1. Ob lub pawg m 1 = 2 kg thiab m 2 = 5 kg nyob rau ntawm lub dav hlau inclined thiab txuas ua ke los ntawm ib txoj hlua raws li pom hauv daim duab. Tus coefficient ntawm kinetic friction ntawm m 1 thiab incline yog 0.2 thiab tus coefficient ntawm kinetic friction ntawm m 2 thiab incline yog 0.1.
(a) Txheeb xyuas lawv qhov kev nrawm
(b) Txheeb xyuas lub zog nruj
Paub:
Qhov hnyav 1 (m 1 ) = 2 kg
Qhov hnyav 2 (m2 ) = 4 kg
Coefficient ntawm kinetic sib txhuam ntawm m 1 thiab inclined plane (μ k1 ) = 0.2
Coefficient ntawm kinetic sib txhuam ntawm m 2 thiab inclined plane (μ k2 ) = 0.1
Kev nrawm vim lub ntiajteb txawj nqus (g) = 9.8 m/s 2
a) Qhov loj thiab kev coj ntawm qhov kev nrawm
w 1 = qhov hnyav 1 = m 1 g = (2 kg)(9.8 m/s 2 ) = 19.6 Newtons
w 1x = w 1 sin 30 o = (19.6 N)(0.5) = 9.8 Newtons
w 1y = w 1 cos 30 o = (19.6 N)(0.87) = 17 Newtons
N 1 = Lub zog ib txwm muaj rau m 1 = w 1y = 17 Newtons
F k1 = Lub zog ntawm kev sib txhuam kinetic ntawm m 1 = μ k1 N 1 = (0.2)(17 N) = 3.4 Newtons
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w 2 = qhov hnyav 2 = m 2 g = (4 kg) (9.8 m/s 2 ) = 39.2 Newtons
w 2x = w 2 sin 60 o = (39.2 N)(0.87) = 34.1 Newtons
w 2y = w 2 cos 60 o = (39.2 N)(0.5) = 19.6 Newtons
N 2 = Lub zog ib txwm muaj rau m 2 = w 2y = 19.6 Newtons
F k2 = Lub zog ntawm kev sib txhuam kinetic ntawm m 2 = μ k2 N 2 = (0.1)(19.6 N) = 1.96 Newtons
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Qhov loj ntawm kev nrawm:
∑ F x = ma x
w 2x > w 1x yog li ntawd qhov kev taw qhia ntawm qhov kev nrawm yog tib yam li qhov kev taw qhia ntawm w 2x.
Cov zog uas taw qhia raws qhov kev nrawm yog cov zoo thiab cov zog uas muaj kev taw qhia sib txawv rau qhov kev nrawm yog cov tsis zoo.
w2x - Fk2 - T2 +T1 - w1x - Fk1 = (m1 +m2) thiabx
w 2x – F k2 – w 1x – F k1 = (m 1 + m 2 ) a x
34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) a x
18.94 N = (6 kg) a x
a x = 18.94 N : 6 kg
a x = 3.16 m/s 2
Qhov loj ntawm qhov kev nrawm = 3.16 m/s 2. Kev taw qhia ntawm qhov kev nrawm = kev taw qhia ntawm T 1 = kev taw qhia ntawm w 2x
b) Qhov loj ntawm lub zog nruj
Siv Newton txoj cai thib ob rau ntawm yam khoom 2:
w 2x – F k2 – T 2 = m 2 a x
34.1 N – 1.96 N – T 2 = (4 kg)(3.16 m/s 2 )
32.14 N – T 2 = 12.64 N
T 2 = 32.14 N – 12.64 N = 19.5 Newtons
Lub zog nruj = T = T 1 = T 2 = 19.5 Newtons
[irp]
2. m 1 = 4 kg, m 2 = 2 kg. Txheeb xyuas (a) qhov loj thiab kev coj ntawm qhov kev nrawm (b) Qhov loj ntawm lub zog nruj uas txuas m 1 thiab m 2 (c) qhov loj ntawm lub zog nruj uas txuas lub pulley thiab lub ru tsev.
