General Principles

Objectives To provide an introduction to the basic quantities and idealizations of mechanics. To review the principles for applying the SI system of units. To examine the standard procedures for performing numerical calculations. To present a general guide for problem solving.

Mechanics: Definition Branch of physical sciences concerned with the state of rest or motion of bodies subjected to forces.

Engineering Mechanics Solid Mechanics Fluid Mechanics Rigid Bodies Deformable Bodies Statics Dynamics

Rigid Body Mechanics Statics – Bodies at rest Dynamics – Accelerated motion of bodies

Basic Quantities Length Time Mass Force meter foot second kilogram slug Force newton pound

Length Needed to locate the position of a point in space and describe the size of a physical system.

Time Conceived as a succession of events. Concepts of STATICS are time independent.

Mass A property of matter by which we can compare the action of one body to another. This property manifests itself as a gravitational attraction between two bodies and provide a qualitative measure of the resistance of matter to a change in velocity.

Force Generally considered as a push of a pull exerted by one body on another. Interaction occurs when there is direct contact between the bodies. Gravitational, electrical and magnetic forces do not require direct contact. Force is characterized by magnitude, direction and point of application.

Idealizations Particle - an object having mass but the size is neglected. Rigid Body - a combination of a large number of particles which remain in a fixed position relative to each other, both before and after the application of a force.

SI Units Modern version of metric system. Base units are length, time and mass, meter (m), second (s), and kilogram (kg) Acceleration of gravity:

SI Units Force is derived quantity measured in unit called a newton

Systems of Units Name SI Length Time Mass Force meter (m) second (s) kilogram (kg) newton (N)

Prefixes for SI units

Dimensional Homogeneity Each of the terms of an equation must be expressed in the same units. s = v t + 1/2 a t 2 s is position in meters v is velocity in m/s a is acceleration in m/s2 t is time in seconds [m] = [m/s]  [s] + [m/s2]  [s2] = [m]

Calculations When performing calculations retain a greater number of digits than the problem data. Engineers usually round off final answer to three significant figures. Intermediate calculations are usually done to four significant figures. Answer can never have more significant figures than given data!

Convert 2 km/h to m/s. How many ft/s is this? SOLUTION: Since 1 km = 1000 m and 1 h = 3600 s, the conversion factors are arranged so that a cancellation of units can be applied. Recall that 1 ft = 0.3038 m

Procedure for Analysis Read the problem carefully and correlate the actual physical situation with the theory studied. Draw necessary diagrams and tables. Apply relevant principles, generally in mathematical form.

Procedure for Analysis Solve the equations algebraically (without numbers) as far as possible, then obtain a numerical answer. Be sure to use a consistent set of units. Report the answer with no more significant figures than the accuracy of the given data. Decide if answer seems reasonable. Think about what the problem taught you!

Important Points Statics is the study of bodies at rest or moving with constant velocity. A particle has mass but a size that can be neglected. A rigid body does not deform under load. Concentrated forces are assumed to act at a point on a body.

Important Points Mass is a property of matter that does not change from one location to another. Weight is the gravitational attraction of the earth on a body or quantity of mass. Its magnitude depends on the location of the mass.

Important Points In the SI system the unit of force is the newton. It is a derived quantity. Mass, length and time are the base quantities. In the SI system prefixes are used to denote large or small numerical quantities of a unit. Be sure that all equations are dimensionally homogeneous.

Equilibrium of a Particle

Objectives To introduce the concept of the free-body diagram for a particle. To show how to solve particle equilibrium problems using the equations of equilibrium.

Definitions A particle is in equilibrium if it is at rest if originally at rest or has a constant velocity if originally in motion. Static equilibrium denotes a body at rest. Newton’s first law is that a body at rest is not subjected to any unbalanced forces.

Static Equilibrium

Static Equilibrium is the vector SUM of all forces acting on the particle.

The Free-Body Diagram To apply equilibrium equations we must account for all known and unknown forces acting on the particle. The best way to do this is to draw a free-body diagram of the particle. This is a sketch that shows the particle “free” from its surroundings with all the forces acting on it.

