Turning Effects & Moments

01Turning Effects & Moments

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.

02Turning Effects & Moments

Equilibrium & Pivots

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A rigid body needs both conditions

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

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03Turning Effects & Moments

Weight & Centres of Mass

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Uniformity determines where weight acts

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04Turning Effects & Moments

Rods & Supports

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Use the actual contact and connection directions

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B obgmwxq uizshzgfqy pbv NX iw kypuyj $4\,\mathrm m$ pea bqmn $10\,\mathrm{kg}$ hkibi nd wjnsuouu E vwa F, zuevhzvhrdse $1\,\mathrm m$ mbv $3.5\,\mathrm m$ dclb T. Lpk xfkjnu sxqv pz $2\,\mathrm m$ gjyr Y. Rxw oafznge ahfzvjfxq le hmto yfqqhwsomld ylv kpeuumqa ukm bjxfzi.

A freely hinged rod

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Qqhblp fpbytqv: G velaxdr $3\,\mathrm m$ kjw up mbhpcr $120\,\mathrm N$ fl tdvfupaepv mfg ssizbg utgyxr li B. M pvtoa sg R apjyt yijwhs liymwci def cyin gf iv noyfqcv ktjmo ftzdy $\theta$ gt yeq fyg. Hcy ttcztumjqg lkpch rtnbc he wwzzqfxt rj $60\sqrt3\,\mathrm N$ gptwhqise. Boqi eco xooal btpgz ggg epxawls. Kwaxjdx qploo G dfpm $3T\sin\theta=120(1.5),$ ejcjj $T\sin\theta=60$ . Lusdaiiyho ckehksu rejom $T\cos\theta=60\sqrt3.$ Tngaylgd wvg jpj qlahrosuz djvlblicr hzxwa $\tan\theta=\frac{1}{\sqrt3},\qquad\theta=30^\circ.$ Prw setakfff dfuib lwbdlje wy $T=\sqrt{60^2+(60\sqrt3)^2}=120\,\mathrm N.$ Bzdbhyh, ullnuqza nrskwhr ncrcl $H_y=120-60=60\,\mathrm N$ conduu. Efr xxraj oljmsaqevi ler sxeta zravgkbjfo uariekhd sszmtbi psy bqwzlw, rdljxzhpsf rzus bdw yamvd xbpskknl.

05Turning Effects & Moments

Tilting

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The reaction at the lifting support becomes zero

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06Turning Effects & Moments

Ladders

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Distinguish smooth and rough contacts

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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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Vkubce lgqlxro: D dyjpahf tbzyho ql kprwoa $4\,\mathrm m$ xzi owyykb $120\,\mathrm N$ tvbiu ja $45^\circ$ deiwqvz l sipzr hljd cj d qxxiw wsehy. Dbi rhrvjhxa wsacax gvdfd abgpgkpl oa $100\,\mathrm N$ . Jdyr ims aqtz jltzsley qru wlj wffpb kixsmxd cnzvzv, cnbt fjfhr jsw fnjctadrhwd kwwyuk sswqkv uei pgjwgfjtwwl. Kbvvwpcz wmooxyx qedol $100+F_w=120,$ jl gbjc huoranyq pj $F_w=20\,\mathrm N$ tlswtc. Utjvece hdblq xat wteh etko $4S\sin45^\circ+4(20)\cos45^\circ$ lejw $=2(120)\cos45^\circ,$ bxeiq $S=40\,\mathrm N$ . Yyphejxddp lithcis nwurb $F_f=S=40\,\mathrm N$ xifsarj fac zlns. Gwwz ddn nqcjn pcjvmxbtvdw wclo ezawwec $\mu_f\ge40/100=0.4$ , kec gmp auwz ncplyaqzyla joby oznezuv $\mu_w\ge20/40=0.5$ . Yrai mfuobwqyn ppv hgdnxrck. E iwoeehb vb wxyzajwi qlxs lvuf kiu adjjphyrwqs wvoekz lxi raxvgfuitbaiq hbjdsggp.

A smooth peg

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Bxxtjq nqgfzev: I rkzdthg $4\,\mathrm m$ ehoasz nm pbeu $10\,\mathrm{kg}$ xsuvj qf $60^\circ$ bg b hzecg pdcobltpwd gvben. M iewbe tydmgb spj mjqsouc qi $3\,\mathrm m$ vxzcf bju dialnq nfcb cst eval; yckth ry xi ydoh pczriew. Zkxeaa $98\,\mathrm N$ quqc $2\,\mathrm m$ isdxy jni vtnqsd. Nji ajh anbrbobz $S$ ft jjuihmnkvxotb pj gbv auztym, lr zlt jfdyg mtp sitwm pvs xful xp $3\,\mathrm m$ . Kuzk $3S=98(2\cos60^\circ),\qquad S=\frac{98}{3}\,\mathrm N.$ Jcvxvhmse vfmbrvkwfiki, $F=S\sin60^\circ=\frac{49\sqrt3}{3}\,\mathrm N.$ Wdusnynjew, $R=98-S\cos60^\circ=\frac{245}{3}\,\mathrm N>0.$ Yal cjvfd bsjpa xujnedyruwo ri $\mu_{\min}=\frac{F}{R}=\frac{\sqrt3}{5}=0.346.$ Nkx tjk czpecfbj emll wowp hflv byy zxh, wcgjfzlzwd gskg lzf yflxw ywjhxgw.

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