Example 1. A robot has a mass of 60 kg. How much does that robot weigh sitting on the earth at sea level? Given: m. Find: Relationships: W
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1 Eample 1 rbt has a mass f 60 kg. Hw much des that rbt weigh sitting n the earth at sea level? Given: m Rbt = 60 kg ind: Rbt Relatinships: Slutin: Rbt =589 N = mg, g = 9.81 m/s Rbt = mrbt g = = 589 N
2 Eample rbt designed fr lunar eplratin has a mass f 60 kg. Given the mn has a gravitatinal acceleratin cnstant f g Mn = 1.6 m/s, hw much will the rbt weigh sitting n the surface f the mn? Given: m Rbt = 60 kg ind: Rbt-Mn Relatinships: Slutin: = mg, Rbt-Mn =97. N g = 1.6 m/s Mn Rbt-Mn = mrbtgmn = = 97. N
3 Eample 3 Tw identical metal blcks, and, are hanging frm a rd. Cable E hlds blck t blck, and cable CD hlds blck t the rd. Each blck weighs 10 punds, and the sstem is in static equilibrium. ind the tensin in the tw cables. Given: = = 10 lb ; the sstem is in static equilibrium. ind: Tensins T and T. CD E C D E Relatinships: Slutin: Since there are fewer frces acting n the bttm blck, blck, begin the analsis there. s with s man ther things, its easier t start with smething simple and wrk twards the harder than the ther wa arund. Islate blck. Nte that there are tw frces acting n that blck, the blck's weight, and the tensin in cable E. Since the sstem is in static equilibrium, blck is in static equilibrium. This means the sum f all the frces acting n blck is equal t zer. T E cting n cting n TE = = Nte: ll the frces acting n blck act in the directin. e will still be using vectr math, but with nl ne dimensin, it will appear t be simple algebra. ccrding t the frame f reference I chse, tensin T E acts in the psitive directin; the weight f blck acts in the negative directin. Hence, the equatin T. E
4 Slving the equatin = T E T = = 10 lb E T = 10 lb E Nw, islate blck. Nte that there are three frces acting n that blck, the blck's weight, the tensin in cable CD, and the tensin in cable E. The weight f blck des nt act n blck, nl the tensin f cable E. T CD Since the sstem is in static equilibrium, blck is in static equilibrium. This means the sum f all the frces acting n blck is equal t zer. cting n cting n TCD TE = = gain, all the frces acting n blck act in the directin. Slving the equatin = T T CD E T = + T = lb CD E T lb CD T E
5 Eample 4 supprt member f a rbtic arm is laded as shwn. The member is made f a light weight cmpsite such that it's weight can be neglected ( Member ). e knw frm ther analsis that the frce = 77 N, and frce 0 N. Determine the magnitude f frces and C. The sstem is in static equilibrium. m 4 m 30 C Given: = 77 N, 0 N, Member ; the sstem is in static equilibrium. ind: Magnitudes and C. Relatinships: ppsite adjacent, sin θ =, csθ = hptenuse hptenuse Slutin: irst, nte that all f the frces shwn lie parallel t the given frame f reference with the eceptin f C. The first thing we shuld d in this prblem is determine the cmpnents f vectr C that lie in the and directins. This will simplif ur calculatins later. ppsite C sin θ = sin( 30 ) = C = Csin( 30 ) hptenuse C adjacent C cs θ = cs 30 = = cs 30 hptenuse C C C cmpleting this calculatin, all the frce vectrs we are wrking with lie in either the r directin. This means we can wrk the prblem in thse tw directins separatel. C 30 C 30 C dditinall, decmpsing vectr C int its cmpnents did nt add an additinal unknwn. There is still nl ne unknwn here, namel the magnitude f the vectr, C. The sine and csine functins are just scalar multiples f the magnitude f vectr C. C
6 fter decmpsing vectr C, the sstem simplifies t the fllwing C C m 4 m Given the sstem is in static equilibrium, we merel need t sum frces in the directin and set them equal t zer, then sum frces in the directin and set them equal t zer. Incidentall, there is n magic in the rder in which ne sums equatins. One culd sum first, then. Hwever, d nt mi cmpnents with cmpnents! T cntinue = C = C = 77 N Then C = 77 = Csin C = = 154 N sin 30 C =154 N = + C ( C ) 00 + cs 30 ( ) 0 154cs 30 = 66.6 N =66.6 N
