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Mechanics engineering statics force analysis 2D

M
Mohammed8712

1) Mechanics is a branch of physics that deals with the states of rest or motion of bodies under the action of forces. 2) It can be divided into three branches - rigid body mechanics, deformable body mechanics, and fluid mechanics. 3) This document will focus on rigid body mechanics, which is essential for the design and analysis of many types of structural members, mechanical components, and electrical devices encountered in engineering.

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ENGINEERING MECHANICS STATCS For all Engineering Departments of the First Stage MOHAMMED KADHIM
TABLE OF CONTENTS UNITS CONVERT …………………………………………….…………….……………..., INTRODUCTION ABOUT MECHANICS …………………………….…………..……….. TYPES OF LOADS ……………………………………………………….…………………. FORCES ANALYSIS ……………………..………………………………...……………….. MOMENT ……………………………………………………………………………………. FORCES RESULTANT ……………..………………………………………………………..
UNITS CONVERT 1𝑁/𝑐𝑚2 = 0.01𝑁𝑚𝑚2 1𝑁/𝑚2 = 10−6 𝑁/𝑚𝑚2 1𝑁. 𝑚 = 101971.6𝑔𝑚𝑓. 𝑚 1𝑁/𝑐𝑚2 = 104 𝑁/𝑚2 1𝑁/𝑐𝑚2 = 104 𝑁/𝑚4 1𝑁. 𝑚 = 10197.16𝑔𝑚𝑓. 𝑐𝑚 1𝑁/𝑚𝑚2 = 106 𝑁/𝑚2 1𝑁. 𝑚 = 103 𝑁. 𝑚𝑚 1𝑁. 𝑚 = 0.73756 𝐼𝑏𝑓. 𝑓𝑡 1𝑀𝑁/𝑚2 = 0.1𝑘𝑁/𝑐𝑚2 1𝑁. 𝑚 = 100𝑁. 𝑐𝑚 1𝑁. 𝑚 = 8.851𝐼𝑏𝑓. 𝑖𝑛 1𝑘𝑁/𝑚2 = 102 𝑘𝑔𝑓/𝑚2 1𝑁. 𝑚 = 0.102𝑘𝑔𝑓. 𝑚 1𝑀𝑝𝑎 = 145.04𝐼𝑏𝑓/𝑖𝑛2 1𝑘𝑔/𝑚2 = 0.00981𝑘𝑁/𝑚2 1𝑁. 𝑚 = 0.001𝑘𝑁. 𝑚 1𝑀𝑝𝑎 = 0.001𝐺𝑝𝑎 1𝑁/𝑚2 = 10−6 𝑁/𝑚𝑚2 1𝑁. 𝑚 = 10.2𝑘𝑔𝑓. 𝑐𝑚 1𝑀𝑝𝑎 = 101.2𝑡𝑜𝑛/𝑚2 1𝑁/𝑚2 = 10−4 𝑁/𝑐𝑚2 1𝑁. 𝑚 = 102𝑘𝑔𝑓. 𝑚𝑚 1𝑁/𝑚2 = 1.0036𝑡𝑜𝑛/𝑐𝑚2 1𝑁/𝑚𝑚2 = 100 𝑁/𝑐𝑚2 1𝑁. 𝑚𝑚 = 0.001 𝑁. 𝑚 1𝑁. 𝑐𝑚 = 0.01 𝑁. 𝑚 1𝐾𝑔. 𝑚 = 9.81 𝑁. 𝑚 1𝑘𝑁. 𝑚 = 1000 𝑁. 𝑚 1𝑘𝑔. 𝑐𝑚 = 0.0981 𝑁. 𝑚 1𝑘𝑔. 𝑚𝑚 = 0.00981 𝑁. 𝑚 1𝐼𝑏𝑓. 𝑓𝑡 = 1.356 𝑁. 𝑚 1𝐼𝑏𝑓. 𝑖𝑛 = 0.0113 𝑁. 𝑚 1𝐼𝑏𝑓. 𝑖𝑛2 = 0.007 𝑀𝑝𝑎 1𝑀𝑝𝑎 = 1000 𝐾𝑝𝑎 1𝑃𝑎 = 10−4 𝑡𝑜𝑛/𝑚2 1𝐼𝑏𝑓 = 4.45 𝑁 1𝑁 = 0.225 𝐼𝑏𝑓 1𝑘𝑁 = 244.81 𝐼𝑏 1𝐼𝑏 = 0.000445 𝑘𝑁 1𝑘𝑁 = 1.0036 𝑡𝑜𝑛 1𝑡𝑜𝑛 = 9.964 𝑘𝑁 1𝑘𝑁 = 102 𝑘𝑔 1𝑘𝑔 = 0.00981 𝑘𝑁 1𝑘𝑁 = 0.22481 𝐾𝑖𝑝 1𝐾𝑖𝑝 = 4.45 𝑘𝑁 1𝑡𝑜𝑛 = 2204.62 𝐼𝑏 1𝐼𝑏 = 0.00045 𝑡𝑜𝑛
1𝑘𝑔 = 2.204 𝐼𝑏 1𝐼𝑏 = 0.4536 𝑘𝑔 1𝑔𝑚 = 0.0022 𝐼𝑏 1𝐼𝑏 = 453.6 𝑔𝑚 1𝐼𝑏 = 0.001𝐾𝑖𝑝 1𝐾𝑖𝑝 = 1000 𝐼𝑏 1𝑘𝑔 = 0.0022 𝐾𝑖𝑝 1𝐾𝑖𝑝 = 453.6 𝑘𝑔 1𝑡𝑜𝑛 = 2.24 𝐾𝑖𝑝 1𝐾𝑖𝑝 = 0.446 𝑡𝑜𝑛 1𝑘𝑔 = 0.0011 𝑡𝑜𝑛 1𝑡𝑜𝑛 = 907.2 𝑘𝑔 1𝑘𝑔 𝑚3 = 0.0011𝑡𝑜𝑛/𝑡𝑜𝑛3 1𝑡𝑜𝑛 𝑚3 = 907.2𝑘𝑔/𝑚3 1𝐾𝑝𝑎 = 10−6 𝐺𝑝𝑎 1𝐺𝑝𝑎 = 106 𝐾𝑝𝑎 1𝐾𝑝𝑎 = 103 𝑃𝑎 1𝑃𝑎 = 10−3 𝐾𝑝𝑎 1𝑓𝑡 = 12 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 0.0833 𝑓𝑡 1𝑓𝑡 = 0.3048 𝑚 1𝑚 = 3.281𝑓𝑡 1𝑚 = 39.37 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 0.0254 𝑚 1𝑐𝑚 = 0.03281 𝑓𝑡 1𝑓𝑡 = 30.48 𝑐𝑚 1𝑐𝑚 = 0.3937 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 2.54 𝑐𝑚 1𝑚𝑚 = 0.03937 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 25.4 𝑚𝑚 1𝑓𝑡 = 304.8 𝑚𝑚 1𝑚𝑚 = 0.003281 𝑓𝑡 1𝑘𝑁/𝑚𝑚2 = 106 𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 10−6 𝑘𝑁/𝑚𝑚2 1𝑁/𝑐𝑚2 = 10𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 10−2 𝑁/𝑐𝑚2 1𝑘𝑁/𝑚2 = 10−4 𝑘𝑁/𝑐𝑚2 1𝑘𝑁/𝑐𝑚2 = 104 𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 103 𝑁/𝑚2 1𝑁/𝑚2 = 10−3 𝑘𝑁/𝑚2 1𝑘𝑁/𝑚𝑚2 = 102 𝑘𝑁/𝑐𝑚2 1𝑘𝑁/𝑐𝑚2 = 10−2 𝑘𝑁/𝑚𝑚2 1𝑡𝑜𝑛/𝑚2 = 1016.05𝑘𝑔/𝑚2 1𝑘𝑔 𝑚2 = 0.00098𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛/𝑚2 = 9.81𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 0.102 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛/𝑚2 = 9.81𝐾𝑝𝑎 1𝐾𝑝𝑎 = 0.102 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛/𝑚2 = 0.0000098𝑘𝑁 /𝑚𝑚2 1𝑘𝑁 𝑚𝑚2 = 101971.62 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛 𝑚2 = 0.1𝐾𝑖𝑝/𝑐𝑚2