Typical Supports Springs Cables and Pulleys

Springs Force Stiffness Coefficient Deformation

Springs Deformation

Cables and Pulleys

Cables and Pulleys Cables are assumed to have the following characteristics : negligible weight they cannot stretch they can only support tension or pulling (you can’t push on a rope) A continuous cable passing over a frictionless pulley must have tension force of a constant magnitude the tension force is always directed in the direction of the cable.

Drawing Free-Body Diagrams Draw Outlined Shape - Imagine the particle isolated or cut “free” from its surroundings Show All Forces - Include “active forces” and “reactive forces” Identify Each Force - Known forces labeled with proper magnitude and direction. Letters used for unknown quantities.

Force Types Active Forces - tend to set the particle in motion. Reactive Forces - result from constraints or supports and tend to prevent motion.

Gravitational Forces q O C.G. W

Contact Forces N f Normal Force Surface Friction Force Surface Plane tangent to the surfaces Normal Force Surface Friction Force Surface Curved Contacting Surfaces

Example -1 The sphere has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, cord CE, and the knot at C

Sphere There are two forces acting on the sphere. These are its weight and the force of cord CE. The weight is: W = 6 kg (9.81 m/s2)=58.9 N.

Sphere FCE Free-Body Diagram 58.9 N

Cord CE There are two forces acting on the cord. These are the force of the sphere, and the force of the knot. A cord is a tension only member..

Cord CE FCE FEC C E Free-Body Diagram

Knot at C There are three forces acting on the knot at C. These are the force of the cord CBA, and the force of the cord CE, and the force of the spring CD.

Knot at C FCBA 60o FCD Free-Body Diagram C FCE

Example -2 Not a Free Body Diagram! 250 kg

Example 3

Example -4

2D Equilibrium Equations Scalar equations of equilibrium require that the algebraic sum of the x and y components of all the forces acting on a particle be equal to zero.

Assume a sense for an unknown force Assume a sense for an unknown force. If the equations yield a negative value for the magnitude then the sense is opposite of what was assumed. F + 10 N = 0 F = -10 N F acts to the left (opposite of direction shown).

Procedure for Analysis Free-Body Diagram Establish the x, y axes in any suitable orientation. Label all known and unknown force magnitudes and directions on the FBD. The sense of an unknown force may be assumed.

Procedure for Analysis Equations of Equilibrium Apply equations of equilibrium. Components of force are positive if directed along a positive axis and negative if directed along a negative axis. If solution yields a negative result the force is in the opposite sense of that shown on the FBD.

Example 5 Determine the tension in cables AB and AD for equilibrium of the 250 kg engine block.

Problem 1 Determine the force in cables AC and AB needed to hold the 20-kg cylinder in equilibrium. Set F = 300 N and d = 1m

Problem 2 The cylinder D has a mass of 20 kg. If a force of F = 100 N is applied horizontally to the ring at A, determine the largest dimension d so that the force in cable AC is zero.

Magnitude of FR can be found by Pythagorean Theorem Force Resultants In all cases we have Magnitude of FR can be found by Pythagorean Theorem * Take note of sign conventions

Problem 3 The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.

(Choose the most appropriate answer) 1. When a particle is in equilibrium, the sum of forces acting on it equals ___ . A) A constant B) A positive number C) Zero D) A negative number E) An integer 2. For a frictionless pulley and cable, tensions in the cables are related as A) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin  T1 T2

QUIZ 100 N ( A ) ( B ) ( C ) 3. Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above? 4. Why? A) The weight is too heavy. B) The cables are too thin. C) There are more unknowns than equations. D) There are too few cables for a 100 kg weight.

QUIZ 5. Select the correct FBD of particle A. A 40 100 kg 30 30 A) 40° F1 F2 C) 30° F F1 F2 D)

QUIZ 6. Using this FBD of Point A, the sum of forces in the x-direction ( FY) is ___ . A) F1 sin 30° – F2 sin 40° = 0 B) F1 cos 30° – F2 sin 40° -981 = 0 C) F1 sin 30° + F2 sin 40° + 981 = 0 D) Non of the above

TUTORIAL

Question 1 Determine the magnitude and direction of the resultant force from two forces acting on the eyebolt in figure below. Θ=30.6o

Question 2 what is the magnitude and orientation of the resultant force for the two forces in the figure below. Θ=55o

Question 3 If the 5 kg block is suspended from the pulley B and the sag of the cord is d=0.15m, determine the force in the cord ABC. Neglect the size of the pulley