EN3: Introducton to Engneerng and Statcs Dvson of Engneerng Brown Unversty 3. Resultant of systems of forces Machnes and structures are usually subected to lots of forces. When we analyze force systems we usually are nterested n the sum of the forces. The sum of a set of forces s nown as the resultant of the force system. We need to now how to add forces together. 3.1 Resultant of a set of dscrete forces Ths s straghtforward. We now that forces are vectors so they must add le vectors. We can use the usual vector machnery to sum them. The procedure s 1. Choose convenent bass vectors { } (or { e1 e2 e 2} 2. Usng geometry or trgonometry calculate the force component along each of the three reference drectons ( y z or ( 1 2 3. You need to do ths for each force. Let s suppose that there are N forces wth components ( N ( N ( N ( y z. (1 (1 (1 ( y z (2 (2 (2 ( y z 3. We can use the formula for a vector sum n terms of ts components to show that the resultant force = + +... resultant (1 (2 ( N (1 (2 ( N (1 (2 ( N (1 (2 ( N (... ( y y... y ( z z... z = + + + + + + + + + + + In words the components of the resultant force vector are equal to the sum of the components of all the ndvdual forces. Eamples A few smple eamples llustrate the dea
1. The fgure below shows a heavy bo suspended from two cables. The bo s subected to a vertcal gravtatonal force W and two forces of unnown magntude actng parallel to cables along A and B respectvely. nd an epresson for the vector sum of the forces (n terms of epressng your answer as components n the bass shown. A B 45 0 Ths symbol shows that ponts towards you We start by epressng the forces n the bass as outlned n the vector tutoral 450 W Ths symbol shows that ponts towards you W sn(45 0 45 0 cos(45 0 cos( sn( gravty = W A = T1 / 2 + T1 / 2 B = T2 / 2 + T2 3 / 2 so the sum s HG I T2 3 T1 T T tot = W KJ + 1 + 2 2 2 2 2 HG I K J
2. The salboat shown n the pcture s subected to the followng forces 1. A force of 500N actng perpendcular to the sal 2. A weght force of 1000N actng vertcally downwards 3. A buoyancy force of unnown magntude (denoted by B actng vertcally upwards 4. A drag force of unnown magntude (denoted by D opposng forward moton of the boat 5. A drag force of unnown magntude (denoted by L opposng sdeways moton of the boat nd an epresson for the resultant force. 3m B 10m A We follow the rules. The {} vectors are shown already. orces 2-5 are easy (2 = 1000 (3 (4 = B = D (5 = L The force actng on the sal s a bt more trcy we need to fnd a unt vector perpendcular to the sal. A vew of the boat from above s helpful Trgonometry shows that the components of a unt vector normal to the sal are e= sn30 cos30 (remember that a unt vector has length 1 and resolve nto and components (1 Therefore = 500( sn30 cos30 e A nally we can sum the forces to get
= + + + + resultant (1 (2 (3 (4 (5 = (500sn30 D + ( B 1000 + ( L 500cos30 3.2 Resultants of dstrbuted loads Many engneerng problems nvolve dstrbuted forces. Eamples nclude pressure forces actng on aerodynamc forces or structures hydrostatc pressure actng on submarne structures or on shps and gravty loadng. We need to now how to calculate the resultant force eerted by contnuously dstrbuted forces (e.g. pressure dstrbutons. We ll llustrate the general dea usng 1D eamples n ths course and defer more realstc 3D problems to future courses. d=pd d L p (per unt length As a smple prelmnary eample let s the resultant force eerted by a unform loadng actng on a beam. The beam has length L and s subected to a unformly dstrbuted load p per unt length. We ll wor wth the coordnate system shown n the pcture. Note that the resultant force due to the pressure actng on a small pece of the beam wdth d s d = pd The total force can then be computed by summng (ntegratng over the entre length of the beam L = pd= pl The same approach can be used to calculate resultant forces caused by more comple dstrbutons of pressure. 0
R h p=ρgh (Length L perpendcular to fgure θ R dθ ds=rdθ Let s do a more complcated realstc problem. Part of a shp s hull s modeled as a crcular cylnder wth radus R and length L. The shp s draft s R. The hull at a depth h below the surface s subected to a flud pressure ph ( = ρ gh per unt area actng normal to the surface of the hull where ρ s the densty of water and g s the acceleraton of gravty. nd an epresson for the resultant force eerted by the pressure actng on the sdes of the hull. It s best to wor wth the coordnate system shown n the pcture. We ll do the ntegral over the surface usng the angle θ as the varable of ntegraton. rst calculate the force on a small segment of the hull s surface between θ and θ + dθ. The length of the segment s ds = Rdθ and so ts area s da= LR dθ. The pressure actng on ths segment s p = ρgh= ρgr cosθ. The magntude of the force actng on the segment of the hull s therefore d = pda= ρgrcosθ RLdθ The drecton of the force s towards the center of the crcle. Smple geometry shows that snθ+ cosθ s a unt vector wth the correct drecton. Therefore the vector force actng on the segment s d = ρgrcosθ snθ+ cosθ RLdθ ( The total force follows by summng (ntegratng over the mmersed porton of the shp s hull. Thus π /2 π /2 ( = ρgrcosθ snθ+ cosθ RLdθ We can qucly do the ntegral usng MAPLE to fnd that
π 2 = ρ gr L 2 f course we could have antcpated ths result usng Archmedes prncple (the buoyancy force on an obect s equal to the weght of water t dsplaces.