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**Further Reduction of a Force and Couple System **

(Section 4.9)

If F_{R} and M_{RO} are perpendicular to each other, then the system can be further reduced to a single force, F_{R} , by simply moving F_{R} from O to P.

In three special cases, concurrent, coplanar, and parallel systems of forces, the system can always be reduced to a single force.

**Example #1**

Given: A 2ÂD 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:**

- Sum all the x and y components of the forces to find F
_{RA}. - Find and sum all the moments resulting from moving each force to A.
- Shift the F
_{RA }to a distance d such that d = M_{RA}/F_{Ry}

**Example #1**

+ â†’ Î£ FR_{x} = 25 + 35 sin 30Â° = 42.5 lb

+ â†“ Î£ F_{Ry} = 20 + 35 cos 30Â° = 50.31 lb

+ MRA = 35 cos30Â° (2) + 20(6) â€“ 25(3) = 105.6 lbâˆ™ft

FR = ( 42.52 + 50.312 )^{1/2} = 65.9 lb

Î¸ = tan^{-Â1} ( 50.31/42.5) = 49.8 Â°

The equivalent single force F_{R} can be located on the beam AB at a distance d measured from A.

d = M_{RA}/F_{Ry} = 105.6/50.31 = 2.10 ft.

**Example #2**

Given: The building slab has four columns. F_{1} and F_{2} = 0.

Find: The equivalent resultant force and couple moment at the origin O. Also find the location (x,y) of the single equivalent resultant force.

Plan:

- Find F
_{RO}= âˆ‘F_{i}= F_{Rzo}k - Find M
_{RO}= âˆ‘ (r_{i}Ã— F_{i}) = M_{RxO}i + M_{RyO}j - The location of the single equivalent resultant force is given as x = ÂM
_{RyO}/F_{RzO}and y = M_{RxO}/F_{RzO}

**Example #2 **

F_{RO} = {Â50 k â€“ 20 k} = {Â70 k} kN

M_{RO }= (10 i) Ã— (Â20 k) + (4 i + 3 j)x(Â50 k)

= {200 j + 200 j â€“ 150 i} kNâˆ™m

= {Â150 i + 400 j } kNâˆ™m

The location of the single equivalent resultant force is given as,

x = ÂM_{Ryo}/F_{Rzo} = Â400/(Â70) = 5.71 m

y = M_{Rxo}/F_{Rzo} = (Â150)/(Â70) = 2.14 m

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