MEC30: STATICS OF RIGID BODIES BY: ENGR. LAMBERTO B. MARCIAL, JR. BASIC PRINCIPLES OF MECHANICS MEC30: STATICS OF RIGID BODIES BASIC PRINCIPLES OF MECHANICS Mechanics • is a branch of the physical sciences that is concerned with the state of rest or motion of bodies that are subjected to the action of forces. Three Branches of Mechanics • Rigid-Body Mechanics • Deformable-Body Mechanics • Fluid Mechanics BASIC PRINCIPLES OF MECHANICS Rigid-Body Mechanics • Is divided into two areas: – Statics – Dynamics Statics Dynamics • deals with the effects and distribution of forces on rigid bodies which are and remain at rest. • is concerned with the accelerated motion of bodies. BASIC PRINCIPLES OF MECHANICS Rigid Body • is a combination of a large number of particles in which all the particles remain at a fixed distance from one another, both before and after applying a load. Particle • is a representative of the body (whole) which has a mass but the size can be neglected BASIC PRINCIPLES OF MECHANICS External or Applied Force • is anything which changes or tends to change the state of motion of a body. Concurrent Forces • are forces whose lines of action are passing through one common point. Parallel Forces • are forces in the same direction whose lines of action never meet. BASIC PRINCIPLES OF MECHANICS Concurrent Forces Parallel Forces BASIC PRINCIPLES OF MECHANICS Resultant Force • is the equivalent single force that represents a system of forces. Equilibrant Force • is also known as the balancing force which has the same magnitude as the resultant force but in opposite direction. BASIC OPERATIONS WITH FORCE SYSTEM Law of Inertia (Newton’s First Law of Motion) • An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Action and Reaction Law (Newton’s Third Law of Motion) • for every action, there is always an opposite and equal reaction. • action and reaction forces are equal but opposite in direction. BASIC OPERATIONS WITH FORCE SYSTEM Superposition Law or Law of Transmissibility • It states that the state of rest or motion of a rigid body is unaltered if a force acting on the body is replaced by another force of the same magnitude and direction but acting anywhere on the body in the line of action of the replaced force. BASIC OPERATIONS WITH FORCE SYSTEM Resultant of Force Systems • the effect of a system of forces on a body is usually expressed in terms of a resultant. Note: A force is a vector quantity which means it is characterized by its magnitude and direction. VECTOR OPERATIONS Vector Addition • All vector quantities obey the parallelogram law of addition . Parallelogram Law • States that the resultant of two forces is the diagonal of the parallelogram formed on the two vector of those forces. PARALLELOGRAM LAW Determining Resultant using Parallelogram Law • The magnitude of the resultant force can be determined using law of cosines, and its direction using the law of sines. The magnitudes of two force components are determined from law of sines. PARALLELOGRAM LAW Sample Problem Determine the resultant and direction of the resultant. Answer: 212.6 N, 54.8o above positive x-axis PARALLELOGRAM LAW Sample Problem At what angle must the 800lb force be applied in order that the resultant the resultant R of the two forces has a magnitude of 2000lb? For this condition, determine the angle β between R and the vertical force. Answer: β = 18.19o COMPONENT OF A FORCE At a given angle: Fy = Fsinθ F θ Fx = Fcosθ COMPONENT OF A FORCE At a given slope: Fy = F(b/c) F c c2 = a2 + b2 b a Fx = F(a/c) COMPONENT OF A FORCE Sign Convention: Fy F c a Fx b positive (+) negative (-) COMPONENT OF A FORCE Sample Problem Determine the resultant and direction of the resultant. Answer: 212.6 N, 54.8o above positive x-axis COMPONENT OF A FORCE Sample Problem At what angle must the 800lb force be applied in order that the resultant the resultant R of the two forces has a magnitude of 2000lb? For this condition, determine the angle β between R and the vertical force. Answer: β = 18.19o DETERMINING THE RESULTANT FORCE Sample Problem Determine the Resultant along x and along y, the Magnitude of the Resultant and Direction of the Resultant of the system shown. Answer: R = 567.28 N, θ = 38.1o above +x-axis DETERMINING THE RESULTANT FORCE Sample Problem Determine the magnitude of the resultant force and its direction measured from the positive x-axis. Answer: R = 1040.82 N, θ = 56.42o above +x-axis DETERMINING THE RESULTANT FORCE Determining θ y (-Rx, +Ry) (+Rx, +Ry) θ θ θ θ (-Rx, -Ry) PROVIDED, x (+Rx, -Ry) 0o < θ < 90o DETERMINING THE RESULTANT FORCE Sample Problem Determine the magnitude of the resultant force and its direction measured from the positive x-axis. Answer: R = 545.79 N, θ = 252.64o from +x-axis SYSTEMS IN EQUILIBRIUM MEC30: STATICS OF RIGID BODIES SYSTEMS IN EQUILIBRIUM STATIC EQUILIBRIUM • When forces acting on an object which is at rest are balanced. • Even if there are forces exerted toward the object, these forces are balanced. Thus, the forces will not incur any movement. Object remains at rest and is at static equilibrium. • At static equilibrium, the resultant is equal to zero (Fr= 0). SYSTEMS IN EQUILIBRIUM CONDITIONS UNDER STATIC EQUILIBRIUM • TRANSLATIONAL EQUILIBRIUM – the sum of forces acting on a body is equal to zero. • • • • Ftotal = F1 + F2 + F3 + …… + Fn = 0 Fx = F1x + F2x + F3x + …… + Fnx = 0 Fy = F1y + F2y + F3y + …… + Fny = 0 Fz = F1z + F2z + F3z + …… + Fnz = 0 SYSTEMS IN EQUILIBRIUM CONDITIONS UNDER STATIC EQUILIBRIUM • ROTATIONAL EQUILIBRIUM – the sum of the moment forces or torque acting on a body is zero. Mtotal = M1 + M2 + M3 + …… + Mn = 0 SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM Determine the tension in cables BA and BC necessary to support the 60kg cylinder. FBC = 475.66 N FBA = 420.43 N We can also use the Force Triangle to solve for FBC and FBA SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM If the mass of cylinder C is 40 kg, determine the mass of cylinder A in order to hold the assembly in the position shown. mA = 20kg SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM Determine the maximum weight of concrete block and the contact force between the block and the incline that can be supported by force P = 500N as shown in the figure. .Note: Assume that the concrete block is resting on a smooth surface. W = 642.79 N CF = 939.69 N SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM Determine the reactions at A, B, C, and D. SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM Determine the reactions at A, B, C, and D. SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM Determine the reactions at A, B, C, and D. SYSTEMS IN EQUILIBRIUM SAMPLE PROBLEM Determine the reactions at A, B, C, and D. RA = 347.44 kN RB = 600 kN RC = 400.89 kN RD = 347.44 kN SEATWORK SEATWORK Resultant of a Force Determine the resultant of the following concurrent forces systems: SEATWORK SYSTEMS IN EQUILIBRIUM Determine the force P and its angle of inclination, θ, that will make the system in equilibrium. SEATWORK SYSTEMS IN EQUILIBRIUM If Block D weighs 300lb and Block B weighs 275lb, determine the weight of Block C and the angle θ for equilibrium.