1𝐾𝑖𝑝 𝑐𝑚2 = 10 𝑡𝑜𝑛/𝑚2 1𝐾𝑠𝑖 = 6.895 𝑀𝑝𝑎 1𝑀𝑝𝑎 = 0.145 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 70306.96 𝑔𝑚/𝑐𝑚2 1𝑔𝑚 𝑐𝑚2 = 0.000014 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 703.07 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛 𝑚2 = 0.0014 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 6.895 𝑁/𝑚𝑚2 1𝑁 𝑚𝑚2 = 0.145 𝐾𝑠𝑖 1𝑘𝑔 𝑚2 = 0.0000014 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 703068.31𝑘𝑔/𝑚2 1𝑘𝑔 𝑚2 = 9.81 𝑃𝑎 1𝐾𝑝𝑎 = 102𝑘𝑔/𝑚2 1𝑃𝑎 = 0.102 𝑘𝑔/𝑚2 1𝑘𝑔 𝑚2 = 0.00981𝐾𝑝𝑎 1𝑘𝑁/𝑚2 = 102 𝑘𝑔/𝑚2 1𝑘𝑔/𝑚2 = 0.00981𝑘𝑁/𝑚2 1𝑘𝑁 𝑐𝑚2 = 1019.2𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛 𝑚2 = 0.000981𝑘𝑛/𝑐𝑚2
INTRODUCTION ABOUT 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. In general, this subject can be subdivided into three branches: Rigid-Body Mechanics, Deformable-Body Mechanics and Fluids Mechanics. We will study Rigid-Body Mechanics where this branch it is very essential for the design and analysis of many types of structural members, mechanical components, or electrical devices encountered in engineering. ‫الميكانيكا‬ ‫ل‬ ‫تخضع‬ ‫التي‬ ‫لألجسام‬ ‫الحركة‬ ‫أو‬ ‫السكون‬ ‫بحالة‬ ‫تهتم‬ ‫التي‬ ‫الفيزيائية‬ ‫العلوم‬ ‫فروع‬ ‫من‬ ‫فرع‬ ‫هي‬ ‫حركة‬ ‫بشكل‬ .‫القوى‬ :‫فروع‬ ‫ثالثة‬ ‫إلى‬ ‫الموضوع‬ ‫هذا‬ ‫تقسيم‬ ‫يمكن‬ ، ‫عام‬ ‫الصلب‬ ‫الجسم‬ ‫ميكانيكا‬ ‫و‬ ، ‫للتشوه‬ ‫القابل‬ ‫الجسم‬ ‫ميكانيكا‬ ‫و‬ ، ‫ميكاني‬ ‫الس‬ ‫كا‬ ‫وائل‬ . ‫ميكانيكا‬ ‫ندرس‬ ‫سوف‬ ‫الهيكلية‬ ‫األعضاء‬ ‫أنواع‬ ‫من‬ ‫العديد‬ ‫وتحليل‬ ‫لتصميم‬ ‫ًا‬‫د‬‫ج‬ ‫ًا‬‫ي‬‫ضرور‬ ‫الفرع‬ ‫هذا‬ ‫يعتبر‬ ‫حيث‬ ‫الصلب‬ ‫الجسم‬ ‫الهند‬ ‫في‬ ‫مواجهتها‬ ‫يتم‬ ‫التي‬ ‫الكهربائية‬ ‫األجهزة‬ ‫أو‬ ‫الميكانيكية‬ ‫المكونات‬ ‫أو‬ ‫س‬ .‫ة‬ Rigid-Body Mechanics is divided into two areas: Static and Dynamics. Statics deals with the equilibrium of bodies. Dynamics is concerned with the accelerated motion of bodies. We will study static because we can consider statics as a special case of dynamics in which the acceleration is zero; however, statics deserve separate treatment in engineering education since many objects are designed with intention that they remain in equilibrium. ‫الصلب‬ ‫الجسم‬ ‫ميكانيكا‬ :‫مجالين‬ ‫إلى‬ ‫مقسمة‬ ‫ثابت‬ ‫و‬ ‫ديناميكي‬ ‫توازن‬ ‫مع‬ ‫تتعامل‬ ‫اإلحصائيات‬ . ‫تهتم‬ ‫الديناميات‬ .‫األجسام‬ ‫بالحر‬ ‫لألجسام‬ ‫المتسارعة‬ ‫كة‬ ‫فيها‬ ‫يكون‬ ‫الديناميكيات‬ ‫من‬ ‫خاصة‬ ‫حالة‬ ‫اإلحصائيات‬ ‫نعتبر‬ ‫ألننا‬ ‫االستاتيكة‬ ‫ندرس‬ ‫سوف‬ . ‫الكائنات‬ ‫من‬ ‫العديد‬ ‫ألن‬ ‫ا‬ً‫نظر‬ ‫الهندسي‬ ‫التعليم‬ ‫في‬ ‫منفصلة‬ ‫معالجة‬ ‫تستحق‬ ‫اإلحصائيات‬ ‫فإن‬ ، ‫ذلك‬ ‫ومع‬ ‫؛‬ ‫ا‬ً‫صفر‬ ‫التسارع‬ ‫تظ‬ ‫أن‬ ‫بقصد‬ ‫مصممة‬ ‫توازن‬ ‫حالة‬ ‫في‬ ‫ل‬ .
