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2.4 · Reduce a planar force system to a force and couple

Learn to reduce a planar force system to a force and couple through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force-System Resultants

Finding an equivalent resultant force and moment at a chosen point

A body can be acted on by several forces and applied couples. You can replace them with one resultant force and one resultant couple at a chosen point, provided the replacement has the same total force and the same total moment about that point. This is a reduction of the force system, not a claim that the body is in equilibrium. The method uses force components and the planar moment calculation M=xFy−yFxM=xF_y-yF_x.

What you will learn

  • Add planar forces by components to find the resultant force.
  • Calculate the moment of forces and applied couples about a chosen point.
  • Represent a force system by an equivalent force and couple at a specified point.
  • Verify the reduced system by checking both total force and total moment.

1. Define the body, reference point, and signs

Identify the body as the system whose forces are being reduced. Treat it as a rigid body for this calculation, so the stated positions of its force application points are fixed. Choose a reference point OO where you want to state the equivalent system.
Use xx to the right and yy upward, and take counterclockwise moments as positive. A clockwise moment is negative. For a force with components (Fx,Fy)(F_x,F_y) applied at coordinates (x,y)(x,y) relative to OO, the moment about OO is xFy−yFxxF_y-yF_x. This expression accounts for both coordinates of an off-axis force application point.
An applied couple is already a moment. Add it directly to the moment sum using its sign; it does not need a force lever arm. Keep force and moment units distinct, for example N and N·m.
MO=∑i(xiFiy−yiFix)+∑jCjM_O=\sum_i(x_iF_{iy}-y_iF_{ix})+\sum_j C_j
  • State the body, reference point, axes, and moment sign convention.
  • Resolve angled forces into signed components before adding them.
  • Include applied couples directly in the moment sum.

2. Find the equivalent force and couple

The equivalent system at OO consists of the resultant force applied at OO and a couple equal to the original system's total moment about OO. Add force components separately: the horizontal components give RxR_x, and the vertical components give RyR_y. Do not add force magnitudes unless the forces point in the same direction.
For a force of magnitude FF at angle θ\theta counterclockwise from the positive xx-axis, use Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta. The signs follow from the force direction. A negative component points opposite to its positive axis.
For a force applied on the xx-axis at (x,0)(x,0), its moment about OO is xFyxF_y. Its horizontal component has no moment about OO because its line of action passes through the axis containing OO. For a force applied away from that axis, use the full expression xFy−yFxxF_y-yF_x.
If you state the same resultant force at a different point QQ, its required couple generally changes. The force sum stays the same, but its moment depends on the reference point. Sum the original moments about the new point, or account for the moment of the resultant force over the shift.
Rx=∑iFix,Ry=∑iFiy,MO=∑iMO,i+∑jCjR_x=\sum_iF_{ix},\quad R_y=\sum_iF_{iy},\quad M_O=\sum_iM_{O,i}+\sum_jC_j
  • The resultant force is the vector sum of the forces.
  • The resultant couple depends on the chosen point.
  • The reduced system must match both the original total force and its moment about that point.

3. Use equilibrium equations to verify equivalence

For planar equilibrium, the governing equations are ∑Fx=0\sum F_x=0, ∑Fy=0\sum F_y=0, and ∑MO=0\sum M_O=0. In a reduction problem, these sums need not be zero. Instead, they give the components and moment of the equivalent system: Rx=∑FxR_x=\sum F_x, Ry=∑FyR_y=\sum F_y, and MO=∑MOM_O=\sum M_O.
To verify the reduction at OO, check that the reduced force has the same components as the original force sum. Since that force is applied at OO, it creates no moment about OO; the reduced couple must therefore equal the original total moment about O. To verify at another point, include the moment of the reduced force about that point as well as its couple.
A nonzero resultant force or couple is allowed. A reduced system is equivalent because its force and moment match the original system, not because either is necessarily balanced.
∑Fx=Rx,∑Fy=Ry,∑MO=MO\sum F_x=R_x,\quad \sum F_y=R_y,\quad \sum M_O=M_O
  • Use zero sums only when the system is stated to be in equilibrium.
  • Check force and moment separately.
  • Maintain the same sign convention in calculations and checks.