tshuaj
w 1 = m 1 g = (4 kg)(9.8 m/s 2 ) = 39.2 Newtons
w 2 = m 2 g = (2 kg)(9.8 m/s 2 ) = 19.6 Newtons
a) Qhov loj thiab kev coj ntawm kev nrawm
∑ F y = ma y
w 1 > w 2 yog li ntawd qhov kev taw qhia ntawm yam khoom yog tib yam li qhov kev taw qhia ntawm qhov hnyav 1 ( w 1 ) . Cov zog uas muaj tib qho kev taw qhia li kev nrawm yog qhov zoo thiab cov zog uas muaj qhov kev taw qhia sib txawv nrog kev nrawm yog qhov tsis zoo.
w 1 – T 1 + T 2 – w 2 = (m 1 + m 2 ) a y
w 1 – w 2 = (m 1 + m 2 ) ay
39.2 N – 19.6 N = (4 kg + 2 kg) ib y
19.6 N = (6 kg) ib xyoos
ay = 19.6 N : 6 kg
ay = 3.26 m/s 2
Qhov loj ntawm kev nrawm = 3.26 m/s 2. Kev taw qhia ntawm kev nrawm = kev taw qhia ntawm w 1.
b) Qhov loj ntawm lub zog nruj uas txuas m 1 thiab m 2
Siv Newton txoj cai thib ob rau m2 :
∑ F y = ma y
w 1 – T 1 = m 1 a y
39.2 N – T 1 = (4 kg)(3.26 m/s 2 )
39.2 N – T 1 = 13.04 N
T 1 = 39.2 N – 13.04 N
T 1 = 26.16 Newtons
Qhov loj ntawm lub zog nruj uas txuas cov khoom = T = T 1 = T 2 = 26.16 Newtons
c) Qhov loj ntawm lub zog nruj uas txuas lub pulley thiab lub ru tsev.
Pulley so:
∑ F y = ma y —— a y = 0
∑ Fy = 0
Cov zog sab saud yog cov zoo, cov zog sab hauv yog cov tsis zoo:
T 3 – T 1 – T 2 = 0
T3 = T1 + T2
T 1 thiab T 2 muaj tib qhov loj , T 1 = T 2 = T = 26.16 N:
T 3 = 2T = 2(26.16 N) = 52.32 Newtons
[irp]
3. Thaiv 1 (m 1 = 10 kg) thiab thaiv 2 (m 2 = 15 kg) txuas nrog ib txoj hlua hla lub pulley tsis muaj kev sib txhuam. Coefficient ntawm kev sib txhuam zoo li qub ntawm thaiv 2 nrog incline = 0.6. Coefficient ntawm kev sib txhuam kinetic ntawm thaiv 2 nrog incline = 0.42. Txheeb xyuas (a) Qhov loj ntawm lub zog tsawg kawg nkaus F siv rau cov khoom kom cov khoom nrawm dua (b) Txheeb xyuas qhov loj ntawm lub zog nruj.
tshuaj
w 1 = Qhov hnyav ntawm lub thaiv 1 = m 1 g = (10 kg) (9.8 m/s 2 ) = 98 Newtons
w 2 = Qhov hnyav ntawm lub thaiv 2 = m 2 g = (15 kg) (9.8 m/s 2 ) = 147 Newtons
w 2y = w 2 cos 30 o = (147 N)(0.87) = 127.89 Newtons
w 2x = w 2 sin 30 o = (147 N)(0.5) = 73.5 Newtons
N 2 = Lub zog ib txwm muaj rau ntawm lub thaiv 2 = w 2y = 127.89 Newtons
F k2 = Lub zog ntawm kev sib txhuam kinetic ntawm lub block 2 = μ k2 N 2 = (0.42)(127.89 N) = 53.7 Newtons
F s2 = Lub zog ntawm kev sib txhuam zoo li qub ntawm lub thaiv 2 = μ s2 N 2 = (0.6)(127.89 N) = 76.7 Newtons
a) Qhov loj ntawm lub zog tsawg kawg nkaus F siv rau cov khoom kom cov khoom nrawm dua
∑ F x = ma x —— a x = 0
∑ F x = 0
Cov zog sab saud thiab cov zog sab xis yog cov zoo, cov zog hauv qab thiab cov zog sab laug yog cov tsis zoo.
