Ukulingana kwemizimba endizeni ethambekele – ukusetshenziswa kwezinkinga nezixazululo zomthetho wokuqala kaNewton

1. Ibhulokhi elingamakhilogremu angu-2 lilele endizeni ethambekele emaceleni nge-engeli engu-37 ^o kuya endaweni evundlile. Thola ubukhulu bamandla angaphandle asetshenziswa ebhulokhini, ukuze ibhulokhi lingasheleli phansi endizeni. (sin 37 ^o = 0.6, cos 37 ^o = 0.8, g = 10 ms ^-2 , µ _k = 0.2)

Kwaziwa:

Isisindo (m) = 2 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi (w) = mg = (2)(10) = 20 amaNewton

Isono 37 ^o = 0.6

I-Cos 37 ^o = 0.8

I-coefficient ye- kinetic friction (µ _k ) = 0.2

Ingxenye ka-y yesisindo (w _y ) = w cos 37 ^o = (20)(0.8) = 16 Newtons

Ingxenye engu-x yesisindo (w _x ) = w sin θ = (20)(sin 37) = (20)(0.6) = 12 Newtons

amandla avamile (N) = w _y = 16 Newtons

Okufunwayo : Amandla angaphandle (F)

Isixazululo :

w_x = 12 uNewton

Amandla okungqubuzana kwe-kinetic (f _k ) = µ _k N = (0.1)(16) = 1.6 AmaNewton

Ubukhulu bamandla angaphandle u-F asetshenziswe ebhulokini :

F + f _k – w _x = 0

F = w _x – f _k

F = 12 – 1.6

F = 10.4 Newton

Amandla angaphandle u-F makhulu kune-10.4 Newtons.

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2. Isisindo sebhulokhi = 2 kg, i-coefficient of static friction µ _s = 0.4 kanye no-θ = 45 ^o . Thola ubukhulu bamandla F ukuze ibhulokhi iqale ukushelela phezulu.

Kwaziwa:

I-coefficient yokungqubuzana okungaguquki (µs ₎ = 0.4

I-engeli (θ) = 45 ^o

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi (m) = 2 kg

Isisindo sebhulokhi (w) = mg = (2 kg)(10 m/s ² ) = 20 kg m/s ² = 20 Newtons

Ingxenye engu-x yesisindo (w _x ) = w sin θ = (20)(sin 45) = (20)(0.5√2) = 10√2 AmaNewton

Ingxenye engu-y yesisindo (w _y ) = w cos θ = (20)(cos 45) = (20)(0.5√2) = 10√2 AmaNewton

Okufunwayo : Ubukhulu bamandla F

Isixazululo:

Ibhulokhi iqala ukushelela phezulu, uma F ≥ w_x + f_s.

Ingxenye yesisindo engu-x:

w _x = 10√2 Newton

ingxenye engu-y yesisindo :

w _y = 10√2 Newton

Amandla avamile :

N = w _y = 10√2 Newton

Amandla okungqubuzana okungaguquki :

f _s = µ _s N = (0,4)(10√2) = 4√2

Ubukhulu bamandla F ukuze ibhloko liqale ukushelela phezulu :

F ≥ w _x + f _s

F ≥ 10√2 + 4 √2

F ≥ 14√2 Newton

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1. Izinhlayiya ezilingana ngokulingana okukodwa
2. Izinhlayiya ezilingana ngezindlela ezimbili
3. Ukulingana kwemizimba exhunywe ngezintambo nama-pulley
4. Ukulingana kwemizimba endizeni ethambekele