The document discusses moments (torques) of forces, moments of couples, and equivalent force-couple systems. It provides:
1) Definitions of the moment of a force as the tendency for rotation about a point, and the moment of a couple as two parallel forces with equal magnitude acting in opposite directions separated by a perpendicular distance.
2) Examples calculating moments of forces about a point using scalar analysis and resolving forces into components.
3) Discussion of replacing multiple forces and couples with an equivalent single force and couple moment at a point, in order to more easily understand their overall effect.
Three key concepts are discussed in the document:
1) Mechanics deals with the static and dynamic behavior of bodies under the influence of forces or torques. This includes rigid bodies, deformable bodies, and fluids.
2) A free body diagram shows all external forces acting on a particle or rigid body and is essential for writing equations of equilibrium.
3) The equilibrium of a particle in 2D involves applying equations that set the sum of forces in the x and y directions equal to zero to solve for unknown forces or angles.
Usaha didefinisikan sebagai hasil kali antara besarnya gaya yang diberikan pada benda dengan besar perpindahan benda tersebut. Daya adalah perubahan energi potensial atau energi kinetik tiap satu satuan waktu, dan didefinisikan sebagai usaha yang dilakukan tiap satuan waktu. Semakin besar daya yang dimiliki suatu benda, semakin besar pula kemampuannya untuk mengubah bentuk energi.
1. Dokumen membahas tentang momentum dan impuls serta berbagai konsep terkait seperti tumbukan, kekekalan momentum, dan hukum Newton kedua.
2. Definisi momentum secara fisika dan matematika dijelaskan, serta contoh soal perhitungan momentum.
3. Jenis-jenis tumbukan diuraikan, termasuk tumbukan lenting sempurna, sebagian, dan tidak lenting.
Radiasi matahari merupakan sumber energi utama untuk proses atmosfer dan iklim bumi. Radiasi matahari berasal dari fusi nuklir di matahari dan berbentuk gelombang elektromagnetik. Berbagai faktor seperti jarak bumi-matahari, sudut datang sinar matahari, dan keberadaan gas rumah kaca mempengaruhi penerimaan dan distribusi radiasi matahari di permukaan bumi.
Teks tersebut membahas tentang usaha, energi, dan daya. Secara singkat, usaha adalah hasil perkalian gaya dan perpindahan, energi adalah kemampuan untuk melakukan usaha, dan daya adalah laju perubahan energi terhadap waktu. Teks tersebut juga membedakan antara gaya konservatif dan non-konservatif berdasarkan ketergantungan usahanya terhadap lintasan.
Three key concepts are discussed in the document:
1) Mechanics deals with the static and dynamic behavior of bodies under the influence of forces or torques. This includes rigid bodies, deformable bodies, and fluids.
2) A free body diagram shows all external forces acting on a particle or rigid body and is essential for writing equations of equilibrium.
3) The equilibrium of a particle in 2D involves applying equations that set the sum of forces in the x and y directions equal to zero to solve for unknown forces or angles.
Usaha didefinisikan sebagai hasil kali antara besarnya gaya yang diberikan pada benda dengan besar perpindahan benda tersebut. Daya adalah perubahan energi potensial atau energi kinetik tiap satu satuan waktu, dan didefinisikan sebagai usaha yang dilakukan tiap satuan waktu. Semakin besar daya yang dimiliki suatu benda, semakin besar pula kemampuannya untuk mengubah bentuk energi.
1. Dokumen membahas tentang momentum dan impuls serta berbagai konsep terkait seperti tumbukan, kekekalan momentum, dan hukum Newton kedua.
2. Definisi momentum secara fisika dan matematika dijelaskan, serta contoh soal perhitungan momentum.
3. Jenis-jenis tumbukan diuraikan, termasuk tumbukan lenting sempurna, sebagian, dan tidak lenting.
Radiasi matahari merupakan sumber energi utama untuk proses atmosfer dan iklim bumi. Radiasi matahari berasal dari fusi nuklir di matahari dan berbentuk gelombang elektromagnetik. Berbagai faktor seperti jarak bumi-matahari, sudut datang sinar matahari, dan keberadaan gas rumah kaca mempengaruhi penerimaan dan distribusi radiasi matahari di permukaan bumi.