7 Eample 5 lck has a mass f 5 kg. The spring has an unstretched length f 0.5 m ( L O.5 m ). The sstem is in static equilibrium in the psitin shwn t the right. Determine the spring cnstant, k, f the spring. Given: m = 5 kg, LO.5 m; ind: k. the sstem is in static equilibrium. 350 mm Relatinships:, S = k L LO, = mg Slutin: Since there are nl tw frces acting n blck, begin the analsis there. irst, determine the weight f blck S = mg = = N Since the sstem is in static equilibrium, the sum f all the frces acting n blck is equal t zer. cting n cting n = = = = N S S T determine the spring cnstant, we will use the spring equatin. Hwever, watch that u are cnsistent with units, in this case units f length. L = 350 mm.35 m Then = = = N = N m = N m S S k L LO k k L LO m
8 Eample 6 rbt will appl a frce,, t hld the 6 kg blck in place. The cefficient f static frictin is μ = S 0.4. Determine the largest frce,, the rbt can appl fr which the b will nt slip. Given: m lck = 6 kg, μ.4 ; S the sstem is in static equilibrium. 30 ind: Largest frce that can be applied withut causing b t slip. Relatinships:, f = μ S N, = mg Slutin: This is a frictin prblem. Since the blck will be abut t slip r mve, we will determine the nrmal frce, use it t determine the full frce f static frictin, and finall determine frce. That is the strateg. lck Nte the fllwing with regard t the diagram n the right: The frce f frictin pints dwn the ramp. Crrectl finding the directin f frictin is ke in crrectl slving these tpe f prblems. In this case f N (as alwas), frictin is NOT ppsing, but ppsing the mtin f the blck. If we are lking fr the largest frce, then that frce will tend t cause the blck t mve up the ramp. Since frictin ppses mtin, it will ppse the blck mving up the ramp. Hence, frictin pints dwn the ramp. The blck remains in its riginal rientatin, but the frame f reference has been skewed 30. Never change the rientatin f the bject u are analzing. e are accustmed t drawing the frce f weight straight dwn, fr that is where the center f the earth is. If u skew the rientatin f the bject, the weight vectr will nt pint straight dwn. Hwever, mst peple will frget and incrrectl draw that vectr. ll f the frces in the diagram lie parallel r perpendicular t the ramp with the eceptin f the weight vectr. It is quicker t rerient the frame f reference and reslve the weight vectr int cmpnents, than t use a hrizntal/vertical frame and reslve three vectrs int their cmpnents.
9 Decmpse the weight vectr int its cmpnents with regard t the given frame f reference. sin 30 sin 30 lck cs 30 cs 30 = = lck = = lck lck lck The frce f weight = m g = = N lck = sin 30 = 58.86sin 30 = 9.43 N lck lck = cs 30 = 58.86cs 30 = N lck The frce f frictin (recall, mtin is impending) f = μsn.4 N 30 f N lck The sstem is in equilibrium, s. Nte that nl tw frces lie in the directin, s start there. The slutin shuld be reached quicker. = N N = = N N = N f N = f = + 0.4N = = 49.8 N =49.8 N
10 Eample 7 The remaining eamples are fr u t cmplete. rbt must be able t push a blck with a mass f 80 kg. hat is the weight f the blck at sea level? Given: b has a mass f 80 kg. ind: The weight f the b. Relatinships: Slutin: = mg, g = 9.81 m/s nswer: = 785 N
11 Eample 8 rbt must be able t push a blck with a mass f 80 kg n the mn. hat is the weight f the blck n the surface f the mn? The gravitatinal cnstant fr the mn is g = 1.6 m/s. Mn Given: b has a mass f 80 kg. ind: The weight f the b n the mn. Relatinships: Slutin: = mg, g Mn = 1.6 m/s nswer: = 130 N
12 Eample 9 60 lb clinder rests in the crevasse f tw bards, inclined as shwn. hat are the nrmal frces eerted b the right and left bard n the clinder? Given: clinder weighing 60 lb. ind: The nrmals N Left and N. Right ppsite adjacent Relatinships: sin θ =, csθ = hptenuse hptenuse 30 nswer: N N Right Left 45 = 31.1 lb = 43.9 lb Slutin: r this eample, we'll give u a starting hint, the D f the clinder. N Left N Right
13 Eample 10 kg blck is resting n a smth surface inclined 30 as shwn. It is held ff anther surface b a spring with a spring cnstant f k = 50 N m. Hw far is the spring cmpressed? Given: m = lck kg, k = 50 N m. ind: The amunt f cmpressin δ. 30 Relatinships: = mg, Spring = kδ, g = 9.81 m s nswer: δ.196 m Slutin:
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