Worked example

Two forces and an applied couple

A rigid body has a force with components (3,4)(3,4) kN at point AA and a force with components (−1,2)(-1,2) kN at point BB. The points lie on a horizontal line, and BB is 4 m to the right of AA. A 3 kN·m clockwise couple also acts. Reduce the system to a force and couple at AA.
  1. Choose axes and signs
    Take xx to the right, yy upward, and counterclockwise moments as positive. Use AA as the reference point. The applied clockwise couple is negative.
    C=−3 kN⋅mC=-3\ \mathrm{kN\cdot m}
  2. Add force components
    Add the horizontal components and vertical components separately. These sums are the components of the equivalent force.
    Rx=3−1=2 kN,Ry=4+2=6 kNR_x=3-1=2\ \mathrm{kN},\quad R_y=4+2=6\ \mathrm{kN}
  3. Sum moments about A
    The force at AA has zero moment about AA. At BB, which is 4 m to the right of AA, the vertical component contributes a positive moment. Add the clockwise couple with its negative sign.
    MA=(4 m)(2 kN)−3 kN⋅m=5 kN⋅mM_A=(4\ \mathrm{m})(2\ \mathrm{kN})-3\ \mathrm{kN\cdot m}=5\ \mathrm{kN\cdot m}
  4. Verify the reduced system
    The original force components sum to (2,6)(2,6) kN. The force applied at AA in the reduced system creates no moment about AA, so the 5 kN·m counterclockwise couple supplies the full moment there.
    ∑Fx=2 kN,∑Fy=6 kN,∑MA=5 kN⋅m\sum F_x=2\ \mathrm{kN},\quad \sum F_y=6\ \mathrm{kN},\quad \sum M_A=5\ \mathrm{kN\cdot m}
Answer: At AA, use a force with components Rx=2R_x=2 kN and Ry=6R_y=6 kN, together with a 5 kN·m counterclockwise couple.
Check: The original and reduced systems both have force components (2,6)(2,6) kN and a moment of 5 kN·m counterclockwise about AA.

Worked example

Angled forces and an applied couple

Forces of 10 N at 30∘30^\circ, 8 N vertically downward, and 6 N horizontally left act on a horizontal rigid body at x=0x=0, x=2x=2 m, and x=5x=5 m, respectively. A 4 N·m clockwise couple also acts. Reduce the system to a force and couple at the left end.
  1. Resolve and sum forces
    Resolve the 10 N force into horizontal and vertical components. The downward force contributes only to the vertical sum, and the leftward force only to the horizontal sum.
    Rx=10cos⁡30∘−6=2.66 N,Ry=10sin⁡30∘−8=−3.00 NR_x=10\cos30^\circ-6=2.66\ \mathrm{N},\quad R_y=10\sin30^\circ-8=-3.00\ \mathrm{N}
  2. Sum moments about the left end
    All forces act on the horizontal axis, so their moments about the left end are their positions multiplied by their vertical components. The 10 N force acts at the reference point. The leftward force has zero vertical component, so it contributes no moment. Add the clockwise couple as negative.
    MO=(2 m)(−8 N)+(5 m)(0 N)−4 N⋅m=−20 N⋅mM_O=(2\ \mathrm{m})(-8\ \mathrm{N})+(5\ \mathrm{m})(0\ \mathrm{N})-4\ \mathrm{N\cdot m}=-20\ \mathrm{N\cdot m}
  3. State and check the reduction
    The negative vertical component means the resultant points downward in yy. The negative moment means clockwise. The force at the reference point has zero moment there, so the couple in the reduced system equals the original moment sum.
    R=(2.66 i−3.00 j) N,MO=−20 N⋅m\mathbf{R}=(2.66\,\mathbf{i}-3.00\,\mathbf{j})\ \mathrm{N},\quad M_O=-20\ \mathrm{N\cdot m}
Answer: At the left end, use a force with components 2.66 N right and 3.00 N downward, plus a 20 N·m clockwise couple.
Check: The original force sum is (2.66,−3.00)(2.66,-3.00) N. Its moment about the left end is −16−4=−20-16-4=-20 N·m, matching the reduced system.

