Question

Determine the magnitude of the resultant force acting on the bracket and its direction measured counterclockwise from the positive u axis. 

Answer 1

first step. calculate the x and y components of the different forces f₁ y f₂

F_x=F.cos($\Theta$)

F_2x=F₂.cos($\Theta$)

F_2x=-300.cos()

50

F_2x=-192,8363lb

F_1x=-F₁.cos($\alpha$)

F_1x=-100.cos()

309

F_1x=-86,6025lb

F_y=F.sin($\theta$)

F_2y=F₂.sin($\theta$)

F_2y=300.sin()

50

F_2y=229,8133lb

F_1y=-F₁.sin($\alpha$)

F_1y=-100lb.sin()

309

F_1y=-50lb

second step. sum the components in x and y to determine the resulting force

$\sum Fx=-F_{2x}-F_{1x}$

Fr= -192, 8363 – 86, 6025

$\sum Fx=-279,4388lb$

$\sum Fy=F_{2y}-F_{1y}$

$\sum Fy=229,8133lb-50lb$

$\sum Fy=179,8133lb$

third step. calculate the resulting force

$R=\sqrt{(\sum Fx)^{2}+(\sum Fy)^{2}}$

$R=\sqrt{(-279,4388lb)^{2}+(179,8133lb)^{2}}$

$R=332,2933lb$

fourth step. calculate the angle of the resulting force

$\beta =tan^{-1}(\frac{\sum Fy}{-\sum Fx})$

$\beta =tan^{-1}(\frac{179,8133}{-279,4388})$

$\beta =-32,7605^{\circ}$

fifth step. its direction measured counterclockwise from the positive u axis is

$\phi =180-\beta$

$\phi =180^{\circ}-32,7605^{\circ}$

$\phi =147,2395^{\circ}$

$332,2933lb\measuredangle 147,2395^{\circ}$

F2= 30000 0 390 2020 30° o= 30 + zoo F-100L6 o=500 R B

Answer 2

Determine the magnitude of the resultant force acting on the bracket and its direction measured clockwise from the positive x axis