Teks tersebut membahas tentang usaha, energi, dan daya. Secara singkat, usaha adalah hasil perkalian gaya dan perpindahan, energi adalah kemampuan untuk melakukan usaha, dan daya adalah laju perubahan energi terhadap waktu. Teks tersebut juga membedakan antara gaya konservatif dan non-konservatif berdasarkan ketergantungan usahanya terhadap lintasan.
Praktikum mengukur intensitas radiasi surya dan lama penyinaran menggunakan alat Campbell Stokes. Dihitung panjang bakar kertas pias selama dua interval waktu untuk menentukan lama penyinaran harian dan rata-rata intensitasnya. Hasilnya adalah lama penyinaran harian 17,5 cm/jam.
Kinematika mempelajari gerak benda tanpa mempertimbangkan penyebabnya. Studi ini meliputi konsep lintasan, kecepatan, percepatan, dan jenis-jenis gerak seperti gerak lurus dan parabola.
Momen gaya adalah besaran yang menunjukkan kecenderungan suatu benda untuk berputar akibat gaya yang diberikan. Momen gaya didefinisikan sebagai hasil kali antara besar gaya dengan lengan momen, yang merupakan jarak tegak lurus antara sumbu putaran dan garis vektor gaya. Momen gaya dapat dihitung menggunakan rumus skalar maupun vektor, dan resultan momen adalah jumlah aljabar dari semua momen yang beker
Modul ini membahas tentang keseimbangan benda terapung berdasarkan hukum Archimedes. Hukum ini menyatakan bahwa benda yang terendam akan mengalami gaya apung sebesar berat cairan yang dipindahkannya. Benda akan mengapung jika beratnya lebih kecil dari gaya apung, tenggelam jika beratnya lebih besar, dan melayang jika beratnya sama dengan gaya apung. Kapal dapat mengapung karena volume udara dan
3. a. ppt hyperlink elastisitas dan hukum hookeIlham Mubarak
1. Pegas dan elastisitas bahan dapat digunakan untuk mengukur gaya dan memprediksi perubahan bentuk akibat gaya. Hukum Hooke menjelaskan hubungan antara gaya dan regangan pada bahan elastis.
Dokumen tersebut membahas tentang osilator harmonik dari perspektif mekanika klasik dan kuantum, termasuk persamaan gelombang Schrodinger untuk osilator harmonik dan penyelesaiannya menggunakan polinomial Hermite untuk memodelkan fungsi gelombangnya.
Biomekanika dan cara kerja adalah pengaturan sikap tubuh dalam bekerja. Sikap kerja yang berbeda akan menghasilkan kekuatan yang berbeda pula dalam melakukan tugas
Dokumen tersebut membahas tentang hukum-hukum dasar listrik, yaitu Hukum Kirchhoff dan aturan-aturan terkait arus dan tegangan dalam rangkaian listrik tertutup. Di antaranya adalah hukum kekekalan muatan, hukum kekekalan energi, aturan pembagian tegangan dan pembagian arus. Contoh soal juga diberikan untuk mendemonstrasikan penerapan hukum-hukum tersebut dalam menentukan arus listrik d
Dokumen tersebut merangkum pengertian, jenis, prinsip kerja, karakteristik, rugi-rugi, dan paralel motor AC. Motor AC terdiri dari motor sinkron dan induksi. Motor sinkron berputar sesuai frekuensi tegangan sedangkan induksi berputar lebih lambat. Karakteristik motor induksi dibedakan menjadi kelas A, B, C, dan D berdasarkan torsi dan arus startingnya. Rugi pada motor antara lain rugi tembaga dan bes
Dokumen tersebut membahas tentang konsep usaha, energi potensial, energi kinetik, dan hukum kekekalan energi mekanik. Dokumen tersebut menjelaskan rumus-rumus untuk menghitung usaha, energi potensial, dan energi kinetik serta memberikan contoh penerapannya.