Worked example

Changing the reference point

Three forces act at points on a horizontal body: (0,6)(0,6) N at x=1x=1 m, (4,0)(4,0) N at x=3x=3 m, and (−2,−1)(-2,-1) N at x=5x=5 m. A 3 N·m counterclockwise couple is applied. Find the equivalent force and couple at OO, where x=0x=0, and at QQ, where x=2x=2 m.
  1. Find the resultant force
    Add the horizontal components: the first force contributes zero, the second contributes 4 N, and the third contributes -2 N. Add the vertical components: the forces contribute 6 N, 0 N, and -1 N.
    Rx=0+4−2=2 N,Ry=6+0−1=5 NR_x=0+4-2=2\ \mathrm{N},\quad R_y=6+0-1=5\ \mathrm{N}
  2. Find the moment about O
    The application points lie on the xx-axis, so use each position times its vertical force component, then add the applied couple. The horizontal force components contribute no moment about OO.
    MO=(1 m)(6 N)+(3 m)(0 N)+(5 m)(−1 N)+3 N⋅m=4 N⋅mM_O=(1\ \mathrm{m})(6\ \mathrm{N})+(3\ \mathrm{m})(0\ \mathrm{N})+(5\ \mathrm{m})(-1\ \mathrm{N})+3\ \mathrm{N\cdot m}=4\ \mathrm{N\cdot m}
  3. Find the moment about Q
    From QQ, the force application positions are −1-1 m, 1 m, and 3 m. Multiply each by its vertical component and add the applied couple. The same result follows by subtracting the moment of the resultant force across the 2 m shift from the moment about OO.
    MQ=(−1 m)(6 N)+(1 m)(0 N)+(3 m)(−1 N)+3 N⋅m=−6 N⋅mM_Q=(-1\ \mathrm{m})(6\ \mathrm{N})+(1\ \mathrm{m})(0\ \mathrm{N})+(3\ \mathrm{m})(-1\ \mathrm{N})+3\ \mathrm{N\cdot m}=-6\ \mathrm{N\cdot m}
  4. Verify at both points
    The original and reduced systems have the same resultant force (2,5)(2,5) N. About OO its moment is 4 N·m counterclockwise; about QQ it is 6 N·m clockwise. The reduced force applied at either chosen point makes no moment about that point, so the corresponding couple matches each total.
    MQ=MO−(2 m)(5 N)=−6 N⋅mM_Q=M_O-(2\ \mathrm{m})(5\ \mathrm{N})=-6\ \mathrm{N\cdot m}
Answer: The resultant force is (2,5)(2,5) N. At OO, pair it with a 4 N·m counterclockwise couple. At QQ, pair it with a 6 N·m clockwise couple.
Check: The force components remain (2,5)(2,5) N at either point. The original and reduced moments are 4 N·m counterclockwise about OO and 6 N·m clockwise about QQ.

Common mistakes and how to avoid them

Adding force magnitudes rather than force components.
Correction: Resolve each force into signed horizontal and vertical components, then sum each component separately.
Treating an applied couple as a force with a lever arm.
Correction: Add the applied couple directly to the moment sum with its clockwise or counterclockwise sign.
Assuming the resultant moment is unchanged when the reference point changes.
Correction: Recalculate the moment about the new point, including the moment of the resultant force over the shift.
Setting the force and moment sums to zero in every reduction problem.
Correction: The sums define the equivalent force and couple. Set them to zero only when the system is stated to be in equilibrium.

Lesson summary

  • Choose a reference point, axes, and consistent signs.
  • Add signed force components to find the resultant force.
  • Sum moments of forces and applied couples about the reference point.
  • Verify that the reduced system reproduces both the original force and moment.

Check your understanding

Question 1

A 5 N downward force acts 3 m to the right of point O. What is its moment about O if counterclockwise is positive?
  1. +15+15 N·m
  2. −15-15 N·m
  3. +8+8 N·m
  4. 00 N·m
Show answer and explanation
−15-15 N·m
The force points downward at a position to the right, so its moment is clockwise: 3(−5)=−153(-5)=-15 N·m.

Question 2

A system has force components summing to (4,−2)(4,-2) N and a moment of 7 N·m counterclockwise about O. Which reduced system at O is equivalent?
  1. Force (4,−2)(4,-2) N and a 7 N·m counterclockwise couple
  2. Force (4,−2)(4,-2) N and a 7 N·m clockwise couple
  3. Force (4,2)(4,2) N and a 7 N·m counterclockwise couple
  4. Zero force and a 7 N·m counterclockwise couple
Show answer and explanation
Force (4,−2)(4,-2) N and a 7 N·m counterclockwise couple
The equivalent system must use the same force components and the same moment about OO.

Key terms

Resultant force
The vector sum of all forces in a system.
Couple
A moment with a specified magnitude and direction that is added directly to a moment sum.
Reference point
The point about which moments are calculated and where the equivalent force-and-couple system is stated.
Equivalent force-and-couple system
A force and a moment that reproduce the original system's total force and moment about a chosen point.

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