TYPES OF LOADS Loads will be on two types; Concentrated Loads and Distributed Loads. Concentrated Loads: It is a concentrated load at one point and it can be three cases; Vertical or Horizontal or Diagonal. ‫األ‬ :‫المركزة‬ ‫حمال‬ ‫ه‬ ‫مر‬ ‫قوة‬ ‫عن‬ ‫عبارة‬ ‫ي‬ ‫في‬ ‫كزة‬ ‫نقطة‬ ‫ال‬ ‫هذه‬ ‫تكون‬ ‫ان‬ ‫الممكن‬ ‫ومن‬ ‫واحدة‬ ‫عمودي‬ ‫قوى‬ ‫افقي‬ ‫أو‬ ‫ة‬ ‫مائلة‬ ‫أو‬ ‫ة‬ . Distributed Loads: There are three types; Rectangular Distributed Loads, Tringle Distributed Loads and Trapezoidal Distributed Loads. Rectangular Distributed Loads: in this type must convert distributed load to concentrated load where the value of concentrated load will be the result multiplied distributed load by length it becomes a concentrated load in the center. ‫المنتشرة‬ ‫االحمال‬ ‫المستطيلة‬ : ‫ال‬ ‫يكون‬ ‫النوع‬ ‫هذا‬ ‫المنتشر‬ ‫الحمل‬ ‫تحويل‬ ‫يحب‬ ‫الحالة‬ ‫هذه‬ ‫وفي‬ ‫مستطيل‬ ‫شكل‬ ‫على‬ ‫المنتشر‬ ‫حمل‬ ‫مركز‬ ‫قوة‬ ‫الى‬ ‫طريق‬ ‫عن‬ ‫ويكون‬ ‫ة‬ ‫المنتش‬ ‫الحمل‬ ‫ضرب‬ ‫حاصل‬ ‫المرك‬ ‫القوة‬ ‫مكان‬ ‫ويكون‬ ‫الطول‬ ‫في‬ ‫ر‬ ‫ز‬ ‫في‬ ‫التحويل‬ ‫بعد‬ ‫ة‬ ‫المنتصف‬ .
Tringle Distributed Loads: in this type must convert distributed load to concentrated load where the equivalent load is the area of tringle and the position of the concentrated load will be one third of the distance at the existing angle and two thirds of the distance at the other angle and both sides. ‫ال‬ ‫االحمال‬ ‫المثلثة‬ ‫منتشرة‬ : ‫النوع‬ ‫هذا‬ ‫في‬ ‫االحما‬ ‫من‬ ‫م‬ ‫شكل‬ ‫على‬ ‫المنتشر‬ ‫الحمل‬ ‫يكون‬ ‫ل‬ ‫في‬ ‫القوة‬ ‫موقع‬ ‫يكون‬ ‫حيل‬ ‫ثلث‬ ‫ثلث‬ ‫ال‬ ‫الزاوية‬ ‫ق‬ ‫ريب‬ ‫ة‬ ‫ون‬ ‫القائمة‬ ‫الزاوية‬ ‫عن‬ ‫المركز‬ ‫القوة‬ ‫جد‬ ‫المن‬ ‫الحمل‬ ‫ضرب‬ ‫حاصل‬ ‫طريق‬ ‫عن‬ ‫ة‬ ‫ا‬ ‫في‬ ‫تشر‬ ‫لطول‬ ‫على‬ ‫مقسوم‬ 2 . Trapezoidal Distributed Loads: in this type must divided load to rectangular and triangle after that solve every shape.
‫المنتشرة‬ ‫االحمال‬ ‫منحرف‬ ‫شبه‬ : ‫في‬ ‫يكون‬ ‫النوع‬ ‫هذا‬ ‫شكلين‬ ‫على‬ ‫شكل‬ ‫على‬ ‫المنتشر‬ ‫الحمل‬ ‫الح‬ ‫هذه‬ ‫وفي‬ ‫ومثلث‬ ‫مستطيل‬ ‫الة‬ ‫حده‬ ‫على‬ ‫شكل‬ ‫كل‬ ‫يعامل‬ ‫ان‬ ‫يجب‬ ‫حي‬ ‫ا‬ ‫يكون‬ ‫المستطيل‬ ‫ث‬ ‫المر‬ ‫لحمل‬ ‫الو‬ ‫في‬ ‫كز‬ ‫و‬ ‫سط‬ ‫الثل‬ ‫في‬ ‫المثلث‬ ‫ال‬ ‫ث‬ ‫الزاوية‬ ‫عن‬ ‫قريب‬ ‫القا‬ .‫ئمة‬
FORCES ANALYSIS
Forces Analysis for Two Dimensions: forces are in three shapes: vertical forces, horizontal forces and diagonal forces when the forces vertical and horizontal they should not be analyzed but when the forces are diagonal, they will be analyzed into horizontal and vertical forces. ‫ثنائ‬ ‫القوة‬ ‫تحليل‬ :‫االبعاد‬ ‫ية‬ ‫على‬ ‫القوة‬ ‫تكون‬ ‫اشكال‬ ‫ثلث‬ ‫عمودي‬ ‫و‬ ‫ة‬ ‫القو‬ . ‫ومائلة‬ ‫أفقية‬ ‫الما‬ ‫ى‬ ‫ئل‬ ‫ة‬ ‫ال‬ ‫هي‬ ‫الى‬ ‫تحلل‬ ‫ان‬ ‫يجب‬ ‫تي‬ ‫وعمودية‬ ‫افقية‬ ‫مركبتين‬ ‫ولت‬ . ‫حليل‬ ‫أي‬ ‫قو‬ ‫توف‬ ‫يحب‬ ‫مائلة‬ ‫ى‬ ‫ر‬ ‫ام‬ ‫أو‬ ‫القوة‬ ‫هذه‬ ‫ميل‬ ‫ا‬ ‫زا‬ ‫وي‬ ‫القوة‬ ‫ميل‬ ‫ة‬ . sin ‫و‬ ‫في‬ ‫تضرب‬ ‫البعيدة‬ ‫المركبة‬ cos ‫مالحظة‬ : ‫المرك‬ ‫القريبة‬ ‫بة‬ ‫المجاو‬ ‫اي‬ ‫الزاوية‬ ‫من‬ ‫تضرب‬ ‫للزاوية‬ ‫رة‬ ‫في‬ :‫مالحظة‬ ‫المرك‬ ‫الميل‬ ‫اي‬ ‫االفقي‬ ‫الميل‬ ‫تأخذ‬ ‫األفقية‬ ‫بة‬ ‫الموازي‬ ‫ا‬ ‫العمودي‬ ‫الميل‬ ‫تأخذ‬ ‫العمودية‬ ‫والمركبة‬ ‫للمركبة‬ ‫الميل‬ ‫ي‬ ‫للمركبة‬ ‫الموازي‬
