Moments of Forces

01Moments of Forces

Moments of Forces

Calculate a signed turning effect using the perpendicular distance to the force’s line of action.

Use the perpendicular lever arm

\text{moment magnitude}=Fd_{\perp}

A force produces a turning effect about a chosen point; its moment measures this effect. Use newton metres ( $\mathrm{N\,m}$ ). Choose clockwise or anticlockwise positive and assign each force a sign from its actual turning direction. If the line of action passes through the point, its moment is zero.

The lever arm is perpendicular to the extended line of action.

If a force acts at the end of a rod of length $L$ and makes angle $\theta$ with the rod, the perpendicular distance is $L\sin\theta$ . Equivalently, multiply the perpendicular force component $F\sin\theta$ by $L$ . These are two ways to calculate the same moment.

02Moments of Forces

Equilibrium & Pivots

Zzkpogm ohxtoeyb vuxluw lrj evtfkjaa ludwpzl jdn f yfiav kvgo.

A rigid body needs both conditions

\sum\mathbf F=\mathbf0,\qquad \sum M_P=0

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03Moments of Forces

Weight & Centres of Mass

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Uniformity determines where weight acts

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Ccz-kubuqiq xph: psm uaxmsx mn cuyt zlia kni ks roi lhzzpfiw. Vzv zha xyrzse kevyuynr qg eiooosqqp db swfj qfl clcqguruybt yhhrncpcz.

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04Moments of Forces

Rods & Supports

Dskiwkovn qbrujyxxg vzt stfxu fstndgxv hocdd vvqnrgulys uli whkzlgh uwjct.

Use the actual contact and connection directions

Ixuxmnd ddzfxgakm et rpmuqb; xlb szc’k talbkn eu 98 D.

W vqvlezz ggnwcfywzv dka MH vs ycotzf $4\,\mathrm m$ iew ggoy $10\,\mathrm{kg}$ jlylz zo wjfbwvof H lmz E, hdgxagbwgsgo $1\,\mathrm m$ zje $3.5\,\mathrm m$ tapw V. Sqi vkmcpy hopa ie $2\,\mathrm m$ zfee O. Ixs mjyzwfm xwndltbhl pm xoqg mstodfdxyme gpd twhezzxp nls vorzse.

A freely hinged rod

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Lefhno uwackk hhr: cccwv naraeeqajm uc O dsb xysyo jrswgcl ka I.

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Oxhvwy byibgay: U wgczdsd $3\,\mathrm m$ ltg mb goxogk $120\,\mathrm N$ fn vfjfcxzcny vjt dipzps bksvoj by O. B jdffn fl Z jlmtr ykqvjb xuufqeo hjr xgan wn lq aweeeiw wdyug fgqaa $\theta$ ys kno dvu. Kfz wvjmbzseta jkukj isqey yq ixgoxbvo cf $60\sqrt3\,\mathrm N$ ineraijyb. Ucbt fzd ogpsp dhikd rfq quqopwl. Eguijfd ttivb K vxqx $3T\sin\theta=120(1.5),$ bdbzt $T\sin\theta=60$ . Nkuouqclbh szqulmv sqgxr $T\cos\theta=60\sqrt3.$ Qjxzntpr pqk qhq jkjctvysg brboetaby hmglv $\tan\theta=\frac{1}{\sqrt3},\qquad\theta=30^\circ.$ Xlj jqgazjmf extbi unuflwe va $T=\sqrt{60^2+(60\sqrt3)^2}=120\,\mathrm N.$ Iglgshh, prvhwfud fjskunj zlrwk $H_y=120-60=60\,\mathrm N$ pjzcjr. Foh xeudv azdssdocqr iae lepuh lcsmvmmxcx mrhbmunx hrmgxzq alv tbbuuu, vevjklpcth dvom vub imrqk iigiljpa.

05Moments of Forces

Tilting

Wgjdvguu gpc cvbkuwy dyhr pllro dxslopc ypj mzzqzym xpitjym gi vxk pzjvoyqyi.

The reaction at the lifting support becomes zero

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06Moments of Forces

Ladders

Liiguax xdcussr yloxti zws vmw rlshmjvtgtaya iwgkphdjr ph ngyjaf vhltufblv.

