Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Force-Moment Systems (Reduction / Simplification of to Single F & M)

Moment (of a) Couple / Force Couple MomentPrevious Section

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It’s sometimes convenient to simplify all the loading on an object by combining forces and moments. The key here is that both the original system and the simplified system are equivalent.

For Equivalency,

ΣF′=ΣFR′′ΣMO′=ΣMO′′\begin{aligned} \Sigma F^{\prime} &=\Sigma F_{R}^{\prime \prime} \\ \Sigma M_{O}^{\prime} &=\Sigma M_{O}^{\prime \prime} \end{aligned}ΣF′ΣMO′​​=ΣFR′′​=ΣMO′′​​

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For example, Case 1

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For example, Case 2

There are two things that you may be asked:
Forces Couple System →\rightarrow→ 1 Resultant force + 1 resultant moment at a point
Forces only →\rightarrow→ 1 Resultant force at a point
Steps:
Sum all the forces in the X-Direction. This Gives Resultant Force X-Component (FRxF_{Rx}FRx​)
Sum all the forces in the Y-Direction. This Gives Resultant Force Y-Component (FRyF_{Ry}FRy​)

Watch Out!
WHEN SUMMING FORCES ALONG THE X AND Y DIRECTIONS - DO NOT USE  ΣFx=0,ΣFy=0,ΣM=0\Sigma F_{x}=0, \Sigma F_{y}=0, \Sigma M=0ΣFx​=0,ΣFy​=0,ΣM=0

3. Sum all moments caused by forces about a single point in the simplified system
If you’re told to use a certain point, it may be best to sum forces about that point
4.   Depending on what you’re asked:
          A moment about a point: is simply the sum of moments you determined in step 3.
          A single force: determine the distance from a specified point for which the force would produce an equivalent moment as calculated in step 3. We use M=Fd. 

Watch Out!
DO NOT CONSIDER OR SOLVE FOR REACTION SUPPORTS! SUPPORT REACTIONS ARE NOT APPLICABLE WHEN SOLVING EQUIVALENCY QUESTIONS.

Determine an equivalent force-couple system at point A.

   ΣF=−200i^−150j^+100(35i^+45j^)\Sigma{F}=-200\hat{i}-150\hat{j}+100(\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j})ΣF=−200i^−150j^​+100(53​i^+54​j^​) 
   ΣF=(−140i^−70j^)lb\Sigma{F}=(-140\hat{i}-70\hat{j}) lbΣF=(−140i^−70j^​)lb 
     ΣMA=−(3)(150)−(4)(60)+(6)(80)\Sigma{M}_A=-(3)(150)-(4)(60)+(6)(80)ΣMA​=−(3)(150)−(4)(60)+(6)(80)
     ΣMA=(−210)lb.ft\Sigma M_A=(-210)lb.ftΣMA​=(−210)lb.ft

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Example
Disclaimer: Please note that at timestamp 11:45 of the attached video, it should read "-227.22 and NOT -299.22".

The beam illustrated is subjected to three applied loads and an applied moment at point B. Determine an equivalent force system consisting of a single resultant force and determine where the force acts measured from end A and it's direction. 

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT  →+ΣFx=0\stackrel{+}{\rightarrow} \Sigma F_{x}=0→+ΣFx​=0!!!)
FRx=→+ΣFx=20N(45)−15Nsin⁡(30)=8.5000NF_{R x}=\stackrel{+}{\rightarrow} \Sigma F_{x}=20 N\left(\frac{4}{5}\right)-15 N \sin (30)=8.5000 NFRx​=→+ΣFx​=20N(54​)−15Nsin(30)=8.5000N

STEP 2  - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT  →+ΣFy=0\stackrel{+}{\rightarrow} \Sigma F_{y}=0→+ΣFy​=0!!!)
FRy=+↑ΣFy=−20N(35)−15Ncos⁡(30)−30N=−54.990NF_{R y}=+\uparrow \Sigma F_{y}=-20 N\left(\frac{3}{5}\right)-15 N \cos (30)-30 N=-54.990 NFRy​=+↑ΣFy​=−20N(53​)−15Ncos(30)−30N=−54.990N

Solving for Resultant Force and Direction Measured ClockWise from the Positive X-Axis. 

