MODULE-1 ENGINEERING MECHANICS PDF

# Physics ## Mechanics ### Statics - is the study of bodies and structures that are in equilibrium. - For a body to be in equilibrium, there must be no net force acting on it. - **True or False?** Statics only deals with objects at rest. **False.** Objects in constant motion (no acceleration),...

# Physics ## Mechanics ### Statics - is the study of bodies and structures that are in equilibrium. - For a body to be in equilibrium, there must be no net force acting on it. - **True or False?** Statics only deals with objects at rest. **False.** Objects in constant motion (no acceleration), is also treated using the principles of equilibrium. ### Dynamics - is the subdivision of mechanics that is concerned with the motion of material objects in relation to the physical factors that affect them: force, mass, momentum, energy. - **Kinematics:** concerned with the geometrically possible motion of a body or system of bodies without consideration of the forces involved (i.e., causes and effects of the motions). - **Kinetics:** concerns the effect of forces and torques on the motion of bodies having mass. ## Scalar and Vector Quantities: - The mathematical quantities that are used to describe the motion of objects can be divided into two categories. - The quantity is either a vector or a scalar. - These two categories can be distinguished from one another by their distinct definitions. ### Scalars - are quantities that are fully described by a magnitude (or numerical value) alone. - Examples: time, volume, area, speed, mass, temperature, distance, energy, work etc. ### Vectors - are quantities that are fully described by both a magnitude and a direction. - Examples: force, acceleration, velocity, momentum, weight etc. - The combination of two or more single vectors is called a resultant vector or resultant. ## Resultant of Forces/Vectors ### Resultant: - Something that results; an outcome. - Issuing or following as a consequence or result. - A single vector that is the equivalent of a set of vectors. - That results; following as a consequence. - Resulting from two or more forces or agents acting together. - **Components of forces** - Resolve the 40 N force into components along x & y axes - $P_{x}$ = 40 cos 35° = 32.77 N - $P_{y}$ = 40 sin 35° = 22.94 N - Complex Mode: P20 = 40∠35° = 32.77 + 22.94i ## Components of Forces - Resolve the 50 N force into components along x & y axes - $P_{x}$ = 50 cos 36.87° = 40 N - $P_{y}$ = 50 sin 36.87° = 30 N - $P_{x}$ = 50 (4/5) = 40 N - $P_{y}$ = 50 (3/5) = 30 N - Complex Mode: P∠a = 50∠(180° – 36.87°) - $P_{zα}$ = -40+ 30i - Resolve the 50 N force into components along x & y axes - $P_{x}$ = 50 sin 40° = 31.14 N - $P_{y}$ = 50 cos 40° = 38.30 N - Complex Mode: 50∠310° = 32.14 - 38.30i, 50∠-50° = 32.14 - 38.30i - Resolve the 100 N force into components along A & B axes - $P_{A}$ = 84.81 N - $P_{B}$ = 79.31 N ## Resultant of Forces - Solve for the resultant and direction - $R_{x}$ = 181.47 N - $R_{y}$ = 251.22 N - $R$ = $\sqrt{R_{x}^{2}+R_{y}^{2}}$ = 301.91 N - tan θ = $R_{y}/R_{x}$ = 251.22/181.47 - θ = 54.157° - Complex Mode: R = 200∠0° + 300∠60° + 220∠140° + 150∠-90° - R = 181.47 + 251.221i - R∠θ= 309.91∠54.157° - Solve for the resultant and direction - R = 200∠0° + 300∠240° + 220∠140° + 150∠-90° - R = -118.53 - 268.394i - R∠θ = 