Problem y A Cable AB is 65 ft long, and 56 ft

Slides:

Advertisements

Similar presentations

Resultant of two forces

Advertisements

Chapter 2 STATICS OF PARTICLES

Right Triangle Trigonometry

ENGR-1100 Introduction to Engineering Analysis

Today’s Objectives: Students will be able to :

Forging new generations of engineers

Forces and moments Resolving forces.

Trigonometry Right Angled Triangle. Hypotenuse [H]

Exam 1, Fall 2014 CE 2200.

CE Statics Lecture 4. CARTESIAN VECTORS We knew how to represent vectors in the form of Cartesian vectors in two dimensions. In this section, we.

Problem lb A couple of magnitude 30 lb

APPLICATION OF VECTOR ADDITION

Equilibrium of a Rigid Body

Force Vectors. Vectors Have both a magnitude and direction Examples: Position, force, moment Vector Quantities Vector Notation Handwritten notation usually.

Force Vectors Principles Of Engineering

Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland ENGI.

POSITION & FORCE VECTORS (Sections ) Today’s Objectives: Students will be able to : a) Represent a position vector in Cartesian coordinate form,

Force System in Three Dimension

Lecture 24 ENGR-1100 Introduction to Engineering Analysis.

DOT PRODUCT (Section 2.9) Today’s Objective:

EQUILIBRIUM OF A PARTICLE IN 2-D Today’s Objectives: Students will be able to : a) Draw a free body diagram (FBD), and, b) Apply equations of equilibrium.

ENGR-1100 Introduction to Engineering Analysis

Vector Components. Coordinates  Vectors can be described in terms of coordinates. 6.0 km east and 3.4 km south6.0 km east and 3.4 km south 1 N forward,

VECTORS in 3-D Space Vector Decomposition Addition of Vectors:

Chapter 2 Statics of Particles

Forces and equilibrium

EXAMPLE 1 Use an inverse tangent to find an angle measure

Examples from Chapter 4.

Trigonometry. Basic Ratios Find the missing Law of Sines Law of Cosines Special right triangles

Vectors. Vectors and Direction Vectors are quantities that have a size and a direction. Vectors are quantities that have a size and a direction. A quantity.

Problem y C 10 ft The 15-ft boom AB has a fixed 6 ft

POSITION & FORCE VECTORS (Sections ) Today’s Objectives: Students will be able to : a) Represent a position vector in Cartesian coordinate form,

Solution of Triangles COSINE RULE. Cosine Rule  2 sides and one included angle given. e.g. b = 10cm, c = 7 cm and  A = 55° or, a = 14cm, b = 10 cm and.

Find the angle between the forces shown if they are in equilibrium.

Trig Ratios SohCahToa Sine = Sin A = ___ Sin C = ___.

DOT PRODUCT In-Class Activities: Check Homework Reading Quiz Applications / Relevance Dot product - Definition Angle Determination Determining the Projection.

MOMENTS Created by The North Carolina School of Science and Math.The North Carolina School of Science and Math Copyright North Carolina Department.

Vector Addition Cummutative Law A B C B A C A + B = C B + A = C A + B = B + A.

FORCES IN SPACE (Noncoplanar System of Forces)

Lecture #3 Additional Methods of Vector Analysis (Ref: Section 2.7 and 2.8, position vectors and force vector directed along a line) R. Michael PE 8/14/2012.

College of Engineering CIVE 1150 Fall Rectangular Components of a Force: Unit Vectors Vector components may be expressed as products of the unit.

A jogger runs 145m in a direction 20

Chapter 2 Statics of Particles. Addition of Forces Parallelogram Rule: The addition of two forces P and Q : A →P→P →Q→Q →P→P →Q→Q += →R→R Draw the diagonal.

Problem y C D A container of weight W = 1165 N is supported O

Force Vectors Phy621- Gillis

5.5 Law of Sines. I. Law of Sines In any triangle with opposite sides a, b, and c: AB C b c a The Law of Sines is used to solve any triangle where you.

1.Total degrees in a triangle: 2.Three angles of the triangle below: 3.Three sides of the triangle below: 4.Pythagorean Theorem: x 2 + y 2 = r A.

THREE-DIMENSIONAL FORCE SYSTEMS In-class Activities: Check Homework Reading Quiz Applications Equations of Equilibrium Concept Questions Group Problem.

Warm-Up Write the sin, cos, and tan of angle A. A BC

Homework Review (Ch. 1 & 2) 1-6 (pg. 15) 1-8 (pg. 15) 2-10 (pg. 28) 2-32 (pg. 39) 2-57 (pg. 42)  Any questions???