Orthogonal Analysis: when the force is perpendicular to the surface it has an inverse slope of the surface. ‫التح‬ :‫المتعامد‬ ‫ليل‬ ‫الم‬ ‫التحليل‬ ‫في‬ ‫تعامد‬ ‫يك‬ ‫الج‬ ‫ون‬ ‫مائ‬ ‫سم‬ ‫وال‬ ‫ل‬ ‫الحال‬ ‫هذه‬ ‫في‬ ‫عليه‬ ‫عمودية‬ ‫قوة‬ ‫فأن‬ ‫ة‬ ‫عن‬ ‫د‬ ‫فأن‬ ‫القوة‬ ‫تحليل‬ ‫المركبة‬ ‫العمودية‬ ‫للج‬ ‫سم‬ ‫م‬ ‫تأخذ‬ ‫الميل‬ ‫قلوب‬ ‫وال‬ ‫مرك‬ ‫الموازية‬ ‫بة‬ ‫ع‬ ‫تأخذ‬ ‫الجسم‬ ‫لى‬ ‫نف‬ ‫ا‬ ‫س‬ ‫لميل‬ . ‫وتكون‬ ‫عبا‬ ‫مركبة‬ ‫كل‬ ‫قيمة‬ ‫ر‬ ‫ة‬ ‫عن‬ ‫الموا‬ ‫الميل‬ ‫في‬ ‫القوة‬ ‫ضرب‬ ‫حاصل‬ ‫لها‬ ‫زي‬ . Non-Orthogonal Analysis: in this type we divided the force or component on the sin ø corresponding to force or component. Law Sin uses: if given (1) Measure two angles and the length of any member. (2) Length two members and measure the angle corresponding to one.
:‫متعامد‬ ‫الغير‬ ‫التحليل‬ ‫التح‬ ‫من‬ ‫النوع‬ ‫هذا‬ ‫في‬ ‫ليل‬ ‫ويت‬ ‫مركبات‬ ‫الى‬ ‫القوى‬ ‫تحلل‬ ‫ال‬ ‫قانون‬ ‫طريق‬ ‫عن‬ ‫ايجادها‬ ‫م‬ (Sin Law) ‫حالتين‬ ‫في‬ ‫الطريقة‬ ‫هذه‬ ‫الى‬ ‫ونلجأ‬ ‫األول‬ ‫الحالة‬ ‫ى‬ ‫ال‬ ‫في‬ ‫يعطي‬ ‫عندما‬ : ‫وضلع‬ ‫زاويتين‬ ‫سؤال‬ :‫الثانية‬ ‫الحالة‬ ‫الس‬ ‫في‬ ‫يعطي‬ ‫عندما‬ ‫ؤ‬ ‫ضلعي‬ ‫ال‬ ‫الضلعين‬ ‫ألحد‬ ‫مقابلة‬ ‫وزاوية‬ ‫ن‬
Examples: 1- Determine a pair of horizontal and vertical components of the 340 Ib force for fig (1)? Solution: 𝑥2 + 𝑦2 = 𝑟2 → 82 + 152 = 𝑟2 → 𝑟 = 17 Fx = 160 Ib , Fy = 300 Ib. 2- Determine a set of horizontal and vertical components of the 200 Ib force for fig (2)?
Solution: 32 + 32 = 𝑟2 → 𝑟 = 5 𝑭𝒙 = 𝟏𝟐𝟎 𝑰𝒃 , 𝑭𝒚 = 𝟏𝟔𝟎 𝑰𝒃 3- Resolve the 115 Ib force of fig into horizontal and vertical components for each of the following values of ø: a) 35°, b) 65°, c) 145°. Solution: a) 𝐹𝑥 = 115 ∗ sin 35 = 65.96 𝐼𝑏 → 𝐹𝑦 = 115 ∗ cos 35 = 94.2 𝐼𝑏 ↓ b) 𝐹𝑥 = 115 ∗ sin 65 = 104.22 𝐼𝑏 → 𝐹𝑦 = 115 ∗ cos 65 = 48.6 𝐼𝑏 ↓ c) 𝐹𝑥 = 115 ∗ sin 145 = 65.96 𝐼𝑏 → 𝐹𝑦 = 115 ∗ cos 145 = −94.2 = 94.2 𝐼𝑏 ↑
4- Resolve the 600 Ib force of fig into components: a shearing component parallel to AB and a normal component perpendicular to AB. Solution: 5- Resolve the 130 Ib force of fig into two nonrectangular components, one having a line of action along AB and the other parallel to CD ?
Solution: 130 𝑆𝑖𝑛 36.9 = 𝐹(𝐴𝐵) 𝑆𝑖𝑛 75.7 → 𝐹(𝐴𝐵) = 209.81 𝐼𝑏 130 𝑆𝑖𝑛 36.9 = 𝐹(𝐶𝐷) 𝑆𝑖𝑛 67.4 = 𝐹(𝐶𝐷) = 200 𝐼𝑏 6- Resolve the 215 Ib force of fig into two components, one having a line of action along AB and the other parallel let to CD? Solution:
215 𝑆𝑖𝑛 75.7 = 𝐴𝐵 𝑆𝑖𝑛 39.4 → 𝐴𝐵 = 149.83 𝐼𝑏 215 𝑆𝑖𝑛 75.7 = 𝐶𝐷 𝑆𝑖𝑛 64.9 → 𝐶𝐷 = 200.92 𝐼𝑏 7- The 305 Ib force of fig act on the box B . Resolve the force into two components one along AC and the other parallel to CD? Solution: 305 𝑆𝑖𝑛 78.2 = 𝐴𝑐 𝑆𝑖𝑛 64.97 → 𝐴𝐶 = 282 .23 𝐼𝑏 305 𝑆𝑖𝑛 78.2 = 𝐶𝐷 𝑆𝑖𝑛 36.9 → 𝐶𝐷 = 187.1 𝐼𝑏
Mechanics engineering statics force analysis 2D

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Mechanics engineering statics force analysis 2D

  • 1. ENGINEERING MECHANICS STATCS For all Engineering Departments of the First Stage MOHAMMED KADHIM
  • 2. TABLE OF CONTENTS UNITS CONVERT …………………………………………….…………….……………..., INTRODUCTION ABOUT MECHANICS …………………………….…………..……….. TYPES OF LOADS ……………………………………………………….…………………. FORCES ANALYSIS ……………………..………………………………...……………….. MOMENT ……………………………………………………………………………………. FORCES RESULTANT ……………..………………………………………………………..
