Bila diketahui m_A = 20 kg ,m_B = 40 kg, dan koefisien gesekan 0,25. Hitunglah Percepatan m_A dan m_B !. ( Massa dan gesekan kedua katrol diabaikan).

Jawaban :

Benda A

ƩFy = 0

N - W_A = 0

N = M_A  x g……………………………………..(1)

ƩFx = M_A x a_A

T₁ – f_ges = M_A x a_A

T₁ = (µ_ges  M_A  x g )+( M_A x a_A )…………..........(2)

Katrol 2

ƩF_katrol = M_katrol x a_katrol

T₂’ – 2T₁ = 0

Dimana T₂’ = T₂

T₂ = 2T₁………………………………………….(3)

Benda B

ƩFy = M_B x a_B

W_B – T₂ = M_B x a_B

W_B – 2T₂ = M_B x a_B

Dimana

T₂ = (W_B – (M_B x a_B)):2 …………………..(4)

                 (2)…………………..(4)

                             T₂ = T₁

(W_B – (M_B x a_B)):2 = (µ_ges  M_A  x g )+( M_A x a_A )

(40* 10 – (40*a_B)):2 = (0,25*20*10 )+( 20*a_A )

(400 - 40 a_B): 2 = 50 +20 a_A

Karena T₂ = 2T_1, maka a_B = a_A : 2

400 – 20 a_A = (50 +20 a_A)*2

400 – 20 a_A = 100 +40 a_A

a_A = 300:60

a_A = 5 m/s²

a_B = a_A : 2

a_B = 2,5 m/s²

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