KINETICS OF RIGID BODIES PROBLEMS
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1 KINETICS OF RIID ODIES PROLEMS
2 PROLEMS 1. The 6 kg frme C nd the 4 kg uniform slender br of length l slide with negligible friction long the fied horizontl br under the ction of the 80 N force. Clculte the tension T in wire C nd the nd components of the force eerted on the br b the pin t. The - plne is verticl. (6/13)
3 PROLEMS rectiliner trnsltion F m M 0 for the whole sstem F m 80 (6 4), 8 m/ s m C = 6 kg, m = 4 kg, clculte the tension T in wire C nd the nd components of the force eerted on the br b the pin t. m g = 4(9.81) = 39.4 N 80 N m C g = 6(9.81) = N FD Ξ m C KD m br F 0 + 3l l M C md 39.4 l 4(8) sin , N, T 7.33 F m, 7.33cos T sin 39.4 T sin 60 4(8), N m g = 39.4 N N FD T l / 60 Ξ l / l / l / KD m l / sin 60 d
4 PROLEMS. The block nd ttched rod hve combined mss of 60 kg nd re confined to move long the 60 guide under the ction of the 800 N pplied force. The uniform horizontl rod hs mss of 0 kg nd is welded to the block t. Friction in the guide is negligible. Compute the bending moment M eerted b the weld on the rod t. (6/)
5 SOLUTION FD Kinetic Digrm m T =60 m totl =60 kg, m rod = 0 kg, compute the bending moment M eerted b the weld on the rod t. N 60 o W=60(9.81) N F 4.84 m m / s (9.81) sin FD of rod KD of rod 60 m 1 =0 M W 1 =0(9.81) N M M m d N m M 0(9.81)0.7 (0)(4.84sin 60)(0.7)
6 PROLEMS 3. The uniform 100 kg log is supported b the two cbles nd used s bttering rm. If the log is relesed from rest in the position shown, clculte the initil tension induced in ech cble immeditel fter relese nd the corresponding ngulr ccelertion of the cbles.
7 SOLUTION +n m=100 kg, log relesed from rest, clculte the initil tension in ech cble nd corresponding ngulr ccelertion of the cbles. +n T T mn t r W=100(9.81) N +t FD.45 rd / s m t When it strts to move, v=0, w = 0 but 0 n w r 0 Fn 0 T T mg cos30 0 T T Ft mt mgsin 30 mt t m / s T M N 1.5 m 0.5 m The motion of the log is curviliner trnsltion. T T sin 60(1.5) T N sin 60(0.5) 0 3T +t T KD * *
8 PROLEMS 4. The prllelogrm linkge is used to trnsfer crtes from pltform to pltform nd is hdrulicll operted. The oil pressure in the clinder is progrmmed to provide smooth trnsition of motion from q = 0 to q = q 0 = p/3 rd given b p p t q 1 cos where t is in seconds. Determine the force t D on the pin when t = 1 s. The crte nd pltform hve combined mss of 00 kg with mss center t. The mss of ech link is smll nd m be neglected. (6/9) 6
9 PROLEMS p q = 0 to q = q 0 = p/3 rd, p t q 1 cos t D for t = 1 s. 6, m totl =00 kg, determine the force m g = 00(9.81) = 196 N t 0 F t F D n F n Ξ m n FD KD r CD1.m 3 q p 1 cos t, sin t, cos t p q p p q p p t 1s, q p rd, q p rd / s, q , m mr q 00(1.)( p /144) 16.6 m/ s t M+ F m d 196(0.6) (0.6cos30) FD 16.6(0.6cos sin 30) F 178 N ( compression) D n n q =30 q =30 q =30
10 PROLEMS 5. The spring is uncompressed when the uniform slender br is in the verticl position shown. Determine the initil ngulr ccelertion of the br when it is relesed from rest in position where the br hs been rotted 30 clockwise from the position shown. Neglect n sg of the spring, whose mss is negligible.
