Question

las fuerzas F1= 6.4i + 8.2j (N); F2= -4.3i + 3.4j (N); F3= 3.9i+5.6j (N) actuan sobre un cuerpo de 500 g. calcula el valor de la aceleracion resultante.

Answer 1

F1= 6.4i + 8.2j (N); F2= -4.3i + 3.4j (N); F3= 3.9i+5.6j (N)R(resultante)=Rx+Ry
R=[ (6.4-4.3+3.9)i +(8.2+3.4+5.6)j ]N
R= (6i +17.2j)N
F=m*a
R=m*a
(6i +17.2j)N=(0.5kg)*a
1/(0.5kg)*(6i+17.2j)kg*m/s^2=a
(12i + 34.4j)m/s^2=a
a=(12i + 34.4j)m/s^2
a=√(12²+34.4²)
a=36.43m/s