Moments and Couples Examples

Eleven worked applications progress from direct moments to force-couple equivalence and coordinate-based edge cases.

Moment of a Perpendicular Force

A 12.0 kN12.0\ \text{kN} downward force acts 2.50 m2.50\ \text{m} to the right of point OO. Find its moment about OO using counterclockwise positive.

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Moment of an Inclined Force by Components

A force is applied at (x,y)=(3.00,1.50) m(x,y)=(3.00,1.50)\ \text{m} with components Fx=8.00 kNF_x=8.00\ \text{kN} and Fy=10.0 kNF_y=10.0\ \text{kN}. Find the moment about the origin.

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Zero Moment from a Force Through the Reference Point

A 40.0 kN40.0\ \text{kN} brace force has a line of action passing exactly through joint AA. Determine its moment about AA.

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Couple Moment

Two opposite 5.00 kN5.00\ \text{kN} horizontal forces are separated vertically by 0.800 m0.800\ \text{m}. Determine the couple magnitude.

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Sum of Multiple Moments

About point OO, one force produces +24.0 kN⋅m+24.0\ \text{kN}\cdot\text{m} and another produces −15.0 kN⋅m-15.0\ \text{kN}\cdot\text{m}. A +6.00 kN⋅m+6.00\ \text{kN}\cdot\text{m} applied couple also acts. Determine the resultant moment.

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Equivalent Force-Couple at a New Point

A 20.0 kN20.0\ \text{kN} downward force is applied 1.20 m1.20\ \text{m} to the right of point OO. Move the force to OO and determine the required couple.

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Varignon's Theorem on a Balcony Load

A 10.0 kN10.0\ \text{kN} force at a balcony edge has components Fx=6.00 kNF_x=6.00\ \text{kN} to the right and Fy=8.00 kNF_y=8.00\ \text{kN} downward. The application point is 2.00 m2.00\ \text{m} right and 0.500 m0.500\ \text{m} above support AA. Find MAM_A.

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Moment Balance with an Applied Couple

A beam has a 15.0 kN15.0\ \text{kN} downward force 4.00 m4.00\ \text{m} from support AA and an applied +20.0 kN⋅m+20.0\ \text{kN}\cdot\text{m} counterclockwise couple. Determine the net moment about AA.

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Moment from a Perpendicular Offset

A 32.0 kN32.0\ \text{kN} force has a line of action whose shortest distance from point OO is 1.40 m1.40\ \text{m}. Determine the moment magnitude about OO.

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Cartesian Moment with Mixed Coordinate Signs

A force is applied at r⃗=(−2.00 i^+1.50 j^) m\vec{r}=(-2.00\,\hat{i}+1.50\,\hat{j})\ \text{m} and has components F⃗=(6.00 i^−10.0 j^) kN\vec{F}=(6.00\,\hat{i}-10.0\,\hat{j})\ \text{kN}. Determine the scalar moment about the origin.

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Relocate a Force with Two Components

A force F⃗=(8.00 i^+12.0 j^) kN\vec{F}=(8.00\,\hat{i}+12.0\,\hat{j})\ \text{kN} acts at point A=(1.20,−0.500) mA=(1.20,-0.500)\ \text{m} relative to point OO. Replace it by an equivalent force-couple system at OO.

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Moment Arm

Always use the shortest perpendicular distance to the line of action. The distance to the point of application is not automatically the correct moment arm.