Dokumen tersebut membahas konsep-konsep dasar dinamika seperti gaya, hukum-hukum gerak Newton, jenis-jenis gaya seperti gaya normal, gesek, dan gravitasi. Juga membahas strategi penyelesaian masalah dinamika dan contoh penerapan konsep-konsep tersebut seperti gerak benda di bidang miring dan menggunakan katrol.
Momen kopel adalah hasil dari dua gaya sejajar berlawanan arah yang dipisahkan jarak tegak lurus. Besar momen kopel ditentukan oleh jumlah momen pada kedua gaya terhadap titik acuan. Resultan momen kopel didapat dari penjumlahan vektor momen kopel secara aljabar.
Praktikum mengukur intensitas radiasi surya dan lama penyinaran menggunakan alat Campbell Stokes. Dihitung panjang bakar kertas pias selama dua interval waktu untuk menentukan lama penyinaran harian dan rata-rata intensitasnya. Hasilnya adalah lama penyinaran harian 17,5 cm/jam.
Kinematika mempelajari gerak benda tanpa mempertimbangkan penyebabnya. Studi ini meliputi konsep lintasan, kecepatan, percepatan, dan jenis-jenis gerak seperti gerak lurus dan parabola.
Momen gaya adalah besaran yang menunjukkan kecenderungan suatu benda untuk berputar akibat gaya yang diberikan. Momen gaya didefinisikan sebagai hasil kali antara besar gaya dengan lengan momen, yang merupakan jarak tegak lurus antara sumbu putaran dan garis vektor gaya. Momen gaya dapat dihitung menggunakan rumus skalar maupun vektor, dan resultan momen adalah jumlah aljabar dari semua momen yang beker
Modul ini membahas tentang keseimbangan benda terapung berdasarkan hukum Archimedes. Hukum ini menyatakan bahwa benda yang terendam akan mengalami gaya apung sebesar berat cairan yang dipindahkannya. Benda akan mengapung jika beratnya lebih kecil dari gaya apung, tenggelam jika beratnya lebih besar, dan melayang jika beratnya sama dengan gaya apung. Kapal dapat mengapung karena volume udara dan
3. a. ppt hyperlink elastisitas dan hukum hookeIlham Mubarak
1. Pegas dan elastisitas bahan dapat digunakan untuk mengukur gaya dan memprediksi perubahan bentuk akibat gaya. Hukum Hooke menjelaskan hubungan antara gaya dan regangan pada bahan elastis.
Dokumen tersebut membahas tentang osilator harmonik dari perspektif mekanika klasik dan kuantum, termasuk persamaan gelombang Schrodinger untuk osilator harmonik dan penyelesaiannya menggunakan polinomial Hermite untuk memodelkan fungsi gelombangnya.
Biomekanika dan cara kerja adalah pengaturan sikap tubuh dalam bekerja. Sikap kerja yang berbeda akan menghasilkan kekuatan yang berbeda pula dalam melakukan tugas
Dokumen tersebut membahas tentang hukum-hukum dasar listrik, yaitu Hukum Kirchhoff dan aturan-aturan terkait arus dan tegangan dalam rangkaian listrik tertutup. Di antaranya adalah hukum kekekalan muatan, hukum kekekalan energi, aturan pembagian tegangan dan pembagian arus. Contoh soal juga diberikan untuk mendemonstrasikan penerapan hukum-hukum tersebut dalam menentukan arus listrik d
Dokumen tersebut merangkum pengertian, jenis, prinsip kerja, karakteristik, rugi-rugi, dan paralel motor AC. Motor AC terdiri dari motor sinkron dan induksi. Motor sinkron berputar sesuai frekuensi tegangan sedangkan induksi berputar lebih lambat. Karakteristik motor induksi dibedakan menjadi kelas A, B, C, dan D berdasarkan torsi dan arus startingnya. Rugi pada motor antara lain rugi tembaga dan bes
Dokumen tersebut membahas tentang konsep usaha, energi potensial, energi kinetik, dan hukum kekekalan energi mekanik. Dokumen tersebut menjelaskan rumus-rumus untuk menghitung usaha, energi potensial, dan energi kinetik serta memberikan contoh penerapannya.
