Fisik11

SOLUCION QUIZ FISICA MECANICA

Primer Punto
T1x =-T1cos 45
T1y = T1sen 45
T2 = 100N
T3x = T3cos 90
T3y = T3sen 90

∑Fx = m.ax
-T1cos 45 + T3cos 90 = 0

∑Fy = m.ayT1sen 45 – T3cos 90 – T2 = 0
T1sen 45 – T3cos 90 = T2

De 1 despejo T3
T3 = T1cos 45
Cos 90

Remplazo T3 en 2

T1sen 45 – T1cos 45.sen 90 = 100N
Cos90
T1sen45.cos90 – T1cos45.sen90 = 100N
Cos 90

T1(sen 45.cos 90 – cos 45.sen 90) = 100N.cos 90

T1 = 100N.cos 90
sen 45.cos 90 – cos 45.sen 90

T1 = 24.13N

RemplazoT1 en T3

T3 = T1cos 45
Cos 90

Formula para T4 según el diagrama de cuerpo libre
-T1cos 45 + T3cos 90 + T4cos60 = 0

Despejo T4

T4 = T1cos 45 – T3cos 90
Cos 60

T4=18.34 – 16.74 = 2.75N
0.58

P1 = T3 + T4 = 109.81N

Segundo Punto

∑fx = m.a
T1 - Fr - P1x = m.a
T1 - Fr – m.gSenᶿ= m.g
T1 = m.a.t µ m1 g cosᶿ

∑fy = 0Fn – P1y = 0

∑fx = m.a
T2 – Fr +T1 = m2a

∑fy = 0
Fn – P2y = 0

∑fy = m.a
T2 – P3 = m.a
T2 =M3a+M3g

T – Fr – m1.g Senᶿ =m.a
T1=m.a+Fr + m1.gsenᶿ

T2 - m3.g =m3.aT2 = m3.a + m3.g

T2 – Fr + T1 = m2.a
M3.a + m3.g - µ.m2.g + m1.a + Fr + m1.g.senᶿ =m2.a
M3.a + m3.g - µ.m2.g + m1.a + µ.m1.g.cosᶿ + m1.g.senᶿ=m2.a
M3.g - µ.m2g + µ.m1.g cosᶿ+ m1.g senᶿ = m2.a - m3.a – m1a
M3.g - µ.m2g + µ.m1.g cosᶿ +m1.g senᶿ = a (m2 – m3 – m1)

M3.g - µ.m2g + µ.m1.g cosᶿ +m1.g senᶿ= a
M2-m3-m1

30k.9,8m/s2– 0,3.10k.9,8m/s2+ 0,3.20k.9,8m/s2.cos37 + 20k.9,8m/s2sen37 = a
10k -30k-20k
10,74m/s2 = a

T1 = m1.a + Fr + M1.g.senᶿ
T1= m1.a + µ.m1.gcosᶿ + m1.gsenᶿ
T1 = 20k.10,74m/s2 + 0,3.20k.9,8m/s2 .cos37 + 20k.9,8m/s2 sen37
T1 = 379.72N

T2 = m3a + m3g
T2 = 30k.10,74m/s2 + 30k.9,8m/s2
T2 = 616,2N