The given problem can be exemplified in the following diagram:
To determine the total force we will determine the horizontal and vertical components of each of the forces.
For the 70 pounds force, we have that the horizontal component is:
[tex]F_{70h}=70\sin 56[/tex]
The vertical component is:
[tex]F_{70v}=-70\cos 56[/tex]
The negative sign is due to the fact that the vertical component is in the negative direction.
Now, we determine the horizontal component of the 50 pounds force:
[tex]F_{50h}=50\sin 72[/tex]
The vertical component is:
[tex]F_{50v}=50\cos 72[/tex]
Now, we add the horizontal components:
[tex]F_h=70\sin 56+50\sin 72[/tex]
Solving the operations:
[tex]F_h=105.59\text{ Pounds}[/tex]
Now, we add the vertical components:
[tex]F_v=-70\cos 56+50\cos 72[/tex]
Solving the operations:
[tex]F_v=-23.69\text{ pounds}[/tex]
Now, the magnitude of the resulting force is given by the following formula:
[tex]F=\sqrt[]{F^2_h+F^2_v}[/tex]
Plugging in the values we get:
[tex]F=\sqrt[]{(105.59)^2+(-23.69)^2}[/tex]
Solving the operations:
[tex]F=108.22\text{ pounds}[/tex]
Therefore, the magnitude of the resulting force is 108.22 pounds.
To determine the angle of the resulting force we use the following formula:
[tex]\theta=\tan ^{-1}(\frac{F_v}{F_h})[/tex]
Plugging in the values we get:
[tex]\theta=\tan ^{-1}(-\frac{23.69}{105.59})^{}[/tex]
Solving the operations:
[tex]\theta=-12.7\text{ degr}ees.\text{ }[/tex]
Therefore, the direction is -12.7°.