3D-Force System. RECTANGULAR COMPONENTS Many problems in mechanics require analysis in three dimensions, and for such problems it is often necessary to.

Copyright © 2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES.

13.5 Law of Cosines Objectives: 1.Solve problems by using the Law of Cosines 2.Determine whether a triangle can be solved by first using the Law of Sines.

Statics (ENGR 2214) Prof. S. Nasseri Statics ENGR 2214 Vectors in three dimensional space.

EXAMPLE 1 Solve a triangle for the AAS or ASA case Solve ABC with C = 107°, B = 25°, and b = 15. SOLUTION First find the angle: A = 180° – 107° – 25° =

Trigonometry Section 7.4 Find the sine and cosine of special angles. Consider the angles 20 o and 160 o Note: sin 20 o = sin160 o and cos 20 o = -cos 160.

Problem The direction of the 75-lb forces may vary, but the angle between the forces is always 50 o. Determine the value of  for which the resultant.

By: Forrest Langley.  In order to solve triangles, you must use Sine, Cosine, and Tangent  Sinx= Opposite/Hypotenuse  Cosx= Adjacent/Hypotenuse  Tanx=

EXAMPLE 1 Use an inverse tangent to find an angle measure Use a calculator to approximate the measure of A to the nearest tenth of a degree. SOLUTION Because.

Basic Info: balanced Forces Objects are balanced only if their net force is zero in both the vertical and horizontal directions Objects are balanced only.

2 - 1 Engineering Mechanics Lecture 2 Composition and Resolution of forces

O ___ A __ (4/5) (4/5) 38.7° (angles are typically rounded to the tenth) 38.7° tanθ = O __ A tanC = 20 ___ 16 C = tan-1(5/4) C ≈ 51.3°

Cartesian Vectors In right-handed Cartesian coordinated system, the right thumb points in the positive z direction when the right hand figures are curled.

Cartesian Vectors In right-handed Cartesian coordinated system, the right thumb points in the positive z direction when the right hand figures are curled.

Cartesian Vectors In right-handed Cartesian coordinated system, the right thumb points in the positive z direction when the right hand figures are curled.

Cartesian Vectors In right-handed Cartesian coordinated system, the right thumb points in the positive z direction when the right hand figures are curled.

Geometry Section 7.7.

Problem B A Two cables are tied together at C and

Presentation transcript:

Problem 2.133 y A Cable AB is 65 ft long, and 56 ft the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx, qy, and qz defining the direction of that force. D a O B 20o 50o z C x

Solving Problems on Your Own x y z 56 ft D B C A 50o 20o a Problem 2.133 Solving Problems on Your Own Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx, qy, and qz defining the direction of that force. 1. Determine the rectangular components of a force defined by its magnitude and direction. If the direction of the force F is defined by the angles qy and f, projections of F through these angles or their components will yield the components of F.

Solving Problems on Your Own x y z 56 ft D B C A 50o 20o a Problem 2.133 Solving Problems on Your Own Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx, qy, and qz defining the direction of that force. 2. Determine the direction cosines of the line of action of a force. The direction cosines of the line of action of a force F are determined by dividing the components of the force by F. cos qx= Fx F cos qy= Fy F cos qz= Fz F

Fy = + F cos qy = (3900 lb)(0.86154) Fy = + 3360 lb Problem 2.133 Solution A Determine the direction cosines of the line of action of a force. 65 ft qy From triangle AOB: cos qy = 56 ft 65 ft = 0.86154 qy = 30.51o F Fy 56 ft Fx O B (a) Fx = _ F sin qy cos 20o = _ (3900 lb) sin 30.51o cos 20o Fx = _1861 lb 20o Fz z x Fy = + F cos qy = (3900 lb)(0.86154) Fy = + 3360 lb Fz = + (3900 lb) sin 30.51o sin 20o Fz = + 677 lb

Fx F = _1861 lb 3900 lb cos qx = _ 0.4771 qx = 118.5o cos qz = Fz F y Problem 2.133 Solution O x B A Fy Fz F Fx 20o qy 65 ft Determine the direction cosines of the line of action of a force. (b) cos qx = Fx F = _1861 lb 3900 lb 56 ft cos qx = _ 0.4771 qx = 118.5o From above (a): qy = 30.5o cos qz = Fz F = + 677 lb 3900 lb = + 0.1736 qz = 80.0o