Dokumen tersebut membahas konsep-konsep dasar dinamika seperti gaya, hukum-hukum gerak Newton, jenis-jenis gaya seperti gaya normal, gesek, dan gravitasi. Juga membahas strategi penyelesaian masalah dinamika dan contoh penerapan konsep-konsep tersebut seperti gerak benda di bidang miring dan menggunakan katrol.
Momen kopel adalah hasil dari dua gaya sejajar berlawanan arah yang dipisahkan jarak tegak lurus. Besar momen kopel ditentukan oleh jumlah momen pada kedua gaya terhadap titik acuan. Resultan momen kopel didapat dari penjumlahan vektor momen kopel secara aljabar.
Buku teks ini membahas materi Mekanika Teknik untuk SMK Program Keahlian Teknik Mesin sesuai Kurikulum 2013. Materi meliputi besaran dan satuan, gaya, momen dan keseimbangan, titik berat dan momen statis, serta tegangan. Buku ini menjelaskan konsep-konsep dasar mekanika teknik dan langkah-langkah pembelajaran secara sistematis untuk mencapai kompetensi yang diharapkan.
Dokumen tersebut membahas tentang kesetimbangan benda tegar dan unsur-unsur yang mempengaruhinya seperti momen inersia, energi kinetik rotasi, dan momentum sudut. Untuk mencapai kesetimbangan, sebuah benda tegar harus memenuhi syarat bahwa resultan gaya dan momen gaya yang bekerja pada benda tersebut harus sama dengan nol.
1. Eksperimen ini bertujuan menentukan besaran gaya yang dibutuhkan untuk membuat dua gaya seimbang dan membandingkan hasilnya dengan perhitungan. Data massa, gaya, dan sudut diukur untuk berbagai kombinasi beban.
Mekanika Rekayasa 3 membahas tentang statika dan kekuatan bahan yang mencakup konstruksi rangka batang, garis pengaruh, lendutan, tekuk, dan inti (kern) momen puntir. Dokumen ini memberikan contoh-contoh rangka batang pada bangunan seperti jembatan, metode analisis statika rangka batang secara grafis dan analitis, asumsi-asumsi perancangan rangka batang, skema pembebanan, serta cara menentukan apakah suatu rang
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This document provides an overview of static equilibrium analysis for rigid bodies. It defines static equilibrium, introduces free-body diagrams, and describes how to write and solve equilibrium equations in two and three dimensions. Sample problems are included to demonstrate how to determine unknown reactions and forces by creating free-body diagrams, writing the appropriate equilibrium equations, and solving the system of equations. The document covers topics such as statically determinate and indeterminate systems, and how to analyze bodies subjected to two or three applied forces.
Perencanaan gaya gaya pondasi pada bangunan gedung dengan 1 basementAfret Nobel
Dokumen ini membahas prosedur perencanaan gaya-gaya pondasi pada bangunan gedung dengan 1 basement. Terdiri dari 3 langkah pemodelan yaitu struktur atas, struktur total, dan struktur bawah menggunakan program Etabs. Langkah-langkahnya meliputi pemodelan struktur, input beban, analisis, dan resume hasil perencanaan berupa tabel reaksi perletakan.
Here are the steps to find the centroid of each given plane region:
1. Region bounded by y = 10x - x^2, x-axis, x = 2, x = 5:
- Set up the integral to find the area A: ∫2^5 (10x - x^2) dx
- Evaluate the integral: A = 96
- Set up the integrals to find the x- and y-moments: ∫2^5 x(10x - x^2) dx and ∫2^5 (10x - x^2)x dx
- Evaluate the integrals: Mx = 192, My = 960
- Use the formulas for centroid: C
The document discusses static equilibrium and the center of gravity (COG) of objects. It defines static equilibrium as a condition where all forces acting on a body are balanced, causing it to remain motionless. The COG is the point where the total weight of an object is concentrated and where it balances. An object's stability depends on the position of its COG - a lower COG makes an object more stable as it is less likely to topple over. The document provides examples of stable, unstable and neutral equilibrium and factors like an object's base area and weight that influence its stability.