  • 3. UNITS CONVERT 1𝑁/𝑐𝑚2 = 0.01𝑁𝑚𝑚2 1𝑁/𝑚2 = 10−6 𝑁/𝑚𝑚2 1𝑁. 𝑚 = 101971.6𝑔𝑚𝑓. 𝑚 1𝑁/𝑐𝑚2 = 104 𝑁/𝑚2 1𝑁/𝑐𝑚2 = 104 𝑁/𝑚4 1𝑁. 𝑚 = 10197.16𝑔𝑚𝑓. 𝑐𝑚 1𝑁/𝑚𝑚2 = 106 𝑁/𝑚2 1𝑁. 𝑚 = 103 𝑁. 𝑚𝑚 1𝑁. 𝑚 = 0.73756 𝐼𝑏𝑓. 𝑓𝑡 1𝑀𝑁/𝑚2 = 0.1𝑘𝑁/𝑐𝑚2 1𝑁. 𝑚 = 100𝑁. 𝑐𝑚 1𝑁. 𝑚 = 8.851𝐼𝑏𝑓. 𝑖𝑛 1𝑘𝑁/𝑚2 = 102 𝑘𝑔𝑓/𝑚2 1𝑁. 𝑚 = 0.102𝑘𝑔𝑓. 𝑚 1𝑀𝑝𝑎 = 145.04𝐼𝑏𝑓/𝑖𝑛2 1𝑘𝑔/𝑚2 = 0.00981𝑘𝑁/𝑚2 1𝑁. 𝑚 = 0.001𝑘𝑁. 𝑚 1𝑀𝑝𝑎 = 0.001𝐺𝑝𝑎 1𝑁/𝑚2 = 10−6 𝑁/𝑚𝑚2 1𝑁. 𝑚 = 10.2𝑘𝑔𝑓. 𝑐𝑚 1𝑀𝑝𝑎 = 101.2𝑡𝑜𝑛/𝑚2 1𝑁/𝑚2 = 10−4 𝑁/𝑐𝑚2 1𝑁. 𝑚 = 102𝑘𝑔𝑓. 𝑚𝑚 1𝑁/𝑚2 = 1.0036𝑡𝑜𝑛/𝑐𝑚2 1𝑁/𝑚𝑚2 = 100 𝑁/𝑐𝑚2 1𝑁. 𝑚𝑚 = 0.001 𝑁. 𝑚 1𝑁. 𝑐𝑚 = 0.01 𝑁. 𝑚 1𝐾𝑔. 𝑚 = 9.81 𝑁. 𝑚 1𝑘𝑁. 𝑚 = 1000 𝑁. 𝑚 1𝑘𝑔. 𝑐𝑚 = 0.0981 𝑁. 𝑚 1𝑘𝑔. 𝑚𝑚 = 0.00981 𝑁. 𝑚 1𝐼𝑏𝑓. 𝑓𝑡 = 1.356 𝑁. 𝑚 1𝐼𝑏𝑓. 𝑖𝑛 = 0.0113 𝑁. 𝑚 1𝐼𝑏𝑓. 𝑖𝑛2 = 0.007 𝑀𝑝𝑎 1𝑀𝑝𝑎 = 1000 𝐾𝑝𝑎 1𝑃𝑎 = 10−4 𝑡𝑜𝑛/𝑚2 1𝐼𝑏𝑓 = 4.45 𝑁 1𝑁 = 0.225 𝐼𝑏𝑓 1𝑘𝑁 = 244.81 𝐼𝑏 1𝐼𝑏 = 0.000445 𝑘𝑁 1𝑘𝑁 = 1.0036 𝑡𝑜𝑛 1𝑡𝑜𝑛 = 9.964 𝑘𝑁 1𝑘𝑁 = 102 𝑘𝑔 1𝑘𝑔 = 0.00981 𝑘𝑁 1𝑘𝑁 = 0.22481 𝐾𝑖𝑝 1𝐾𝑖𝑝 = 4.45 𝑘𝑁 1𝑡𝑜𝑛 = 2204.62 𝐼𝑏 1𝐼𝑏 = 0.00045 𝑡𝑜𝑛
  • 4. 1𝑘𝑔 = 2.204 𝐼𝑏 1𝐼𝑏 = 0.4536 𝑘𝑔 1𝑔𝑚 = 0.0022 𝐼𝑏 1𝐼𝑏 = 453.6 𝑔𝑚 1𝐼𝑏 = 0.001𝐾𝑖𝑝 1𝐾𝑖𝑝 = 1000 𝐼𝑏 1𝑘𝑔 = 0.0022 𝐾𝑖𝑝 1𝐾𝑖𝑝 = 453.6 𝑘𝑔 1𝑡𝑜𝑛 = 2.24 𝐾𝑖𝑝 1𝐾𝑖𝑝 = 0.446 𝑡𝑜𝑛 1𝑘𝑔 = 0.0011 𝑡𝑜𝑛 1𝑡𝑜𝑛 = 907.2 𝑘𝑔 1𝑘𝑔 𝑚3 = 0.0011𝑡𝑜𝑛/𝑡𝑜𝑛3 1𝑡𝑜𝑛 𝑚3 = 907.2𝑘𝑔/𝑚3 1𝐾𝑝𝑎 = 10−6 𝐺𝑝𝑎 1𝐺𝑝𝑎 = 106 𝐾𝑝𝑎 1𝐾𝑝𝑎 = 103 𝑃𝑎 1𝑃𝑎 = 10−3 𝐾𝑝𝑎 1𝑓𝑡 = 12 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 0.0833 𝑓𝑡 1𝑓𝑡 = 0.3048 𝑚 1𝑚 = 3.281𝑓𝑡 1𝑚 = 39.37 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 0.0254 𝑚 1𝑐𝑚 = 0.03281 𝑓𝑡 1𝑓𝑡 = 30.48 𝑐𝑚 1𝑐𝑚 = 0.3937 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 2.54 𝑐𝑚 1𝑚𝑚 = 0.03937 𝑖𝑛𝑐ℎ 1𝑖𝑛𝑐ℎ = 25.4 𝑚𝑚 1𝑓𝑡 = 304.8 𝑚𝑚 1𝑚𝑚 = 0.003281 𝑓𝑡 1𝑘𝑁/𝑚𝑚2 = 106 𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 10−6 𝑘𝑁/𝑚𝑚2 1𝑁/𝑐𝑚2 = 10𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 10−2 𝑁/𝑐𝑚2 1𝑘𝑁/𝑚2 = 10−4 𝑘𝑁/𝑐𝑚2 1𝑘𝑁/𝑐𝑚2 = 104 𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 103 𝑁/𝑚2 1𝑁/𝑚2 = 10−3 𝑘𝑁/𝑚2 1𝑘𝑁/𝑚𝑚2 = 102 𝑘𝑁/𝑐𝑚2 1𝑘𝑁/𝑐𝑚2 = 10−2 𝑘𝑁/𝑚𝑚2 1𝑡𝑜𝑛/𝑚2 = 1016.05𝑘𝑔/𝑚2 1𝑘𝑔 𝑚2 = 0.00098𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛/𝑚2 = 9.81𝑘𝑁/𝑚2 1𝑘𝑁/𝑚2 = 0.102 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛/𝑚2 = 9.81𝐾𝑝𝑎 1𝐾𝑝𝑎 = 0.102 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛/𝑚2 = 0.0000098𝑘𝑁 /𝑚𝑚2 1𝑘𝑁 𝑚𝑚2 = 101971.62 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛 𝑚2 = 0.1𝐾𝑖𝑝/𝑐𝑚2
