2.2

Para:

f2= 20
f1=30
y una masa de 8kg

Σ Fx = m a x =(30 cos 60° - 20 sen 20°) N =8Kg ax
Σ Fx= (15- 6.84)N = 8 Kg
8.15N = 8Kg

Σ Fy= m a y =(30 sen 60° + 20 cos 20°) N =8Kg ay
Σ Fy= (25.98 +18.79)N = 8 Kg
44.77N = 8Kg

F = (masa) * (aceleracion)
el despeje se expresa como:
a= F/m

ax= 8.15N /8 Kg = 1.01 N/kg
ay= 44.77N/8Kg= 5.5 N/Kg

a=√(ax) ^2 + (ay) ^2

a= √(1.01 N/Kg)^2 + (5.5) N/Kg)^2

a=√1.0201 N2 Kg^2 + 30.25 N2 Kg^2

a=√ 31.2701 N^2/Kg`2


a=5.591 N/ Kg

Angulo :

-1 =5.5/1.01


tan-1= 5.44

θ = 79.58 = 79°35' 2.3