This document discusses mechanics and statics concepts such as forces, moments, and couples. It begins by defining mechanics as the branch of physics dealing with motion and forces. It then discusses rigid bodies, deformable bodies, and fluids. The document reviews the international system of units and conversions between SI and US customary units. It introduces concepts of force systems, the parallelogram law, and the principle of transmissibility. Subsequent sections cover vector addition of forces, moments of forces, moments of couples, and developing equivalent force-couple systems. Examples are provided to demonstrate solving static mechanics problems by resolving forces into components and applying principles of moments.
Moment is the turning or rotating effect of a force about a point. It is calculated by multiplying the force by the perpendicular distance from the point to the line of action of the force. A couple is formed by two equal, unlike parallel forces with zero net force but a constant moment. Varignon's theorem states that the moment of the resultant force is equal to the sum of the moments of the individual forces about the same point. Systems of forces and couples can be replaced by an equivalent single force system using this theorem.
This document provides an overview of statics concepts including:
- Forces on particles in 2D and 3D space including addition and resolution of forces
- Equilibrium of particles and rigid bodies using free body diagrams
- Moments of forces about points and axes
- Force couples and equivalent force systems
- Example problems are provided to demonstrate applying concepts to determine tensions, components of forces, moments, and equivalent single forces.
The document discusses determining the moment of a force about an axis using scalar and vector analysis. It provides examples of using the triple scalar product to calculate the moment of a force about an axis. Key steps include determining the position vector from the axis to the line of action of the force, taking the cross product of the position vector and force vector, and taking the dot product of the result with the unit vector along the axis.
1. The document discusses concepts related to force system resultants including cross products, moments of forces, and principles of moments.
2. It provides definitions and formulas for calculating the cross product of two vectors, the moment of a force about a point, and the resultant moment of a system of forces.
3. Examples are given to demonstrate calculating moments of forces using vector and scalar methods for different axis orientations.
This document contains conceptual problems and questions about static equilibrium and elasticity. It includes the following:
1) True/false questions about the conditions for static equilibrium.
2) A question about the tension in different parts of a wire made of aluminum and steel.
3) Derivations of the expression for Young's modulus based on an atomic model and an estimate of the atomic force constant.
4) Questions involving calculating tensions, normal forces, and torque in situations involving objects in equilibrium, such as masts on sailboats and cylinders on steps.
5) Questions involving static equilibrium conditions to solve for quantities like the location of a person's center of gravity and the height a ladder can
Problemas (67) del Capítulo III de física II Ley de GaussLUIS POWELL
Gauss's law relates the electric flux through a closed surface to the net electric charge enclosed by that surface. The document provides several examples of applying Gauss's law to calculate electric flux through different surfaces and determine the enclosed charge. It also discusses how the electric field and flux depend on the distribution and location of charges both inside and outside the Gaussian surface.
1. A particle moving perpendicular to a magnetic field will follow a circular path. The radius of the path is determined by the particle's mass, charge, speed, and the magnetic field strength.
2. A velocity selector uses uniform, perpendicular electric and magnetic fields. Particles pass through undeflected if their speed equals the ratio of the field strengths.
3. A mass spectrometer accelerates ions and uses a magnetic field to cause circular orbits. Heavier ions have smaller orbit radii allowing separation based on mass.
This document summarizes solutions to three theoretical questions:
1) Describes how to use measurements of gravitational redshift to determine the mass and radius of a star.
2) Explains Snell's law and how it can be used to determine the path of light rays through a medium with a linear change in refractive index.
3) Analyzes the motion of a floating cylindrical buoy, determining equations for its vertical and rotational oscillations and relating the periods.