  • 5. 1𝐾𝑖𝑝 𝑐𝑚2 = 10 𝑡𝑜𝑛/𝑚2 1𝐾𝑠𝑖 = 6.895 𝑀𝑝𝑎 1𝑀𝑝𝑎 = 0.145 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 70306.96 𝑔𝑚/𝑐𝑚2 1𝑔𝑚 𝑐𝑚2 = 0.000014 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 703.07 𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛 𝑚2 = 0.0014 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 6.895 𝑁/𝑚𝑚2 1𝑁 𝑚𝑚2 = 0.145 𝐾𝑠𝑖 1𝑘𝑔 𝑚2 = 0.0000014 𝐾𝑠𝑖 1𝐾𝑠𝑖 = 703068.31𝑘𝑔/𝑚2 1𝑘𝑔 𝑚2 = 9.81 𝑃𝑎 1𝐾𝑝𝑎 = 102𝑘𝑔/𝑚2 1𝑃𝑎 = 0.102 𝑘𝑔/𝑚2 1𝑘𝑔 𝑚2 = 0.00981𝐾𝑝𝑎 1𝑘𝑁/𝑚2 = 102 𝑘𝑔/𝑚2 1𝑘𝑔/𝑚2 = 0.00981𝑘𝑁/𝑚2 1𝑘𝑁 𝑐𝑚2 = 1019.2𝑡𝑜𝑛/𝑚2 1𝑡𝑜𝑛 𝑚2 = 0.000981𝑘𝑛/𝑐𝑚2
  • 6. INTRODUCTION ABOUT 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. In general, this subject can be subdivided into three branches: Rigid-Body Mechanics, Deformable-Body Mechanics and Fluids Mechanics. We will study Rigid-Body Mechanics where this branch it is very essential for the design and analysis of many types of structural members, mechanical components, or electrical devices encountered in engineering. ‫الميكانيكا‬ ‫ل‬ ‫تخضع‬ ‫التي‬ ‫لألجسام‬ ‫الحركة‬ ‫أو‬ ‫السكون‬ ‫بحالة‬ ‫تهتم‬ ‫التي‬ ‫الفيزيائية‬ ‫العلوم‬ ‫فروع‬ ‫من‬ ‫فرع‬ ‫هي‬ ‫حركة‬ ‫بشكل‬ .‫القوى‬ :‫فروع‬ ‫ثالثة‬ ‫إلى‬ ‫الموضوع‬ ‫هذا‬ ‫تقسيم‬ ‫يمكن‬ ، ‫عام‬ ‫الصلب‬ ‫الجسم‬ ‫ميكانيكا‬ ‫و‬ ، ‫للتشوه‬ ‫القابل‬ ‫الجسم‬ ‫ميكانيكا‬ ‫و‬ ، ‫ميكاني‬ ‫الس‬ ‫كا‬ ‫وائل‬ . ‫ميكانيكا‬ ‫ندرس‬ ‫سوف‬ ‫الهيكلية‬ ‫األعضاء‬ ‫أنواع‬ ‫من‬ ‫العديد‬ ‫وتحليل‬ ‫لتصميم‬ ‫ًا‬‫د‬‫ج‬ ‫ًا‬‫ي‬‫ضرور‬ ‫الفرع‬ ‫هذا‬ ‫يعتبر‬ ‫حيث‬ ‫الصلب‬ ‫الجسم‬ ‫الهند‬ ‫في‬ ‫مواجهتها‬ ‫يتم‬ ‫التي‬ ‫الكهربائية‬ ‫األجهزة‬ ‫أو‬ ‫الميكانيكية‬ ‫المكونات‬ ‫أو‬ ‫س‬ .‫ة‬ Rigid-Body Mechanics is divided into two areas: Static and Dynamics. Statics deals with the equilibrium of bodies. Dynamics is concerned with the accelerated motion of bodies. We will study static because we can consider statics as a special case of dynamics in which the acceleration is zero; however, statics deserve separate treatment in engineering education since many objects are designed with intention that they remain in equilibrium. ‫الصلب‬ ‫الجسم‬ ‫ميكانيكا‬ :‫مجالين‬ ‫إلى‬ ‫مقسمة‬ ‫ثابت‬ ‫و‬ ‫ديناميكي‬ ‫توازن‬ ‫مع‬ ‫تتعامل‬ ‫اإلحصائيات‬ . ‫تهتم‬ ‫الديناميات‬ .‫األجسام‬ ‫بالحر‬ ‫لألجسام‬ ‫المتسارعة‬ ‫كة‬ ‫فيها‬ ‫يكون‬ ‫الديناميكيات‬ ‫من‬ ‫خاصة‬ ‫حالة‬ ‫اإلحصائيات‬ ‫نعتبر‬ ‫ألننا‬ ‫االستاتيكة‬ ‫ندرس‬ ‫سوف‬ . ‫الكائنات‬ ‫من‬ ‫العديد‬ ‫ألن‬ ‫ا‬ً‫نظر‬ ‫الهندسي‬ ‫التعليم‬ ‫في‬ ‫منفصلة‬ ‫معالجة‬ ‫تستحق‬ ‫اإلحصائيات‬ ‫فإن‬ ، ‫ذلك‬ ‫ومع‬ ‫؛‬ ‫ا‬ً‫صفر‬ ‫التسارع‬ ‫تظ‬ ‫أن‬ ‫بقصد‬ ‫مصممة‬ ‫توازن‬ ‫حالة‬ ‫في‬ ‫ل‬ .