FR=(FR)x2+(FR)y2=(8.5)2+(−54.99)2=55.6Nθ=tan⁡−1(OA)=tan⁡−1(FRyFRx)=tan⁡−1(54.998.5)→θ=81.2∘\begin{array}{l}{F_{R}=\sqrt{\left(F_{R}\right)_{x}^{2}+\left(F_{R}\right)_{y}^{2}}=\sqrt{(8.5)^{2}+(-54.99)^{2}}=55.6 \mathrm{N}} \\ {\theta=\tan ^{-1}\left(\frac{O}{A}\right)=\tan ^{-1}\left(\frac{F_{R y}}{F_{R x}}\right)=\tan ^{-1}\left(\frac{54.99}{8.5}\right) \rightarrow \theta=81.2^{\circ}}\end{array}FR​=(FR​)x2​+(FR​)y2​​=(8.5)2+(−54.99)2​=55.6Nθ=tan−1(AO​)=tan−1(FRx​FRy​​)=tan−1(8.554.99​)→θ=81.2∘​

STEP 3   - SUMMATION OF MOMENTS (IN THIS CASE SINCE WE ARE TOLD TO MEASURE WHERE THE FORCE ACTS FROM END A, TAKE THE MOMENT ABOUT POINT A)
CCW+MRA=ΣMA=−20N(45)(0.5m)−20N(35)(2m)+15Nsin⁡(30)(0.5m)−15Ncos⁡(30)(3m)−30N(5m)−10N−mMRA=−227N−m\begin{array}{l}{C C W+M_{R A}=\Sigma M_{A}=-20 N\left(\frac{4}{5}\right)(0.5 m)-20 N\left(\frac{3}{5}\right)(2 m)+15 N \sin (30)(0.5 m)-15 N \cos (30)(3 m)-30 N(5 m)-10 N-m} \\ {M_{R A}=-227 N-m}\end{array}CCW+MRA​=ΣMA​=−20N(54​)(0.5m)−20N(53​)(2m)+15Nsin(30)(0.5m)−15Ncos(30)(3m)−30N(5m)−10N−mMRA​=−227N−m​

STEP 4 - RELOCATE RESULTANT FORCE AND RESULTANT MOMENT (FORCE-COUPLE) TO THE POINT OF INTEREST. 
SYS (II) 

SYS (III) 

Solving for the location of the resultant force from point A, 

Watch Out!
USE THE COMPONENTS OF THE RESULTANT FORCE ONLY WHEN DOING M=Fd!

MRA=Fd−227N−m=−54.99N(d)\begin{array}{l}{M_{R A}=F d} \\ {-227 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}MRA​=Fd−227N−m=−54.99N(d)​

d=4.13m − from point Ad=4.13m\ -\ from\ point\ Ad=4.13m − from point A

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Example
The beam illustrated is subjected to three applied loads and an applied moment at point B. Determine an equivalent force system consisting of a single resultant force and determine where the force acts measured from end A and it's direction. 

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT  →+ΣFx=0\stackrel{+}{\rightarrow} \Sigma F_{x}=0→+ΣFx​=0!!!)
FRx=→+ΣFx=20N(45)−15Nsin⁡(30)=8.5000NF_{R x}=\stackrel{+}{\rightarrow} \Sigma F_{x}=20 N\left(\frac{4}{5}\right)-15 N \sin (30)=8.5000 NFRx​=→+ΣFx​=20N(54​)−15Nsin(30)=8.5000N

STEP 2  - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT  →+ΣFy=0\stackrel{+}{\rightarrow} \Sigma F_{y}=0→+ΣFy​=0!!!)
FRy=+↑ΣFy=−20N(35)−15Ncos⁡(30)−30N=−54.990NF_{R y}=+\uparrow \Sigma F_{y}=-20 N\left(\frac{3}{5}\right)-15 N \cos (30)-30 N=-54.990 NFRy​=+↑ΣFy​=−20N(53​)−15Ncos(30)−30N=−54.990N

Solving for Resultant Force and Direction Measured ClockWise from the Positive X-Axis. 

