2. calculate the net force & acceleration (with direction!) on the object in the diagram below:

Explanation:

Step1: Resolve the forces into x - and y - components

For the 5 N force: $F_{1x}=5\cos45^{\circ}=\frac{5}{\sqrt{2}}\text{ N}$ and $F_{1y}=5\sin45^{\circ}=\frac{5}{\sqrt{2}}\text{ N}$ For the 10 N force: $F_{2x}=10\cos30^{\circ}= 5\sqrt{3}\text{ N}$ and $F_{2y}=10\sin30^{\circ}=5\text{ N}$ The gravitational force $F_g = mg$, where $m = 5\text{ kg}$ and $g=9.8\text{ m/s}^2$, so $F_g=5\times9.8 = 49\text{ N}$ in the - y direction.

Step2: Calculate the net force in the x - direction

$F_{netx}=F_{1x}+F_{2x}=\frac{5}{\sqrt{2}}+5\sqrt{3}\approx\frac{5}{1.414}+5\times1.732\approx3.54 + 8.66=12.2\text{ N}$

Step3: Calculate the net force in the y - direction

$F_{nety}=F_{1y}+F_{2y}-F_g=\frac{5}{\sqrt{2}}+5 - 49\approx3.54+5 - 49=-39.46\text{ N}$

Step4: Calculate the magnitude of the net force

$F_{net}=\sqrt{F_{netx}^2+F_{nety}^2}=\sqrt{(12.2)^2+(- 39.46)^2}=\sqrt{148.84 + 1557.09}=\sqrt{1705.93}\approx41.3\text{ N}$

Step5: Calculate the direction of the net force

$\theta=\arctan\left(\frac{F_{nety}}{F_{netx}}\right)=\arctan\left(\frac{-39.46}{12.2}\right)\approx - 73.0^{\circ}$ (measured counter - clockwise from the positive x - axis)

Step6: Calculate the acceleration

According to Newton's second law $F = ma$, so $a=\frac{F_{net}}{m}=\frac{41.3}{5}=8.26\text{ m/s}^2$ in the direction $\theta\approx - 73.0^{\circ}$

Answer:

The net force is approximately $41.3\text{ N}$ at an angle of approximately $-73.0^{\circ}$ (counter - clockwise from the positive x - axis). The acceleration is $8.26\text{ m/s}^2$ at an angle of approximately $-73.0^{\circ}$ (counter - clockwise from the positive x - axis).