293.402∠ - 113.827° ## Two forces P and Q acts on a hook - P = 155 N and α = 25°. - If Q = 85 N, find Θ (degrees) such that the resultant of these two forces is horizontal to the right. - If Q = 220 N, find Θ (degrees) such that the resultant acts vertically downward. - If the resultant force is 210 N acting downwards to the right at an angle of 20° with the horizontal, what is the value of Q in N? - If the resultant force is 210 N acting downward to the right at an angle of 20° with the horizontal, what is the value of Θ in degrees? - $R_{y}$ = ∑Fy =0 - 155 sin 25°-85 sin θ = 0 - sin Θ = 155 sin 25°/85 - Θ = 50.76° - $R_{x}$ = ∑Fx = 0 - 155 cos 25° +220 cos θ = 0 - cos Θ = -155 cos 25°/220 - Θ = 129.683° - $R_{x}$ = ∑Fx - 210 cos 20° = Qx +155 cos 25° - Qx = 56.858 N - $R_{y}$ = ∑Fy - -210 sin 20° = -Qy +155 sin 25° - Qy = 137.33 N - Q = $\sqrt{Q_{x}^{2}+Q_{y}^{2}}$ = 148.635 N - tan θ = $Q_{y}/Q_{x}$ = 67.51° - $R_{x}$ = R cos 12° = ∑Fx - R cos 12° = 200 cos 60° +250 cos θ - R = 102.234 +255.585 cos θ - $R_{y}$ = R sin 12° = ∑Fy - R sin 12° = 200 sin 60° -250 sin θ - R = 833.07 -1202.434 sin θ - 102.234 + 255.585 cos θ = 833.07 - 1202.434 sin θ - Θ = 24.48° - R = 102.234 + 255.585 cos 24.48° = 334.84 N - $R_{y}$' = ∑Fy' = 0 - 200 sin 48° -250 sin(θ +12°) = 0 - Θ = 24.48° - R = Rx' = ∑Ex' - R = 200 cos 48° +250 cos(24.48° + 12°) - R= 334.84 N - 250² = R² + 200² -2(R*200) cos 48° - R= 334.85 N - sin(θ + 12°)/ 200 = sin 48°/250 - Θ = 24.48° ## In physics, moment of force (often just moment) is a measure of its tendency to cause a body to rotate about a specific point or axis. - The moment of a force about a point is the product of the force and its perpendicular distance to the point. - M = Force × Distance - Find the moment of the given force about point O. Mo = 20 (2.5) = 50 kN-m - Find the moment of the given force about point 0. Mo = -10 (1.5) = -15 N-m - Find the moment of the given force about point O. Mo = 10 sin 30° (y) -10 cos 30° (x) - x = 1.5 cos 10° - y = 1.5 sin 10° - Mo = -11.49 N-m - Find the moment of the given forces about point O. - Mo = 16 cos 60° (5 sin 60°) - 16 sin 60° (5 cos 60° + 5) -20 cos 40° (5 sin 40°) - 20 sin 40° (5-5 cos 40°) - Mo = -133.561 kN-m - Find the moment of the given forces about point O. - Mo = -(16 sin 60° + 20 sin 40°)(5) - Mo = -133.561 kN-m - Find the moment of the given forces about point B. - MB = -60 sin 18.435°(2) - 60 cos 18.435° (1) - MB-94.87 kN-m - MB-60 sin 18.435°(3 + 2) - MB-94.87 kN-m - The principle of transmissibility states that the point of application of a force can be moved anywhere along its line of action without changing the external reaction forces on a rigid body. ## Resultant of Non-Concurrent Forces - Determine the magnitude and position of the resultant of the given forces. - R = 8+4 +3 +2 (4) = 23 kN - Rx = 8(2) + 4(6.5) + 3(10) + 2(4)(2+3+2) - x = 128/23 = 5.565 m - Using the Statistics Mode (Mode 3 2) - Sum of FREQ is 23 - Determine the moment at A and the magnitude and position of the resultant of the given forces. - Fr = 42 - MA = 165 - Determine the magnitude and position of the resultant force acting on the dam shown. - tan(β) = α - β = 90-59.03624347 - α = 59.03624347 - R = 710+ (1100 + 1600)∠ – 90° +250∠(180° + 30.9638°) - R = 495.627 kN, Ry = 2828.624 kN - R = 2871.72 kN, θ = 80.062° - x = 18810 / 2828.624 = 6.65 m