Showing posts with label statics. Show all posts
Showing posts with label statics. Show all posts
CE Tips

Statics: CE Nov. 2010 - Two tension wires are supported by a concrete block


Two tension wires are supported by a concrete block attached to an anchor ring as shown in the figure.


1. Which of the following most nearly gives the resultant force in the anchor ring?
       a.  7.28 KN                                   c.  9.07 KN
       b.  10.42 KN                                 d.  5.16 KN


2. Which of the following most nearly gives the angle that the resultant force makes with the horizontal?

       a.  43.67°                                      c.  39.55°
       b.  53.17°                                      d.  48.72



For #1

→+ΣFx     (horizontal)
Rx = -7cos15° + 3.5cos30° = -3.7304 =  -3.73 KN


↑+ΣFy     (vertical)
Ry = 7sin15° + 3.5sin30° = 3.5617 =  3.562 KN


Use the Pythagorean Theorem
R² = Rx² + Ry²
R² = (-3.73)² + (3.562)²
R = 5.16 KN


For #2

tan θ = |Ry/Rx|
tan θ = |3.562/-3.73|
θ = 43.67°



Text Box: WText Box: 3.50 KNText Box: 7 KNText Box: 15°          30°
Text Box: WText Box: 3.50 KNText Box: 7 KNText Box: 15°          30° θ = |tan-1(3.562/-3.73)|
θ = 43.67°

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Engineering Mechanics: STATICS (Solution Manual)




Over the past 50 years, Meriam & Kraige's Engineering Mechanics: Statics has established a highly respected tradition of excellence--a tradition that emphasizes accuracy, rigor, clarity, and applications. Now in a Sixth Edition, this classic text builds on these strengths, adding a comprehensive course management system, Wiley Plus, to the text, including an e-text, homework management, animations of concepts, and additional teaching and learning resources. New sample problems, new homework problems, and updates to content make the book more accessible. 

The Sixth Edition continues to provide a wide variety of high quality problems that are known for their accuracy, realism, applications, and variety motivating students to learn and develop their problem solving skills. To build necessary visualization and problem-solving skills, the Sixth Edition continues to offer comprehensive coverage of drawing free body diagrams-- the most important skill needed to solve mechanics problems.













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Engineering Mechanics 3Ed - K. L. Kumar



Engineering is an activity concerned with the creation of new systems for the benefit of mankind. The process of creativity proceeds by way of research, design and development; new systems emerge from innovation and systems may be constituted by mechanical, electromechanical, hydraulic. thermal or other elements. Creation of new systems is thus basic to all engineering. The Living Webster Encyclopedic Dictionary aptly defines engineering as the art of excelling a partial application of scientific knowledge.

lt is important to understand the difference between engineering and science. Science is concerned with a systematic understanding and gathering of the facts, laws and principles governing natural phenomena. Engineering, on the other hand, is an art of utilisation of the established facts, laws and principles to create certain desired phenomena. The activities of science and engineering arc thus mutually opposite. Both may proceed through similar ways and means of analysis and synthesis but are oppositely directed. The training of scientists and engineers should be correspondingly designed for their respective objectives.












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Statics: Determine the fourth force


The 100-lb resultant of four forces together with the three of those four forces is shown. 
Determine the fourth force (P).



Rx = 100 cos45º = 70.71 lb
Ry = 100 sin45º = 70.71 lb


→+ΣFx = 70.71     
-70cos60º - 120cos60º + 50cos50º + Px = 70.71
Px = 133.571 lb


↑+ΣFy = 70.71
-70sin60º - 120sin60º -50sin50º + Py = 70.71
Py = 152.313 lb

P2 = Px2 + Py2
P2 = (133.571)2 + (152.313)2 
P = 202.58 lb



for theta (θ)
tan(θ) = 152.313 / 133.571
θ = 48.75º







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Statics: A truss structure is used to hold power lines as illustrated



A truss structure is used to hold power lines as illustrated. The power lines provide forces skewed to the tower. Calculate the reaction forces at the base of the tower.


Solution
↻+ (clockwise-positive)
→+ (right-positive)
↑+ (upward-positive)



→+ΣFx = 0    assume Ax is postive
Ax - 300sin30º + 400sin25º = 0
Ax = -19.05N
Ax = 19.05N (←)


↻+ΣMA = 0
-By(4) + 400cos25º (4+2) + 400sin25º (10) -300cos30º (2) -300sin30º (10)=0
By = 461.50N


↻+ΣMB = 0    assume Ay is positive (↑+)

Ay(4) - 300cos30º (2+4) - 300sin30º (10) + 400cos25º (2) + 400sin25º (10) = 0
Ay = 160.83N 












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Statics: A mother would like to balance her son on the teetor totter



A mother would like to balance her son on the teetor totter. Calculate the distance "d" so that the system is in equilibrium. Also calculate the reaction at point O.

Solution

↻+ (clockwise positive)

↻+ΣMo = 0
-120lb (d) + 43lb (4ft) = 0
d = 1.433 ft

→+ΣFx = 0
Ox = 0

↑+ΣFy = 0
-120lb + Oy - 43lb = 0
Oy = 163 lb










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