Answer:
The magnitude of the resultant force is:
[tex]R=100.3 \:pound[/tex]
The direction is:
[tex]\theta=1.2^{\circ}[/tex]
Step-by-step explanation:
Let's find the components of each vector is x and y-directions first.
Sum of x-component vector forces.
[tex]F_{tot-x}=F_{1}cos(45)+F_{2}cos(30)[/tex]
[tex]F_{tot-x}=50cos(45)+75cos(30)[/tex]
[tex]F_{tot-x}=100.3 \: pound[/tex]
Sum of y-component vector forces.
[tex]F_{tot-y}=-F_{1}sin(45)+F_{2}sin(30)[/tex]
[tex]F_{tot-y}=-50sin(45)+75sin(30)[/tex]
[tex]F_{tot-y}=2.1 \: pound[/tex]
The magnitude of the resultant force is:
[tex]R=\sqrt{100.3^{2}+2.1^{2}}[/tex]
[tex]R=100.3 \:pound[/tex]
The direction is:
[tex]tan(\theta)=\frac{2.1}{100.3}[/tex]
[tex]\theta=arctan(\frac{2.1}{100.3})[/tex]
[tex]\theta=1.2^{\circ}[/tex]
I hope it helps you!
