problem 1.9 - determine the magnitude and direction of the resultant force acting on the cantilevered beam…

Explanation:

Step1: Resolve forces into x - components

For the 8 kN force: $F_{8x}=8\cos40^{\circ}\text{ kN}\approx 6.13\text{ kN}$ (right - hand direction is positive). For the 15 kN force: $F_{15x}=15\times\frac{4}{5}=12\text{ kN}$ (right - hand direction is positive). The 11 kN force has no x - component. So, $R_x = 6.13+12=18.13\text{ kN}$.

Step2: Resolve forces into y - components

For the 8 kN force: $F_{8y}=- 8\sin40^{\circ}\text{ kN}\approx - 5.14\text{ kN}$ (down - ward direction is negative). For the 15 kN force: $F_{15y}=15\times\frac{3}{5}=9\text{ kN}$ (up - ward direction is positive). The 11 kN force is in the downward direction, so $F_{11y}=-11\text{ kN}$. Then $R_y=-5.14 + 9-11=-7.14\text{ kN}$.

Step3: Calculate the magnitude of the resultant force

Using the Pythagorean theorem $R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(18.13)^{2}+(-7.14)^{2}}\approx\sqrt{328.6969 + 50.9796}=\sqrt{379.6765}\approx19.49\text{ kN}$.

Step4: Calculate the direction of the resultant force

Using $\theta=\arctan\left(\frac{|R_y|}{R_x}\right)=\arctan\left(\frac{7.14}{18.13}\right)\approx\arctan(0.3949)\approx21.5^{\circ}$ below the positive x - axis.

Answer:

Magnitude: approximately $19.49\text{ kN}$, Direction: approximately $21.5^{\circ}$ below the positive x - axis.