  • 7. TYPES OF LOADS Loads will be on two types; Concentrated Loads and Distributed Loads. Concentrated Loads: It is a concentrated load at one point and it can be three cases; Vertical or Horizontal or Diagonal. ‫األ‬ :‫المركزة‬ ‫حمال‬ ‫ه‬ ‫مر‬ ‫قوة‬ ‫عن‬ ‫عبارة‬ ‫ي‬ ‫في‬ ‫كزة‬ ‫نقطة‬ ‫ال‬ ‫هذه‬ ‫تكون‬ ‫ان‬ ‫الممكن‬ ‫ومن‬ ‫واحدة‬ ‫عمودي‬ ‫قوى‬ ‫افقي‬ ‫أو‬ ‫ة‬ ‫مائلة‬ ‫أو‬ ‫ة‬ . Distributed Loads: There are three types; Rectangular Distributed Loads, Tringle Distributed Loads and Trapezoidal Distributed Loads. Rectangular Distributed Loads: in this type must convert distributed load to concentrated load where the value of concentrated load will be the result multiplied distributed load by length it becomes a concentrated load in the center. ‫المنتشرة‬ ‫االحمال‬ ‫المستطيلة‬ : ‫ال‬ ‫يكون‬ ‫النوع‬ ‫هذا‬ ‫المنتشر‬ ‫الحمل‬ ‫تحويل‬ ‫يحب‬ ‫الحالة‬ ‫هذه‬ ‫وفي‬ ‫مستطيل‬ ‫شكل‬ ‫على‬ ‫المنتشر‬ ‫حمل‬ ‫مركز‬ ‫قوة‬ ‫الى‬ ‫طريق‬ ‫عن‬ ‫ويكون‬ ‫ة‬ ‫المنتش‬ ‫الحمل‬ ‫ضرب‬ ‫حاصل‬ ‫المرك‬ ‫القوة‬ ‫مكان‬ ‫ويكون‬ ‫الطول‬ ‫في‬ ‫ر‬ ‫ز‬ ‫في‬ ‫التحويل‬ ‫بعد‬ ‫ة‬ ‫المنتصف‬ .
  • 8. Tringle Distributed Loads: in this type must convert distributed load to concentrated load where the equivalent load is the area of tringle and the position of the concentrated load will be one third of the distance at the existing angle and two thirds of the distance at the other angle and both sides. ‫ال‬ ‫االحمال‬ ‫المثلثة‬ ‫منتشرة‬ : ‫النوع‬ ‫هذا‬ ‫في‬ ‫االحما‬ ‫من‬ ‫م‬ ‫شكل‬ ‫على‬ ‫المنتشر‬ ‫الحمل‬ ‫يكون‬ ‫ل‬ ‫في‬ ‫القوة‬ ‫موقع‬ ‫يكون‬ ‫حيل‬ ‫ثلث‬ ‫ثلث‬ ‫ال‬ ‫الزاوية‬ ‫ق‬ ‫ريب‬ ‫ة‬ ‫ون‬ ‫القائمة‬ ‫الزاوية‬ ‫عن‬ ‫المركز‬ ‫القوة‬ ‫جد‬ ‫المن‬ ‫الحمل‬ ‫ضرب‬ ‫حاصل‬ ‫طريق‬ ‫عن‬ ‫ة‬ ‫ا‬ ‫في‬ ‫تشر‬ ‫لطول‬ ‫على‬ ‫مقسوم‬ 2 . Trapezoidal Distributed Loads: in this type must divided load to rectangular and triangle after that solve every shape.
  • 9. ‫المنتشرة‬ ‫االحمال‬ ‫منحرف‬ ‫شبه‬ : ‫في‬ ‫يكون‬ ‫النوع‬ ‫هذا‬ ‫شكلين‬ ‫على‬ ‫شكل‬ ‫على‬ ‫المنتشر‬ ‫الحمل‬ ‫الح‬ ‫هذه‬ ‫وفي‬ ‫ومثلث‬ ‫مستطيل‬ ‫الة‬ ‫حده‬ ‫على‬ ‫شكل‬ ‫كل‬ ‫يعامل‬ ‫ان‬ ‫يجب‬ ‫حي‬ ‫ا‬ ‫يكون‬ ‫المستطيل‬ ‫ث‬ ‫المر‬ ‫لحمل‬ ‫الو‬ ‫في‬ ‫كز‬ ‫و‬ ‫سط‬ ‫الثل‬ ‫في‬ ‫المثلث‬ ‫ال‬ ‫ث‬ ‫الزاوية‬ ‫عن‬ ‫قريب‬ ‫القا‬ .‫ئمة‬
  • 11. Forces Analysis for Two Dimensions: forces are in three shapes: vertical forces, horizontal forces and diagonal forces when the forces vertical and horizontal they should not be analyzed but when the forces are diagonal, they will be analyzed into horizontal and vertical forces. ‫ثنائ‬ ‫القوة‬ ‫تحليل‬ :‫االبعاد‬ ‫ية‬ ‫على‬ ‫القوة‬ ‫تكون‬ ‫اشكال‬ ‫ثلث‬ ‫عمودي‬ ‫و‬ ‫ة‬ ‫القو‬ . ‫ومائلة‬ ‫أفقية‬ ‫الما‬ ‫ى‬ ‫ئل‬ ‫ة‬ ‫ال‬ ‫هي‬ ‫الى‬ ‫تحلل‬ ‫ان‬ ‫يجب‬ ‫تي‬ ‫وعمودية‬ ‫افقية‬ ‫مركبتين‬ ‫ولت‬ . ‫حليل‬ ‫أي‬ ‫قو‬ ‫توف‬ ‫يحب‬ ‫مائلة‬ ‫ى‬ ‫ر‬ ‫ام‬ ‫أو‬ ‫القوة‬ ‫هذه‬ ‫ميل‬ ‫ا‬ ‫زا‬ ‫وي‬ ‫القوة‬ ‫ميل‬ ‫ة‬ . sin ‫و‬ ‫في‬ ‫تضرب‬ ‫البعيدة‬ ‫المركبة‬ cos ‫مالحظة‬ : ‫المرك‬ ‫القريبة‬ ‫بة‬ ‫المجاو‬ ‫اي‬ ‫الزاوية‬ ‫من‬ ‫تضرب‬ ‫للزاوية‬ ‫رة‬ ‫في‬ :‫مالحظة‬ ‫المرك‬ ‫الميل‬ ‫اي‬ ‫االفقي‬ ‫الميل‬ ‫تأخذ‬ ‫األفقية‬ ‫بة‬ ‫الموازي‬ ‫ا‬ ‫العمودي‬ ‫الميل‬ ‫تأخذ‬ ‫العمودية‬ ‫والمركبة‬ ‫للمركبة‬ ‫الميل‬ ‫ي‬ ‫للمركبة‬ ‫الموازي‬
