Imibuzo emi-3 mayelana ne-Rope tension equation
1. Isithombe esingezansi sibonisa amabhlogo amathathu, okungu-A, B no-C atholakala endizeni ebushelelezi evundlile. Uma isisindo A = 1 kg, isisindo B = 2 kg kanye nesisindo C = 2 kg kanye no-F = 10 N, khona-ke nquma isilinganiso sokucindezeleka entanjeni phakathi kuka-A no-B nokucindezeleka entanjeni phakathi kuka-B no-C.
Kwaziwa:
Isisindo se-A (m A ) = 1 kg
Isisindo B (m B ) = 2 kg
Isisindo se-C (mC ) = 2 kg
Amandla okudonsa (F) = 10 N
Kufunwa: T AB : T BC
Isixazululo:
Bala ukusheshisa kwesistimu usebenzisa ifomula kaNewton's Second Law:
ΣF = ma
F = (m A + m B + m C ) a
10 = (1 + 2 + 2) a
10 = 5 a
a = 10/5
a = 2 m/s 2
Sebenzisa ifomula yokucindezela intambo ukuze ubale i-T AB
ΣF = ma
T AB = m A a = 1 (2) = 2 amaNewton
Sebenzisa ifomula yokucindezela intambo ukuze ubale i-T BC
ΣF = ma
T BC = (m A + m B ) a = (1 + 2) (2) = (3)(2) = 6 AmaNewton
2. Into A enesisindo esingu-6 kg kanye nento B enesisindo esingu-3 kg zixhunywe ngentambo njengoba kuboniswe. Uma i-coefficient of friction ingu-0.3 kanye no-g = 10 m/s 2 , nquma ukusheshisa kwento kanye nokucindezeleka ezintanjeni zebhulokhi ngayinye.
Kuyaziwa:
Isisindo sento A (m A ) = 6 kg
Isisindo sento B (m B ) = 3 kg
I-Coefficient yokungqubuzana kwebhulokhi A (µk ) = 0.3
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Isisindo sebhulokhi A (w A ) = m A g = (6)(10) = 60 N
Amandla ajwayelekile kubhulokhi A (N A ) = w A = 60 N
Isisindo sebhulokhi B (w B ) = m B g = (3)(10) = 30 N
Okufunwayo: Ukusheshisa kwesistimu (a) kanye nokucindezeleka entanjeni (T)
Isixazululo:
Bala amandla okungqubuzana kwe-kinetic okungukuthi amandla okungqubuzana lapho ibhulokhi A lihamba:
F k = µ k N A = (0,3)(60) = 18 Amathoni amasha
Bala ukusheshisa kwesistimu (a):
ΣF = ma
wB – F k = (m A + m B ) a
30 – 18 = (6 + 3) a
12 = 9 a
a = 12 / 9 = 1,3 m/s 2
Bala ukucindezeleka kwentambo kubhulokhi A (T A ):
ΣF = ma
T A – F k = (m A ) a
I-T A – 18 = (6)(1,3)
I-T A – 18 = 7,8
TA = 7,8 + 18 = 25,8 AmaNewton
Bala ukucindezeleka entanjeni ku-beam B (T B ):
ΣF = ma
w B – T B = m B (a)
30 – T B = 3 (1,3)
30 – T B = 3,9
I-T B = 30 – 3,9
T B = 26,1 amaNewton
3. Izinto ezimbili u-A no-B ezinesisindo esingu-5 kg kanye no-3 kg zixhunywe nge-pulley engenakuphikiswana. Amandla u-P asetshenziswa ku-pulley ngokuya phezulu. Uma amabhlogo womabili eqale ephumule phansi, kuyini ukusheshisa kwebhlokhi u-A, uma ubukhulu buka-P bungu-60 N?
Thola futhi ukucindezeleka entanjeni kumabhulokhi A no-B.
Kuyaziwa:
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Isisindo se-A (m A ) = 5 kg
Isisindo sebhulokhi A (w A ) = m A g = (5)(10) = 50 AmaNewton
Isisindo B (m B ) = 3 kg
Isisindo sebhulokhi B (w B ) = m B g = (3)(10) = 30 Newtons
Amandla P = 60 N
Okufunwayo: Ukusheshisa uhlelo lwemishayo u-A no-B (a) kanye nokucindezeleka entanjeni kumishayo u-A (T A ) kanye nomshayo u-B (T B )
Isixazululo:
Bala ukusheshisa kwesistimu usebenzisa ifomula yoMthetho Wesibili kaNewton.
ΣF = ma
w A – w B = (m A + m B ) a
50 – 30 = (5 + 3) a
20 = 8 a
a = 20/8
a = 2,5 m/s 2
Sebenzisa ifomula yamandla okucindezela ukuze ubale ukucindezela entanjeni
Ukucindezeleka entanjeni ebhulokini A:
ΣF = ma
w A – T A = m A a
50 – TA = 5 (2,5)
50 – TA = 12,5
I-T A = 50 – 12,5 = 37,5 AmaNewton
Ukucindezeleka entanjeni ebhulokini B:
ΣF = ma
T B – w B = m B a
I-T B – 30 = 3 (2,5)
I-T B – 30 = 7,5
I-T B = 7,5 + 30 = 37,5 AmaNewton
