Question

. Calcula el módulo del vector resultante de dos vectores fuerza de 9 [N] y 12 [N] concurrentes en un punto o, cuyas direcciones forman un ángulo de a) 30˚ b) 45˚ y c) 90˚

Answer 1

Answer:

a) 20.29N

b) 19.42N

c) 15N

Explanation:

To find the magnitude of the resultant vector you can consider an axis in the middle of the vector, from which you can calculate the components of the vectors by using the angles given:

a) for 30°

$F_1=[9cos(15\°)\hat{i}+9sin(15\°)\hat{j}]N\\\\F_1=[8.69\hat{i}+2.32\hat{j}]N\\\\F_2=[12cos(15\°)\hat{i}-12sin(15\°)\hat{j}]N\\\\F_2=[11.59\hat{i}-3.10\hat{j}]N\\\\F=F_1+F_2=20.28N\hat{i}-0.78N\hat{j}\\\\|F|=\sqrt{(20.28N)^2+(0.78N)^2}=20.29N$

F = 20.29N

b)  for 45°

$F_1=[9cos(22.5\°)\hat{i}+9sin(22.5\°)\hat{j}]N\\\\F_1=[8.31\hat{i}+3.44\hat{j}]N\\\\F_2=[12cos(22.5\°)\hat{i}-12sin(22.5\°)\hat{j}]N\\\\F_2=[11.08\hat{i}-4.59\hat{j}]N\\\\F=F_1+F_2=19.39N\hat{i}-1.15\hat{j}\\\\|F|=\sqrt{(19.39N)^2+(1.15N)^2}=19.42N$

F= 19.42N

c) for 90°

for this case you can consider that the direction of both vectors are the y and x axis of the Cartesian plane:

$F_1=9N\hat{i}\\\\F_2=12N\hat{j}\\\\F=\sqrt{(9N)^2+(12N)^2}=15N$

F=15N