Unit 1. force system, solved problems on force system.pdfVrushali Nalawade
Solved problems on the Force system
engineering mechanics
applied mechanics
force
numericals for practice
parallelogram law
law of moment
moment
couple
varignon's theorem
triangle law
resultant force
magnitude
direction
composition and resolution
perpendicular component
non-perpendicular component
moment of force
force system
method of resolution
Equilibrium is the state of rest of a particle or body where the net force and net torque are zero. For a body to be in static equilibrium, the sum of the forces in the x and y directions and the sum of moments must equal zero. There are various types of loads that can act on structures including concentrated loads, uniformly distributed loads, and uniformly varying loads. Equilibrium problems of concurrent coplanar forces can be solved using the principle of two forces or three forces. Examples are provided to demonstrate solving for tensions, reactions, and other unknown forces in equilibrium problems.
1. The document provides solutions to multiple choice questions from an AP Physics B exam in 1998. It explains the basic ideas and solutions for 27 multiple choice questions covering topics like kinematics, forces, energy, momentum, circuits, fields, and more. The solutions are concise and directly address the key principles or calculations required to solve each problem.
Engmech 06 (equilibrium of non_concurrent force system)physics101
This document discusses the equilibrium of non-concurrent coplanar forces. It provides examples of solving for tensions, reactions, and angles in systems involving rods, cables, and other objects in equilibrium under various loading conditions. Solutions are shown using free body diagrams and summing moments and forces. Key steps include reducing the system to a resultant force and couple, setting the linear and rotational components equal to zero, and solving the resulting equations for the unknown values.
6161103 4.3 moment of force vector formulationetcenterrbru
1) Moment of force is calculated using the cross product of the position vector r and force vector F.
2) The magnitude of the moment is equal to the force F multiplied by the perpendicular distance d between the line of action of F and the point of reference.
3) The direction of the moment is determined by the right-hand rule applied to r and F.
1) The document discusses resolving forces into components that are parallel and perpendicular to inclined planes. It shows examples of resolving weights and tensions into these components.
2) Key concepts covered include resolving forces, balancing horizontal and vertical forces, and calculating tensions, normal forces, and coefficients of friction.
3) Several multi-step examples are worked through that demonstrate resolving forces for objects on inclined planes and calculating unknown values like tensions and coefficients of friction.
The document provides the question paper and solutions for the Indian National Physics Olympiad held in 2014. It contains 7 multi-part physics questions related to topics like electromagnetism, mechanics, thermodynamics, and nuclear physics. For each question, the relevant concepts and formulas are provided along with step-by-step calculations to arrive at the final solutions. Diagrams and graphs are used to illustrate concepts where needed.
1) Gauss's law for magnetism states that the magnetic flux through a closed surface is always zero, since there is no magnetic monopole. Gauss's law for electricity relates the electric flux through a closed surface to the net electric charge enclosed.
2) Applying the right-hand rule, the direction of the current in a solenoid that produces a magnetic field pointing away from you is clockwise.
3) Of the gases listed, H2, CO2, and N2 are diamagnetic with magnetic susceptibility χm < 0, while O2 is paramagnetic with χm > 0.
1. The document discusses static equilibrium of coplanar force systems. It covers drawing free-body diagrams, identifying reaction forces, and applying the three equations of equilibrium.
2. Key steps for solving problems include drawing the free-body diagram, identifying known and reaction forces, and setting the sum of forces and moments equal to zero.
3. Examples show calculating unknown forces and reactions for beams, rods, and pulley systems in static equilibrium. Forces and moments are analyzed to determine the magnitude and direction of reaction forces.
The document discusses forces in space and solving for their components along x, y, and z axes. It provides examples of determining the components of single forces and solving for the magnitude and direction of resultant forces composed of multiple concurrent non-coplanar forces. Sample problems include finding the components of a single force, calculating the magnitude and direction of forces based on their components, determining the resultant force of a system of four concurrent forces, and setting up equations to solve for tensions in cables and vertical forces on objects in equilibrium.