F – F k2 – w 2x – w 1 – T 2 + T 1 = 0
F – F k2 – w 2x – w 1 = 0
F = F k2 + w 2x + w 1
F = 53.7 N + 73.5 N + 98 N
F = 225.2 Newtons
b) Qhov loj ntawm lub zog nruj
Siv Newton txoj cai ntawm kev txav mus los rau ntawm thaiv 1:
∑ F y = ma y —— a y = 0
∑ Fy = 0
T 1 – w 1 = 0
T 1 = w 1 = 98 Newtons
Siv Newton txoj cai ntawm kev txav mus los rau ntawm thaiv 2:
F – F k2 – w 2x – T 2 = 0
T 2 = F – F k2 – w 2x
T 2 = 225.2 N – 53.7 N – 73.5 N
T 2 = 98 Newtons
Qhov loj ntawm lub zog nruj = T 1 = T 2 = T = 98 Newtons
[irp]
4. Thaiv 1 (m 1 = 16 kg) nyob rau ntawm ib qho chaw kab rov tav thiab thaiv 2 (m 2 = 12 kg) nyob rau ntawm ib qho chaw du thiab nkhaus, txuas nrog los ntawm ib txoj hlua uas hla dhau ib lub pulley me me, tsis muaj kev sib txhuam. Thaiv 3 (m 3 = 5 kg) nyob rau ntawm thaiv 2. Tus coefficient ntawm kev sib txhuam kinetic ntawm thaiv 2 thiab qhov chaw kab rov tav yog 0,4. Tus coefficient ntawm kev sib txhuam static ntawm thaiv 2 nrog thaiv 3 yog 0,3.
(a) Thaum lub kaw lus raug tso tawm ntawm qhov chaw so, lub thaiv 3 thiab lub thaiv 2 tseem swb ua ke?
(b) Yog tias muaj thaiv 3, qhov kev nrawm ntawm thaiv 1 thiab thaiv 2 yog dab tsi?
Tshuaj:
a) Thaum lub kaw lus raug tso tawm ntawm qhov chaw so, lub thaiv 3 thiab lub thaiv 2 tseem swb ua ke?
w 1 = Qhov hnyav ntawm lub thaiv 1 = m 1 g = (16 kg) (9.8 m/s 2 ) = 156.8 Newtons
w 1x = w 1 sin 60 o = (156.8 N)(0.87) = 136.4 Newtons
w 1y = w 1 cos 60 o = (156.8 N)(0.5) = 78.4 Newtons
N 1 = Lub zog ib txwm uas siv rau ntawm lub thaiv 1 los ntawm lub dav hlau inclined = w 1y = 78.4 Newtons
w 3 = Qhov hnyav ntawm lub thaiv 3 = m 3 g = (5 kg) (9.8 m/s 2 ) = 49 Newtons
N 23 = Lub zog ib txwm uas lub thaiv 2 siv rau ntawm lub thaiv 3 = w 3 = 49 Newtons
N 32 = Lub zog ib txwm uas lub thaiv 3 siv rau ntawm lub thaiv 2 = N 23 = w 3 = 49 Newtons
(N 23 thiab N 32 yog khub ua haujlwm-ua haujlwm )
F s23 = Lub zog ntawm kev sib txhuam static uas siv rau ntawm lub thaiv 3 los ntawm lub thaiv 2 = μ s N 23 = (0.3)(49 N) = 14.7 Newtons
F s32 = Lub zog ntawm kev sib txhuam static uas siv rau ntawm lub thaiv 2 los ntawm lub thaiv 3 = F s 23 = 14.7 Newtons
(F s23 thiab F s32 yog khub ua haujlwm-ua haujlwm )
w 2 = Qhov hnyav ntawm lub thaiv 2 = m 2 g = (12 kg) (9.8 m/s 2 ) = 117.6 Newtons
N 2 = Lub zog ib txwm uas siv rau ntawm yam khoom 2 los ntawm qhov chaw kab rov tav = w 2 + N 32 = 117.6 Newtons + 49
Newton = 166.6 Newton
F k2 = Lub zog ntawm kev sib txhuam kinetic ntawm lub block 2 = μ k N 2 = (0.4)(166.6 N) = 66.64 Newtons
Siv Newton txoj cai ntawm kev txav mus los rau ntawm lub block 3:
∑ F x = ma x
F s23 = m 3 a x
—–> Fs23 = μs N23 = μs w3 = μs m3 g
μsm3g = m3ax
μsg = a x
a x = (0.3)(9.8 m/s2 ) = 2.94 m/ s2
Qhov kev nrawm siab tshaj plaws ntawm lub thaiv 3 kom lub thaiv 3 thiab lub thaiv 2 tseem swb ua ke yog 2.94 m/ s2.