11 PROLEMS Spring uncompressed when uniform slender br in verticl position, determine initil ngulr ccelertion of br when relesed from rest in position where the br hs been rotted 30 clockwise from the position shown. Unstrecthed length of the spring: 5 l o (l / 4) l l O t W +t 60 o O +n O n 30 o. 60 o FD F spring l l spring 30 o When q=30 o, length of the spring: When q=30 o, spring force: +t m t +n m n KD I F spring mrw 3 l spring 5 k l + M I m d O mgcos60 l F mw k m l 0 4 t l 3 spring l kl l g l (in compression) ml m l t 4 l 4
12 PROLEMS 6. The 65 kg thin rod is held b cbles nd C. If cble C suddenl breks loose determine the initil ngulr ccelertion of the rod nd the tension in cble. 30 cm C 40 cm 40 cm
13 PROLEMS / i I m = 65 kg, cble C suddenl breks loose, determine the initil ngulr ccelertion of the rod nd the tension in cble. / ww r / r / 0 ( strts fromrest) k 0.3 j 0.4i 0.3 i 0.4 j /, / wc w C r / C r 0 ( strts fromrest) Ck 0.4i 0.4 C j i j 0.3 i 0.4 j 0.4 j 1 1 ml (0.8) j 3.47 C 0.4 kgm C / C / 0.4 C C t t 40 cm n n / n 0 n t t 40 cm r r / / w 0 30 cm
14 PROLEMS 0.6T T M F F + I m m T T N C, m = 65 kg, cble C suddenl breks loose, determine the initil ngulr ccelertion of the rod nd the tension in cble. 3 T(0.4) 3.47C, C T 5 4 T , 0.8T 19.5, 5 3 T 65(9.81) C T 7.53 rd / s,, C 3.473T C rd / s 0.041T C C 40 cm T 40 cm 30 cm Ξ C I C 40 cm m 30 cm 40 cm mg FD m KD
15 PROLEMS 7. Crnk rottes with n ngulr velocit of w = 6 rd/s nd ngulr ccelertion of = rd/s. Roller C cn slide long the circulr slot within the fied plte. For the position shown, ngulr velocit nd ngulr ccelertion of rod C re w C =.9 rd/s (counterclockwise) nd C =37.47 rd/s (counterclockwise) The msses of uniform brs nd C re m = kg nd m C =5 kg. Mss of the roller C nd friction cn be neglected. Determine the rections b the pins nd C. w C =.9 rd/s (ccw) m C =5 kg, C =37.47 rd/s (ccw) L =300 mm L C =500 mm w = 6 rd/s = rd/s m = kg r =150 mm 45 o 37 o C
16 PROLEMS Mss of the roller C nd friction cn be neglected. Determine the rections b the pins nd C. w C =.9 rd/s (ccw) m C =5 kg, C =37.47 rd/s (ccw) L =300 mm L C =500 mm w = 6 rd/s = rd/s m = kg r =150 mm 45 o 37 o C
17 PROLEMS 8. In the mechnism shown, member is being rotted with constnt ngulr velocit of w = 10 rd/s b torque (not shown in the figure). Member sets member C in motion (mss of member C is 6 kg), which then cuses ger D with mss of 3 kg to move. The rdius of grtion of the ger with respect to center C is 00 mm. The rdius of the ger is given s r = 50 mm. For the instnt shown, determine the forces cting on pins C nd. w
18 PROLEMS w = 10 rd/s (cst), m C = 6 kg, m D = 3 kg, k ger = 00 mm, r = 50 mm, Determine forces cting on pins C nd. w
19 PROLEMS 9. The nonhomogeneous 0 kg wheel with the mss center t hs rdius of grtion bout of 0 mm. The wheel rolls down the 0 o rough incline without slipping. In the position shown, the wheel hs n ngulr velocit of 3 rd/s. Clculte the norml force nd the friction force cting on the wheel from the surfce t this position mm 50 mm
20 SOLUTION O FD mg F f enerl Motion N = KD m O / O 0.5i k 0.075i i j F m mgsin 0 Ff F f F m N mg cos N M I N( 0.075) Ff (0.5) I I k m 0(0.0) kgm o r k 3 k 0.075i F N 75 mm rd / s N N 50 mm
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