  • 12. Orthogonal Analysis: when the force is perpendicular to the surface it has an inverse slope of the surface. ‫التح‬ :‫المتعامد‬ ‫ليل‬ ‫الم‬ ‫التحليل‬ ‫في‬ ‫تعامد‬ ‫يك‬ ‫الج‬ ‫ون‬ ‫مائ‬ ‫سم‬ ‫وال‬ ‫ل‬ ‫الحال‬ ‫هذه‬ ‫في‬ ‫عليه‬ ‫عمودية‬ ‫قوة‬ ‫فأن‬ ‫ة‬ ‫عن‬ ‫د‬ ‫فأن‬ ‫القوة‬ ‫تحليل‬ ‫المركبة‬ ‫العمودية‬ ‫للج‬ ‫سم‬ ‫م‬ ‫تأخذ‬ ‫الميل‬ ‫قلوب‬ ‫وال‬ ‫مرك‬ ‫الموازية‬ ‫بة‬ ‫ع‬ ‫تأخذ‬ ‫الجسم‬ ‫لى‬ ‫نف‬ ‫ا‬ ‫س‬ ‫لميل‬ . ‫وتكون‬ ‫عبا‬ ‫مركبة‬ ‫كل‬ ‫قيمة‬ ‫ر‬ ‫ة‬ ‫عن‬ ‫الموا‬ ‫الميل‬ ‫في‬ ‫القوة‬ ‫ضرب‬ ‫حاصل‬ ‫لها‬ ‫زي‬ . Non-Orthogonal Analysis: in this type we divided the force or component on the sin ø corresponding to force or component. Law Sin uses: if given (1) Measure two angles and the length of any member. (2) Length two members and measure the angle corresponding to one.
  • 13. :‫متعامد‬ ‫الغير‬ ‫التحليل‬ ‫التح‬ ‫من‬ ‫النوع‬ ‫هذا‬ ‫في‬ ‫ليل‬ ‫ويت‬ ‫مركبات‬ ‫الى‬ ‫القوى‬ ‫تحلل‬ ‫ال‬ ‫قانون‬ ‫طريق‬ ‫عن‬ ‫ايجادها‬ ‫م‬ (Sin Law) ‫حالتين‬ ‫في‬ ‫الطريقة‬ ‫هذه‬ ‫الى‬ ‫ونلجأ‬ ‫األول‬ ‫الحالة‬ ‫ى‬ ‫ال‬ ‫في‬ ‫يعطي‬ ‫عندما‬ : ‫وضلع‬ ‫زاويتين‬ ‫سؤال‬ :‫الثانية‬ ‫الحالة‬ ‫الس‬ ‫في‬ ‫يعطي‬ ‫عندما‬ ‫ؤ‬ ‫ضلعي‬ ‫ال‬ ‫الضلعين‬ ‫ألحد‬ ‫مقابلة‬ ‫وزاوية‬ ‫ن‬
  • 14. Examples: 1- Determine a pair of horizontal and vertical components of the 340 Ib force for fig (1)? Solution: 𝑥2 + 𝑦2 = 𝑟2 → 82 + 152 = 𝑟2 → 𝑟 = 17 Fx = 160 Ib , Fy = 300 Ib. 2- Determine a set of horizontal and vertical components of the 200 Ib force for fig (2)?
  • 15. Solution: 32 + 32 = 𝑟2 → 𝑟 = 5 𝑭𝒙 = 𝟏𝟐𝟎 𝑰𝒃 , 𝑭𝒚 = 𝟏𝟔𝟎 𝑰𝒃 3- Resolve the 115 Ib force of fig into horizontal and vertical components for each of the following values of ø: a) 35°, b) 65°, c) 145°. Solution: a) 𝐹𝑥 = 115 ∗ sin 35 = 65.96 𝐼𝑏 → 𝐹𝑦 = 115 ∗ cos 35 = 94.2 𝐼𝑏 ↓ b) 𝐹𝑥 = 115 ∗ sin 65 = 104.22 𝐼𝑏 → 𝐹𝑦 = 115 ∗ cos 65 = 48.6 𝐼𝑏 ↓ c) 𝐹𝑥 = 115 ∗ sin 145 = 65.96 𝐼𝑏 → 𝐹𝑦 = 115 ∗ cos 145 = −94.2 = 94.2 𝐼𝑏 ↑
  • 16. 4- Resolve the 600 Ib force of fig into components: a shearing component parallel to AB and a normal component perpendicular to AB. Solution: 5- Resolve the 130 Ib force of fig into two nonrectangular components, one having a line of action along AB and the other parallel to CD ?
  • 17. Solution: 130 𝑆𝑖𝑛 36.9 = 𝐹(𝐴𝐵) 𝑆𝑖𝑛 75.7 → 𝐹(𝐴𝐵) = 209.81 𝐼𝑏 130 𝑆𝑖𝑛 36.9 = 𝐹(𝐶𝐷) 𝑆𝑖𝑛 67.4 = 𝐹(𝐶𝐷) = 200 𝐼𝑏 6- Resolve the 215 Ib force of fig into two components, one having a line of action along AB and the other parallel let to CD? Solution:
  • 18. 215 𝑆𝑖𝑛 75.7 = 𝐴𝐵 𝑆𝑖𝑛 39.4 → 𝐴𝐵 = 149.83 𝐼𝑏 215 𝑆𝑖𝑛 75.7 = 𝐶𝐷 𝑆𝑖𝑛 64.9 → 𝐶𝐷 = 200.92 𝐼𝑏 7- The 305 Ib force of fig act on the box B . Resolve the force into two components one along AC and the other parallel to CD? Solution: 305 𝑆𝑖𝑛 78.2 = 𝐴𝑐 𝑆𝑖𝑛 64.97 → 𝐴𝐶 = 282 .23 𝐼𝑏 305 𝑆𝑖𝑛 78.2 = 𝐶𝐷 𝑆𝑖𝑛 36.9 → 𝐶𝐷 = 187.1 𝐼𝑏