Lecture 3 - Mechanical Equilibrium PDF

ENGENG 207A 207A Physics for Engineers 1A LECTURE 3 Mechanical Equilibrium FACULTY OF ENGINEERING UNIVERSITY OF SANTO TOMAS Part 1 OBJECTIVES INTENDED LEARNING OUTCOMES...

ENGENG 207A 207A Physics for Engineers 1A LECTURE 3 Mechanical Equilibrium FACULTY OF ENGINEERING UNIVERSITY OF SANTO TOMAS Part 1 OBJECTIVES INTENDED LEARNING OUTCOMES 2 ❑ To study the first condition for equilibrium ❑ To be familiar with the forces in equilibrium ❑ To understand and calculate the center of gravity for objects FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS INTRODUCTION WHAT DO WE MEAN BY EQUILIBRIUM? 3 ⮚ If we find all forces and torques in equilibrium, we can understand situations we see every day - ladders firmly against a wall, bridges, even hanging pictures. FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS THE CONCEPT OF FORCE CATEGORIES OF FORCES 4 A force may be a push or pull upon an object which exists as a result of an interaction with another object. SI Unit: Newton (N), where 𝟏 𝑵 = 𝟏 𝒌𝒈 ∙ 𝒎/𝒔𝟐 1 CONTACT FORCES Involve physical contact between two objects 2 FIELD FORCES Forces that act through empty space FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS NEWTON’S LAWS OF MOTION THE FIRST LAW OF MOTION 5 Referred to as the Law of Inertia Inertia is the tendency of an object to resist any attempt to change its velocity In the absence of external forces, an object at rest remains at rest and an object in motion remains in motion with constant velocity. FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS NEWTON’S LAWS OF MOTION THE THIRD LAW OF MOTION 6 If two objects interact, the force FA exerted by object A on object B is equal in magnitude and opposite in A FA direction to the force FB exerted by object B on object A. 𝑭𝑨 = −𝑭𝑩 FB B FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS THE EQUILIBRIUM MODEL FREE-BODY DIAGRAM (FBD) 7 A diagram that shows the relative magnitude and direction of all forces acting upon an object in a given situation. 1 GRAVITATIONAL FORCE (Fg = mg) Attractive force exerted by the Earth on an object that is directed towards the center of the Earth. 2 NORMAL FORCE (N) The force exerted by a surface to an object whenever they are in contact. It is always perpendicular to the surface. 3 TENSION (T) A pulling force that is exerted by a rope, string, chain, or cable on an object. It always lies along the direction of the string. FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS THE EQUILIBRIUM MODEL CONDITIONS OF STATIC EQUILIBRIUM 8 An object is said to be in complete static equilibrium when it is at rest and satisfies the two conditions of equilibrium. 1 TRANSLATIONAL EQUILIBRIUM When the sum of all forces (net force) acting on an object is zero 2 ROTATIONAL EQUILIBRIUM When the sum of all moments (net moment) acting on an object is zero FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS INTRODUCTION WHAT DO WE MEAN BY EQUILIBRIUM? 9 EQUILIBRIUM when a body is at rest or moving with constant velocity in an inertial frame of reference STATIC DYNAMIC (at rest) (in motion) FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM FORCES AND TORQUES 10 ⮚ The sum of all forces present in the x, y, and z directions are each distinctly equal to zero. ⮚ The sum of all torques for any given point are equal to zero. ෍ 𝐅Ԧ = 𝟎 ෍𝛕 = 𝟎 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM FORCES AND TORQUES 11 ⮚ This object is in Static Equilibrium: Equilibrium conditions: First condition satisfied: Net force = 0, so the object at rest has no tendency to start moving as a whole. Second condition satisfied: Net torque about the axis = 0, so the object at rest has no tendency to start rotating. Axis of rotation FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM FORCES AND TORQUES 12 ⮚ This object has no tendency to accelerate as a whole, but it has a tendency to start rotating. First condition satisfied: Net force = 0, so the object at rest has no tendency to start moving as a whole. Second condition NOT satisfied: There is a net clockwise torque about the axis, so the object at rest will start rotating clockwise. FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM FORCES AND TORQUES 13 ⮚ This object has a tendency to accelerate as a whole, but no tendency to start rotating. First condition NOT satisfied: There is a net upward force, so the object will start moving upward. Second condition satisfied: Net torque about the axis = 0, so the object at rest has no tendency to start rotating. FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 14 ⮚ Following Newton’s 1st law of motion, For a system to be at rest, the sum of forces acting on a system must add up to zero. ෍ 𝐅Ԧ = 𝟎 And since force is a vector, the components of the net force must each be zero. σ 𝐅Ԧ𝐱 = 𝟎 σ 𝐅Ԧ𝐲 = 𝟎 σ 𝐅Ԧ𝐳 = 𝟎 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 15 ⮚ EXAMPLE 1: Which of the following situations satisfy both the first condition for equilibrium (net force = 0) and the second condition for equilibrium (net torque = 0)? A. an automobile crankshaft turning at an increasing angular speed in the engine of a parked car B. a seagull gliding at a constant angle below the horizontal and at a constant speed C. a thrown baseball that does not rotate as it sails through the air D. more than one of the above E. none of the above FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 16 ⮚ EXAMPLE 1: Which of the following situations satisfy both the first condition for equilibrium (net force = 0) and the second condition for equilibrium (net torque = 0)? A. an automobile crankshaft turning at an increasing angular speed in the engine of a parked car B. a seagull gliding at a constant angle below the horizontal and at a constant speed C. a thrown baseball that does not rotate as it sails through the air D. more than one of the above E. none of the above FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 17 ⮚ EXAMPLE 2: A metal advertising sign (weight w) is suspended from the end of a massless rod of length L. The rod is supported at one end by a hinge at point P and at the other end by a cable at an angle θ from the horizontal. What is the tension in the cable? A. T = w sin θ B. T = w cos θ C. T = w / sin θ D. T = w / cos θ E. none of the above FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 18 ⮚ EXAMPLE 2: A metal advertising sign (weight w) is suspended from the end of a massless rod of length L. The rod is supported at one end by a hinge at point P and at the other end by a cable at an angle θ from the horizontal. What is the tension in the cable? A. T = w sin θ B. T = w cos θ C. T = w / sin θ D. T = w / cos θ E. none of the above FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 19 ⮚ EXAMPLE 3: A metal advertising sign (weight w) is suspended from the end of a massless rod of length L. The rod is supported at one end by a hinge at point P and at the other end by a cable at an angle θ from the horizontal. Which of these