JAWABAN SOAL LATIHAN MEKANIKA TEKNIK (1)

JAWABAN SOAL LATIHAN MEKANIKA TEKNIK (1)

1 (a).

Cara grafis...

Cara analisis...

Diketahui sebagai berikut :

A = 40 N ; α_a = 20⁰

B = 60 N ; α_b = 20⁰ + 25⁰ = 45⁰

Maka :

Ax = A cos α_a = 40 cos 20 = 37,59 N

Ay = A sin α_a = 40 sin 20 = 13,68 N

Bx = B cos α_b = 60 cos 45 = 42,43 N

By = B sin α_b = 60 sin 45 = 42,43 N

1 (b).

Cara grafis

Cara analisis

Diketahui sebagai berikut :

A = 450 lb ; α_a = 35⁰

B = 300 lb ; α_b = 40⁰ + 35⁰ = 75⁰

Maka :

Ax = A cos α_a = 450 cos 35 = 368,62 lb

Ay = A sin α_a = 450 sin 35 = 258,11 lb

Bx = B cos α_b = 300 cos 75 = 77,64 lb

By = B sin α_b = 300 sin 75 = 289,78 lb

1. (c)

Cara grafis...

Cara analisis....

Diketahui sebagai berikut :

A = 5 kN ; α_a = 65⁰

B = 3,5 kN ; α_b = 30⁰

Maka :

Ax = A cos α_a = 5 cos 65 = 2,1131 kN

Ay = A sin α_a = 5 sin 65 = 4,5315 kN

Bx = B cos α_b = 3,5 cos 30 = 3,0311 kN

By = B sin α_b = 3,5 sin 30 = 1,7500 kN (-, arah kebawah)

2 a.

Diketahui :

F = 500 N; α = 35⁰

Ditanyakan komponen gaya Fx dan Fy...?

Fx = F cos 35 = 500 cos 35 = 409,5760 N

Fy = F sin 35 = 500 sin 35 = 286,7882 N

2b.

Diketahui :

A = 60 lb ; α_a = 35⁰

B = 45 lb ; α_b = 20⁰ + 35⁰ = 55⁰

C = 75 lb ; α_c = 50⁰

Ax = A cos α_a = 60 cos 35 = 49,15 lb

Ay = A sin α_a = 60 sin 35 = 34,41 lb

Bx = B cos α_b = 45 cos 55 = 25,81 lb

By = B sin α_b = 45 sin 55 = 36,86 lb

Cx = C cos α_c = 75 cos 50 = 48,21 lb

Cx = C sin α_c = 75 sin 50 = 57,45 lb

∑Fx = Ax + Bx + Cx = 49,15 + 25,81 + 48,21 = 123,17 lb

∑Fy = Ay + By – Cy = 34,41 + 36,86 – 57,45 = 13,82 lb

2c.

Diketahui :

A = 350 N ; α_a = 25⁰

B = 800 N ; α_b = 45⁰ + 25⁰ = 70⁰

C = 600 N ; α_c = 60⁰

Ax = A cos α_a = 350 cos 25 = 317,21 N

Ay = A sin α_a = 350 sin 25 = 147,92 N

Bx = B cos α_b = 800 cos 70 = 273,62 N

By = B sin α_b = 800 sin 70 = 751,75 N

Cx = C cos α_c = 600 cos 60 = 300,00 N

Cy = C sin α_c = 600 sin 60 = 519,62 N

∑Fx = Ax + Bx – Cx = 317,21 + 273,62 – 300,00 = 290,83 N

∑Fy = Ay + By – Cy = 147,92 + 751,75 – 519,62 = 380,05 N

2d.

Diketahui :

α_a = 50⁰ dan α_b = 30⁰

Besar beban = 400 lb

Pertanyaanya berapa besar tegangan pada tali AC dan tali BC...?

Y_A = Y_B = Y

Y_A = AC sin 50  maka AC = Y_A / sin 50

Y_B = BC sin 30 maka BC = Y_B / sin 30

  

Y = (400 x 0,3830)/ 1,2660

Y = 121,0111

AC = Y / sin 50 = 121,0111 / sin 50 = 157,9687 lb

BC = Y / sin 30 = 121,0111 / sin 30 = 242,0221 lb

AC + BC = 400

157,9687 + 242,0221 = 399,9908 ~ 400 lb

2e.

Tg α_a = 450/280= 58,109208⁰

Tg α_b = 450/600 = 36,869898⁰

AC + BC = 330 N

X_A = AC sin 58,109208

X_B = BC sin 36,869898

X_A = X_B = X

X = (330 x 0,509434 )/1,449057

X = 116,015602

AC = X / sin 58,109208 = 116,015602 / sin 58,109208 = 136,640598 N

BC = X / sin 36,869898 = 116,015602 / sin 36,869898 = 193,359335 N

AC + BC = 330

136,640598 + 193,359335 = 329,999933 N ~ 330....Ok...

Jika ada yang mau koreksi silahkan untuk saling mengisi...terimakasih semoga bermanfaat....