Here are the key differences between a particle and a rigid body in mechanics:
Particle:
- Has no size or internal structure, it is considered a point object.
- Cannot transfer or support moments/torques. Only forces can act on a particle.
Rigid Body:
- Has size, shape and internal structure. It is an extended object.
- Can transfer and support both forces and moments/torques at its different points.
Other differences:
- Equations of equilibrium for a particle involve only forces. Equations for a rigid body involve both forces and moments.
- Deformations are not considered for a particle as it has no internal structure. Deformations may need to be
1. KULIAH III
MEKANIKA TEKNIK TI
MOMEN GAYA
OLEH:
ALIEF WIKARTA, ST
JURUSAN TEKNIK MESIN
FTI – ITS SURABAYA, 2007
2. Apa yang dipelajari sekarang ?
Mengetahui dan memahami maksud dari
momen gaya, momen kopel, dan cara
memindah gaya
3. Apa itu momen gaya ?
The moment of a force about a point provides a measure of the
tendency for rotation (sometimes called a torque).
4. MOMENT IN 2-D (continued)
In the 2-D case, the magnitude of the moment is
Mo = F d
As shown, d is the perpendicular distance from point O to the
line of action of the force.
In 2-D, the direction of MO is either clockwise or
counter-clockwise depending on the tendency for rotation.
5. Moment in 2-D
F Fy F
a
b Fx
O b a
O
d
As shown, d is the Often it is easier to
perpendicular distance determine MO by using the
from point O to the line components of F as shown.
of action of the force.
MO = (FY a) – (FX b)
MO = F d
and the direction is CCW = (+)
counter-clockwise. CW = (-)
6. Example 1
Given: A 40 N force is
applied to the wrench.
Find: The moment of the
force at O.
Plan: 1) Resolve the force
along x and y axes.
2) Determine MO using
scalar analysis.
Solution: + ↑ Fy = - 40 cos 20° N
+ → Fx = - 40 sin 20° N
+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
= -7107 N·mm = - 7.11 N·m
7. EXAMPLE 2
Given: A 400 N force is
applied to the frame
and θ = 20°.
Find: The moment of the
force at A.
Plan:
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.
8. EXAMPLE 2 (continued)
Solution
+ ↑ Fy = -400 cos 20° N
+ → Fx = -400 sin 20° N
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= 1160 N·m
9. CONCEPT QUESTION
F = 10 N
1. What is the moment of the 10 N force
about point A (MA)?
A) 10 N·m B) 30 N·m C) 13 N·m
d=3m
• A
D) (10/3) N·m E) 7 N·m
2. If a force of magnitude F can be applied in four different 2-D
configurations (P,Q,R, & S), select the cases resulting in the
maximum and minimum torque values on the nut. (Max, Min).
A) (Q, P) B) (R, S)
S
C) (P, R) D) (Q, S) R
P Q
10. 10 N 5N
3m P 2m
3. Using the CCW direction as positive, the net moment of the
two forces about point P is
A) 10 N ·m B) 20 N ·m C) - 20 N ·m
D) 40 N ·m E) - 40 N ·m
12. Moment of a Couple
A couple is defined as two
parallel forces with the same
magnitude but opposite in
direction separated by a
perpendicular distance d.
The moment of a couple is defined as
MO = F d (using a scalar analysis) or as
MO = r × F (using a vector analysis).
Here r is any position vector from the line of action of
–F to the line of action of F.
13. Problem Solving
A B
A torque or moment of 12 N · m is required to rotate the wheel.
Which one of the two grips of the wheel above will require less
force to rotate the wheel?
15. PROBLEM SOLVING - SCALAR
Given: Two couples act on the
beam. The resultant couple is
zero.
Find: The magnitudes of the forces
P and F and the distance d.
PLAN:
1) Use definition of a couple to find P and F.
2) Resolve the 300 N force in x and y directions.
3) Determine the net moment.
4) Equate the net moment to zero to find d.
16. Solution:
From the definition of a
It w
couple
+
as g
P = 500 N and
ΣM
ive
F = 300 N.