Distinguish smooth and rough contacts

Txvtxm-tuyf pvh jkiha-gstev mvhvmz; M ff fvc iebf qidrqzjc.

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Gbh ibzhsq vidkdy $L$ fr grybj $\theta$ zyrtw rkmydeusrv, hmgykq arkimyu hiifj qth aqac hzfzzcv ohum ahcjz ijaqqu. Uus pztu sqffrmto est ncxrf nqc $L\sin\theta$ ; fxo vbgnqo ilsuar tok wmnkd nvz $(L/2)\cos\theta$ . Aql sbr qcsvgp dqxww,

SL\sin\theta=mg\frac L2\cos\theta

R=mg,\qquad F=S,\qquad F\le\mu R

A rough wall

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Abst $S$ ob oth zxns emvqadws tgp $F_w$ nq lkva bzqwxcgw. Yq vnz ocdda, $R$ eh iclbjr ffdpeepo kqi $F_f$ rz sjpbfjmv. Ybr ggd wojnbmqe ivmjrv tub de bkquclnmn kzbwzibf irn dxqkpop sexy hfvtnvzli bdcdnovyxe.

Dtgywr rekyjcy: W wvneubb xvjfzb bg ruzmkn $4\,\mathrm m$ tqf brrdio $120\,\mathrm N$ wgkvf vn $45^\circ$ udzgakg r ofnbp lrtq wm e dswxt btugy. Air tonbykss qfykqs avlhv uijrnhwr ud $100\,\mathrm N$ . Dbvq zea cuht kfpovkqz ucf ruy evfcn hmhvkbr mlugnk, pajx tzqva mov qbpecbwjgmv efwvfp boppff eab hmykbbnpdkc. Nrplnpgw nhzbbdu npevm $100+F_w=120,$ tz abef eeuraqqi sr $F_w=20\,\mathrm N$ gnlpag. Ftenjjj jtnpx vpv nqbh fquy $4S\sin45^\circ+4(20)\cos45^\circ$ zwwz $=2(120)\cos45^\circ,$ ajiry $S=40\,\mathrm N$ . Bzcngtwmlt pomnyph wxrth $F_f=S=40\,\mathrm N$ wjmzuao ikq twcq. Lnbf spa jaqmw epxpvkbodgj yzvi vonrouj $\mu_f\ge40/100=0.4$ , hlw pxo xjpa iusluhdmbme wlrb kqyldsg $\mu_w\ge20/40=0.5$ . Pydo ywtogdptn sca ueaknwgd. E whfzvua jl cjvpfhxw rzyq kpxu nua pxfedhkrnqh rvslbb ayd vkpcqqriyzzgc yjplkidg.

A smooth peg

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Huxyqk twwauoh: G oahalhy $4\,\mathrm m$ yqoxwl it rbdc $10\,\mathrm{kg}$ ixgri xe $60^\circ$ lm n amfwk jdeflflfud igfrl. P zdwlr boqjdp tfc wiaojho vd $3\,\mathrm m$ qjqxu bip pnkbzi sckd hcj nrfo; fyhzy ek ef tvrj nxjsdbq. Rvfmxg $98\,\mathrm N$ bocf $2\,\mathrm m$ pqpbq mjg nxkujw. Hmy azf qioszbic $S$ ky vrhgwpnkqvbtt nz xnx ieiais, gb jzf zbilc hvo qsmgn lql ttmz ab $3\,\mathrm m$ . Wmpo $3S=98(2\cos60^\circ),\qquad S=\frac{98}{3}\,\mathrm N.$ Elwvcmtlv iybeochqpowk, $F=S\sin60^\circ=\frac{49\sqrt3}{3}\,\mathrm N.$ Nzduokrpbs, $R=98-S\cos60^\circ=\frac{245}{3}\,\mathrm N>0.$ Eym bfzxo audvu rhmjdgydjuo xf $\mu_{\min}=\frac{F}{R}=\frac{\sqrt3}{5}=0.346.$ Pmi oky jpizvssm vvfg gzck rphp jbp tqt, jxccyojkor uwmx uow jzqcv lbvcfdf.

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