forces is least? A. the weight of the sign B. the tension in the cable C. the vertical force component exerted on the rod by hinge P D. two or more of these are tied for greatest FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 20 ⮚ EXAMPLE 3: A metal advertising sign (weight w) is suspended from the end of a massless rod of length L. The rod is supported at one end by a hinge at point P and at the other end by a cable at an angle θ from the horizontal. Which of these forces is least? A. the weight of the sign B. the tension in the cable C. the vertical force component exerted on the rod by hinge P D. two or more of these are tied for greatest FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 21 ⮚ EXAMPLE 4: An automobile engine has a weight of 3150 N. This engine is being positioned above an engine compartment. To position the engine, a worker is using a rope. Find the tension in the supporting cable T1 and the tension in the positioning rope T2? FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 22 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS THE EQUILIBRIUM MODEL: TRANSLATIONAL EQUILIBRIUM EXAMPLE # 01 23 As shown in the figure and given that the mass of block m is 20.0 kg, determine the following: a) tension on cable 1 b) tension on cable 2 c) tension on cable 3 d) tension on cable 4 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS SOLUTION 24 STEP 1: FBD Construction FBD at Knot A FBD at Block m 𝚺𝑭𝒙 = 𝟎 𝚺𝑭𝒚 = 𝟎 𝒎 = 𝟐𝟎 𝒌𝒈 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS SOLUTION 25 STEP 2: Summation of forces and solving for the unknowns FBD at Block m 𝚺𝑭𝒙 = 𝟎: +𝑇1 cos 30𝑜 − 𝑇2 sin 40𝑜 + 0 = 0 𝚺𝑭𝒚 = 𝟎: +𝑇1 sin 30𝑜 + 𝑇2 cos 40𝑜 − 𝐹𝑔 = 0 𝑚 𝐹𝑔 = 𝑚𝑏𝑙𝑜𝑐𝑘 𝑔 = 20.0 𝑘𝑔 9.81 = 𝟏𝟗𝟔. 𝟐 𝑵 𝑠2 𝚺𝑭𝒙 = 𝟎: +𝑇1 cos 30𝑜 − 𝑇2 sin 40𝑜 = 0 𝚺𝑭𝒚 = 𝟎: +𝑇1 sin 30𝑜 + 𝑇2 cos 40𝑜 = 196.2 𝑁 Solving simultaneously, 𝑻𝟏 = 𝟏𝟐𝟖. 𝟎𝟔 𝑵 𝑻𝟐 = 𝟏𝟕𝟐. 𝟓𝟒 𝑵 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS SOLUTION 26 STEP 2: Summation of forces and solving for the unknowns FBD at Knot A 𝚺𝑭𝒙 = 𝟎: +𝑇2 sin 40𝑜 + 𝑇3 cos 45𝑜 − 𝑇4 = 0 𝑻𝟐 = 𝟏𝟕𝟐. 𝟓𝟒 𝑵 𝚺𝑭𝒚 = 𝟎: −𝑇2 cos 40𝑜 + 𝑇3 sin 45𝑜 = 0 𝑇3 sin 45𝑜 = (172.54 𝑁) cos 40𝑜 𝑻𝟑 = 𝟏𝟖𝟔. 𝟗𝟐 𝑵 𝚺𝑭𝒙 = 𝟎: 𝑇4 = (172.54 𝑁) sin 40𝑜 + (186.92) cos 45𝑜 𝑻𝟒 = 𝟐𝟒𝟑. 𝟎𝟖 𝑵 𝑻𝟏 = 𝟏𝟐𝟖. 𝟎𝟔 𝑵 𝑻𝟐 = 𝟏𝟕𝟐. 𝟓𝟒 𝑵 𝑻𝟑 = 𝟏𝟖𝟔. 𝟗𝟐 𝑵 𝑻𝟒 = 𝟐𝟒𝟑. 𝟎𝟖 𝑵 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 27 ⮚ EXAMPLE 5: A block resting on a plane inclined at an angle, 𝜃, of 42o from the horizontal is being held in place by a steel cable. The tension on the cable wire is 67.0 N. a) Determine the normal force N and the force due to gravity 𝑭𝒈. b) What is the mass of the block in kilograms? FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 28 ⮚ EXAMPLE 5: SOLUTION STEP 1 - FBD construction at block Regular x-y plane Shifted x’-y’ plane 𝜽 = 𝟒𝟐° 𝑻 = 𝟔𝟕. 𝟎 𝑵 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS CONDITIONS FOR EQUILIBRIUM 1st CONDITION FOR EQUILIBRIUM 29 ⮚ EXAMPLE 5: SOLUTION STEP 2: Summation of forces and solving for the unknowns 𝚺𝑭𝒙′ = 𝟎: 𝑇 − 𝐹𝑔 sin 42𝑜 = 0 𝑻 = 𝟔𝟕. 𝟎 𝑵 𝚺𝑭𝒚′ = 𝟎: 𝑁 − 𝐹𝑔 cos 42𝑜 a) Determine the normal force N and the force due to gravity 𝑭𝒈. 𝚺𝑭𝒙′ = 𝟎: 67.0 𝑁 − 𝐹𝑔 sin 42𝑜 = 0 𝑭𝒈 = 𝟏𝟎𝟎. 𝟏𝟑 𝑵 𝚺𝑭𝒚′ = 𝟎: 𝑁 − (100.13 𝑁) cos 42𝑜 𝑵 = 𝟕𝟒. 𝟒𝟏 𝑵 b) What is the mass of the block in kilograms? 𝐹𝑔 = 𝑚𝑏𝑙𝑜𝑐𝑘 𝑔 𝑚 100.13 𝑁 = 𝑚𝑏𝑙𝑜𝑐𝑘 9.81 𝒎𝒃𝒍𝒐𝒄𝒌 = 𝟏𝟎. 𝟐𝟏 𝒌𝒈 𝑠2 FACULTY OF ENGINEERING | UNIVERSITY OF SANTO TOMAS ENGENG 207A 207A Physics for Engineers 1A LECTURE 3 Mechanical Equilibrium FACULTY OF ENGINEERING UNIVERSITY OF SANTO TOMAS Part 1