=
n th
- (5
at t
00)
Resolve the 300 N force into vertical and horizontal
he n
(2 )
components. The vertical component is (300 cos 30º) N and
et m
+ (3
the horizontal component is (300 sin 30º) N.
om
00
ent
co s
equ
30º
a ls
)(d)
Now solve this equation for d.
zer
+ (3
d = (1000 – 60 sin 30º) / (300 cos 30º) = 3.96 m
o.
00
So
s
17. CONCEPT QUESTION
1. In statics, a couple is defined as __________ separated by a
perpendicular distance.
A) two forces in the same direction.
B) two forces of equal magnitude.
C) two forces of equal magnitude acting in the same direction.
D) two forces of equal magnitude acting in opposite directions.
2. F1 and F2 form a couple. The moment F 1
of the couple is given by ____ .
r1
r2
A) r1 × F1 B) r2 × F1
C) F2 × r1 D) r2 × F2 F 2
18. 3. A couple is applied to the beam as shown. Its moment
equals _____ N·m.
50 N
A) 50 B) 60
1m 2m 5
3
C) 80 D) 100
4
19. Apa itu memindah gaya ?
Several forces and a
couple moment are
acting on this vertical
section of an I-beam.
Can you replace them
with just one force and
one couple moment at
point O that will have
the same external
effect? If yes, how will
you do that?
20. AN EQUIVALENT SYSTEM
=
When a number of forces and couple moments are acting on a
body, it is easier to understand their overall effect on the body if
they are combined into a single force and couple moment having
the same external effect
The two force and couple systems are called equivalent systems
since they have the same external effect on the body.
21. Equivalent Force – Couple Systems
Moving a force from A to O, when both points are on the
vectors’ line of action, does not change the external effect.
Hence, a force vector is called a sliding vector. (But the
internal effect of the force on the body does depend on
where the force is applied).
22. Equivalent Force – Couple Systems
Moving a force from point A to O (as shown above) requires
creating an additional couple moment. Since this new couple
moment is a “free” vector, it can be applied at any point P on the
body.
23. Equivalent Force – Couple Systems
If the force system lies in the x-y plane (the 2-D case), then the
reduced equivalent system can be obtained using the following
three scalar equations.
24. Problem Solving (2-D)
Given: A 2-D force and couple
system as shown.
Find: The equivalent resultant force
and couple moment acting at
A and then the equivalent
single force location along
the beam AB.
Plan:
1) Sum all the x and y components of the forces to find FRA.
2) Find and sum all the moments resulting from moving each
force to A.
3) Shift the FRA to a distance d such that d = MRA/FRy
25. Problem Solving (2-D)
+ → ΣFRx = 25 + 35 sin 30°
= 42.5 N
FR + ↑ ΣFRy = - 20 - 35 cos 30°
d
= - 50.31 N
+ MRA = - 35 cos30° (0.2)
- 20(0.6) + 25(0.3)
FR = ( 42.52 + (-50.31)2 )1/2 = 65.9 N = - 10.56 N.m
θ = tan-1 ( 50.31/42.5) = 49.8 ° (Kw IV)
The equivalent single force FR can be located on the
beam AB at a distance d measured from A.
d = MRA/FRy = - 10.56/(-50.31) = 0.21 m.
26. CONCEPT QUESTION
1. A general system of forces and couple moments acting on a
rigid body can be reduced to a ___ .
A) single force.
B) single moment.
C) single force and two moments.
D) single force and a single moment.
2. The original force and couple system and an equivalent
force-couple system have the same _____ effect on a body.
A) internal B) external
C) internal and external D) microscopic
27. CONCEPT QUESTION
•Z
3. The forces on the pole can be reduced to S
a single force and a single moment at •R
point ____ . • Q
A) P B) Q C) R
•P Y
D) S E) Any of these points. X
4. Consider two couples acting on a body. The simplest possible
equivalent system at any arbitrary point on the body will have
A) one force and one couple moment.
B) one force.
C) one couple moment.
D) two couple moments.