light and the tension forces of the strings. The construction of a correct free-body diagram is an essential step in applying Newton’s laws. An incorrect diagram will most likely lead to incorrect answers! Fig. 16 is an example of an FBD which shows the external forces acting on the body necessary for the analysis. Fig. 16b shows the forces acting on the traffic light and Fig.16c shows the forces acting on the cable knot.

Fig. 16 Objects that are either at rest or moving with constant velocity are said to be in equilibrium. Because of 𝑎 = 0 𝑚/𝑠 2 , Newton’s second law applied to an object in equilibrium gives ∑𝐹= 0

Find T3 from Fig. 16b, using the condition of equilibrium: ∑ 𝐹 = 𝑇3 − 𝐹𝑔 = 0 → 𝑇3 = 𝐹𝑔

𝑻𝟑 = 𝟏. 𝟎𝟎𝒙𝟏𝟎𝟐𝑵 Note the direction of the force, positive are those applied upward and rightwards while negative are those downward and leftwards. 𝑇1 and 𝑇2 are two-dimensional forces which means you need to solve for the x- and y- components. For the summation of forces along the x-direction, there are two component forces acted by 𝑇1 and 𝑇2 : 𝑇1𝑥 = 𝑇1 𝑐𝑜𝑠𝜃 = −𝑇1 (cos 37.0°) 𝑇2𝑥 = 𝑇2 𝑐𝑜𝑠𝜃 = 𝑇2 (cos 53.0°) 9

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∑ 𝑇𝑥 = −𝑇1 (cos 37.0°) + 𝑇2 (cos 53.0°) = 0

→ 𝑒𝑞. 1

While in the y-direction, there are three component forces acted by 𝑇1 , 𝑇2 and weight 𝐹𝑔 : 𝑇1𝑦 = 𝑇1 𝑠𝑖𝑛𝜃 = 𝑇1 (sin 37.0°) 𝑇2𝑦 = 𝑇2 𝑠𝑖𝑛𝜃 = 𝑇2 (sin 53.0°) 𝑇3 = −1.00𝑥 102 𝑁 ∑ 𝑇𝑦 = 𝑇1 (sin 37.0°) + 𝑇2 (sin 53.0°) − 1.00𝑥 102 𝑁 = 0

Equating eq. 1 and 2 to solve 𝑇1 and 𝑇2 −𝑇1 (cos 37.0°) + 𝑇2 (cos 53.0°) = 0 𝑇1 (sin 37.0°) + 𝑇2 (sin 53.0°) − 1.00𝑥 102 𝑁 = 0 𝑻𝟏 = 𝟔𝟎. 𝟏𝑵 and 𝑻𝟐 = 𝟕𝟗. 𝟗 𝑵

→ 𝑒𝑞. 2

Example 3 A sled is tied to a tree on a frictionless, snow-covered hill shown in Fig. 17. If the sled weighs 77.0 N, find the magnitude of the tension force exerted by the rope on the sled and that of the normal force 𝑛 exerted by the hill on the sled.

Fig. 17 Fig. 17b is the FBD for this force analysis. The object is at equilibrium, therefore, apply Newton’s second law. For the summation of forces along the x-axis, ∑ 𝐹𝑥 = +𝑇 − 𝐹𝑔 𝑠𝑖𝑛30 = 0

𝑇 − 77.0𝑁 𝑠𝑖𝑛30° = 0 𝑻 = 𝟑𝟖. 𝟓 𝑵 Notice that sine function is used to solve for the x-component, that is because of the angle 30° is opposite to the x-component of the weight force. For the summation of forces along the y-axis, ∑ 𝐹𝑦 = +𝑛 − 𝐹𝑔 𝑐𝑜𝑠30 ° = 0

Example 4

𝑛 − 77.0𝑁 𝑐𝑜𝑠30 ° = 0 𝒏 = 𝟔𝟔. 𝟕 𝑵 10

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A car of mass m is on an icy driveway inclined at an angle 𝜃 = 20.0°, as in Fig. 18a.

Fig. 18 a. Determine the acceleration of the car, assuming the incline is frictionless. Since the force causes the car to move, the summation of forces is equal to 𝑚𝑎. The acceleration happens only along x-direction, hence ∑ 𝐹𝑥 = 𝐹𝑔 sin 20.0 ° = 𝑚𝑎𝑥

(𝑚𝑔) sin 20.0 ° = 𝑚𝑎𝑥 𝑚 (9.80 2 ) sin 20.0 ° = 𝑎𝑥 𝑠 𝒂𝒙 = 𝟑. 𝟑𝟓 𝒎/𝒔𝟐 b. If the length of the driveway is 25.0 m and the car starts from rest at the top, how long does it take to travel to the bottom? 1 𝑥𝑓 = 𝑥𝑖 + 𝑣𝑖 𝑡 + 𝑎𝑡 2 2 1 𝑚 25.0 𝑚 = 0 + 0 + (3.35 2 ) (𝑡 2 ) 2 𝑠 𝒕 = 𝟑. 𝟖𝟔 𝒔 c. What is the car’s speed at the bottom? 𝑚 𝑚 𝒎 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 = 0 + (3.35 2 ) (3.86 𝑠) = 𝟏𝟐. 𝟗 𝑠 𝑠 𝒔 Forces of Friction A moving object on a surface or through a viscous medium experiences an opposite resistance force called friction. There are two types of friction: static 𝑓𝑠 and kinetic 𝑓𝑘 . A frictional force of an object exerted by its surface is proportional to its normal force 𝑛. 𝑓≤𝜇𝑛 Where 𝜇 is the proportionality constant called the coefficient of static friction 𝜇𝑠 or the coefficient of kinetic friction 𝜇𝑘 , depending on...