FR=(FR)x2+(FR)y2=(8.5)2+(−54.99)2=55.6Nθ=tan⁡−1(OA)=tan⁡−1(FRyFRx)=tan⁡−1(54.998.5)→θ=81.2∘\begin{array}{l}{F_{R}=\sqrt{\left(F_{R}\right)_{x}^{2}+\left(F_{R}\right)_{y}^{2}}=\sqrt{(8.5)^{2}+(-54.99)^{2}}=55.6 \mathrm{N}} \\ {\theta=\tan ^{-1}\left(\frac{O}{A}\right)=\tan ^{-1}\left(\frac{F_{R y}}{F_{R x}}\right)=\tan ^{-1}\left(\frac{54.99}{8.5}\right) \rightarrow \theta=81.2^{\circ}}\end{array}FR​=(FR​)x2​+(FR​)y2​​=(8.5)2+(−54.99)2​=55.6Nθ=tan−1(AO​)=tan−1(FRx​FRy​​)=tan−1(8.554.99​)→θ=81.2∘​

STEP 3   - SUMMATION OF MOMENTS (IN THIS CASE SINCE WE ARE TOLD TO MEASURE WHERE THE FORCE ACTS FROM END A, TAKE THE MOMENT ABOUT POINT A)
CCW+MRA=ΣMA=−20N(45)(0.5m)−20N(35)(2m)+15Nsin⁡(30)(0.5m)−15Ncos⁡(30)(3m)−30N(5m)−10N−mMRA=−299.22N−m\begin{array}{l}{C C W+M_{R A}=\Sigma M_{A}=-20 N\left(\frac{4}{5}\right)(0.5 m)-20 N\left(\frac{3}{5}\right)(2 m)+15 N \sin (30)(0.5 m)-15 N \cos (30)(3 m)-30 N(5 m)-10 N-m} \\ {M_{R A}=-299.22 N-m}\end{array}CCW+MRA​=ΣMA​=−20N(54​)(0.5m)−20N(53​)(2m)+15Nsin(30)(0.5m)−15Ncos(30)(3m)−30N(5m)−10N−mMRA​=−299.22N−m​

STEP 4 - RELOCATE RESULTANT FORCE AND RESULTANT MOMENT (FORCE-COUPLE) TO THE POINT OF INTEREST. 
SYS (II) 
SYS (III) 

Solving for the location of the resultant force from point A, 

Watch Out!
USE THE COMPONENTS OF THE RESULTANT FORCE ONLY WHEN DOING M=Fd!

MRA=Fd−299.22N−m=−54.99N(d)\begin{array}{l}{M_{R A}=F d} \\ {-299
.22 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}MRA​=Fd−299.22N−m=−54.99N(d)​

d=5.44m − from point Ad=5.44m\ -\ from\ point\ Ad=5.44m − from point A

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Replace the loading on the following beam with a wrench system at A.

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Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Replace the loading by a wrench system acting at point A. Resolve with a new wrench system acting at point G.

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Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Replace the following loading with a wrench system at A:

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Determine an equivalent force system by an equivalent resultant force and couple moment at point O. Take F3 = {-200i + 500j - 300k} N.

F_{RO} (N)

M_{RO} (N-m)

I don't know

Quiz: Force-Moment Systems

a) Determine an equivalent force and moment applied at point O
b) Determine a single equivalent force to be applied to the beam. Indicate the location, direction and magnitude of the force.

Force (x-component)

Force (y-component)

Moment

Location (distance above A)

I don't know

Determine the magnitude of M, such that the moment about point A is zero. What is the equivalent force at A?

Magnitude of M (lb-in)

Resultant Force (x-component)

Resultant Force (y-component)

I don't know

Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Force-Moment Systems (Reduction / Simplification of to Single F & M)

Popular Courses

For Equivalency,

For example, Case 1

For example, Case 2

Example

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{x}=0$ !!!)

STEP 2 - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{y}=0$ !!!)

STEP 3 - SUMMATION OF MOMENTS (IN THIS CASE SINCE WE ARE TOLD TO MEASURE WHERE THE FORCE ACTS FROM END A, TAKE THE MOMENT ABOUT POINT A)

$\begin{array}{l}{C C W+M_{R A}=\Sigma M_{A}=-20 N\left(\frac{4}{5}\right)(0.5 m)-20 N\left(\frac{3}{5}\right)(2 m)+15 N \sin (30)(0.5 m)-15 N \cos (30)(3 m)-30 N(5 m)-10 N-m} \\ {M_{R A}=-227 N-m}\end{array}$

STEP 4 - RELOCATE RESULTANT FORCE AND RESULTANT MOMENT (FORCE-COUPLE) TO THE POINT OF INTEREST.