Tam sim no peb xam qhov loj ntawm lub kaw lus qhov kev nrawm tom qab raug tso tawm ntawm qhov so.
Qhov kev taw qhia ntawm qhov kev hloov chaw ntawm lub block = qhov kev taw qhia ntawm qhov kev nrawm ntawm lub block = qhov kev taw qhia ntawm T2 = qhov kev taw qhia ntawm w1x.
∑ F x = ma x
w1x - T1 +T2 - Fk2 - Fs32 +Fs23 = (m1 +m2 +m3) thiabx
w 1x – F k2 = (m 1 + m 2 + m 3 ) a x
136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) a x
69.76 N = (33 kg) a x
a x = 2.11 m/s 2
a x yog qhov zoo, txhais tau tias qhov kev taw qhia ntawm qhov kev hloov chaw ntawm lub block lossis qhov kev taw qhia ntawm qhov kev nrawm yog tib yam li qhov kev taw qhia ntawm T2 lossis qhov kev taw qhia ntawm w1x.
Qhov loj ntawm qhov kev nrawm yog 2.11 m/s2 , qis dua 2.94 m/s2 yog li peb tuaj yeem xaus lus tias thaiv 3 thiab thaiv 2 tseem swb ua ke tom qab raug tso tawm ntawm qhov chaw so.
b) Qhov loj ntawm qhov kev nrawm ntawm lub thaiv 1 thiab lub thaiv 2
∑ F x = ma x
w 1x – F k2 = (m 1 + m 2 ) a x
—–> Fk2 = μk N2 = μk w2 = μk m2 g = (0.4)(12 kg)(9.8 m/s)2) = 47.04 Newtons
136.4 N – 47.04 N = (16 kg + 12 kg) a x
89.36 N = (28 kg) a x
a x = 89.36 N : 28 kg = 3.19 m/s 2
[wpdm_package id='493′]
- Qhov hnyav thiab qhov hnyav
- ib txwm muaj zog
- Newton txoj cai thib ob ntawm kev txav mus los
- Lub zog sib txhuam
- Kev txav mus los ntawm qhov chaw kab rov tav yam tsis muaj kev sib txhuam
- Kev txav ntawm ob lub cev nrog tib qhov kev nrawm ntawm qhov chaw ntxhib kab rov tav nrog lub zog sib txhuam
- Kev txav mus los ntawm lub dav hlau inclined tsis muaj lub zog sib txhuam
- Kev txav mus los ntawm lub dav hlau inclined nrog lub zog sib txhuam
- Kev txav mus los hauv lub elevator
- Kev txav ntawm lub cev yog txuas nrog los ntawm cov hlua thiab pulleys
- Ob lub cev uas muaj tib lub zog ntawm kev nrawm
- Kev sib hloov ntawm ib txoj kab tiaj tiaj - dynamics ntawm kev txav mus los ncig
- Kev sib hloov ntawm cov kab nkhaus - dynamics ntawm kev txav mus los ncig
- Kev txav mus los sib xws hauv lub voj voog kab rov tav
- Lub zog centripetal hauv kev txav mus los sib npaug