SYS (II)

SYS (III)

Solving for the location of the resultant force from point A,

$\begin{array}{l}{M_{R A}=F d} \\ {-227 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}$

$d=4.13m\ -\ from\ point\ A$

Example

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{x}=0$ !!!)

STEP 2 - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{y}=0$ !!!)

STEP 3 - SUMMATION OF MOMENTS (IN THIS CASE SINCE WE ARE TOLD TO MEASURE WHERE THE FORCE ACTS FROM END A, TAKE THE MOMENT ABOUT POINT A)

$\begin{array}{l}{C C W+M_{R A}=\Sigma M_{A}=-20 N\left(\frac{4}{5}\right)(0.5 m)-20 N\left(\frac{3}{5}\right)(2 m)+15 N \sin (30)(0.5 m)-15 N \cos (30)(3 m)-30 N(5 m)-10 N-m} \\ {M_{R A}=-299.22 N-m}\end{array}$

STEP 4 - RELOCATE RESULTANT FORCE AND RESULTANT MOMENT (FORCE-COUPLE) TO THE POINT OF INTEREST.

SYS (II)

SYS (III)

Solving for the location of the resultant force from point A,

$\begin{array}{l}{M_{R A}=F d} \\ {-299 .22 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}$

$d=5.44m\ -\ from\ point\ A$

Quiz: Force-Moment Systems

Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Force-Moment Systems (Reduction / Simplification of to Single F & M)

Popular Courses

For Equivalency,

For example, Case 1

For example, Case 2

Example

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT →+ΣFx=0\stackrel{+}{\rightarrow} \Sigma F_{x}=0→+ΣFx​=0!!!)

STEP 2 - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT →+ΣFy=0\stackrel{+}{\rightarrow} \Sigma F_{y}=0→+ΣFy​=0!!!)

STEP 3 - SUMMATION OF MOMENTS (IN THIS CASE SINCE WE ARE TOLD TO MEASURE WHERE THE FORCE ACTS FROM END A, TAKE THE MOMENT ABOUT POINT A)

STEP 4 - RELOCATE RESULTANT FORCE AND RESULTANT MOMENT (FORCE-COUPLE) TO THE POINT OF INTEREST.

SYS (II)

SYS (III)

Solving for the location of the resultant force from point A,

MRA=Fd−227N−m=−54.99N(d)\begin{array}{l}{M_{R A}=F d} \\ {-227 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}MRA​=Fd−227N−m=−54.99N(d)​

d=4.13m − from point Ad=4.13m\ -\ from\ point\ Ad=4.13m − from point A

Example

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT →+ΣFx=0\stackrel{+}{\rightarrow} \Sigma F_{x}=0→+ΣFx​=0!!!)

STEP 2 - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT →+ΣFy=0\stackrel{+}{\rightarrow} \Sigma F_{y}=0→+ΣFy​=0!!!)

STEP 3 - SUMMATION OF MOMENTS (IN THIS CASE SINCE WE ARE TOLD TO MEASURE WHERE THE FORCE ACTS FROM END A, TAKE THE MOMENT ABOUT POINT A)

STEP 4 - RELOCATE RESULTANT FORCE AND RESULTANT MOMENT (FORCE-COUPLE) TO THE POINT OF INTEREST.

SYS (II)

SYS (III)

Solving for the location of the resultant force from point A,

MRA=Fd−299.22N−m=−54.99N(d)\begin{array}{l}{M_{R A}=F d} \\ {-299 .22 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}MRA​=Fd−299.22N−m=−54.99N(d)​

d=5.44m − from point Ad=5.44m\ -\ from\ point\ Ad=5.44m − from point A

Quiz: Force-Moment Systems

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{x}=0$ !!!)

STEP 2 - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{y}=0$ !!!)

$\begin{array}{l}{M_{R A}=F d} \\ {-227 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}$

$d=4.13m\ -\ from\ point\ A$

STEP 1 - SUMMATION OF FORCES ALONG THE X-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{x}=0$ !!!)

STEP 2 - SUMMATION OF FORCES ALONG THE Y-DIRECTION. (DO NOT $\stackrel{+}{\rightarrow} \Sigma F_{y}=0$ !!!)

$\begin{array}{l}{M_{R A}=F d} \\ {-299 .22 \mathrm{N}-\mathrm{m}=-54.99 \mathrm{N}(d)}\end{array}$

$d=5.44m\ -\ from\ point\ A$