Question

Determine the resultant force in polar notation for each of these

Answer 1

(a). In the first case, as both the forces are in same direction,

Thus, the angle between the forces is zero.

The resultant of the forces is,

`R=\sqrt[]{F^2_1+F^2_2+2F_1F_2\cos (\theta)}`

Substituting the known values,

`\begin{gathered} R=\sqrt[]{3^2+4^2+(2*3*4*\cos (0^(\circ)))_{}} \\ R=\sqrt[]{9+16+24} \\ R=\sqrt[]{49} \\ R=7\text{ N} \end{gathered}`

Thus, the resultant form of the force is,

`(r,\theta)=(7,180^(\circ))`

(b). In this case,

One force is along horizontal direction and other force is along the vertical direction.

Thus, the angle between these two forces is 90 degree.

The resultant of the two forces is,

`\begin{gathered} R=\sqrt[]{3^2+4^2+(2*3*4*\cos (90^{\circ_{}})} \\ R=\sqrt[]{9+16+(2*3*4*0)} \\ R=\sqrt[]{9+16+0} \\ R=\sqrt[]{25} \\ R=5\text{ N} \end{gathered}`

Thus, the resultant force is 5 N.

The direction of the resultant force is,

`\begin{gathered} \tan (\theta^(\prime))=(-4)/(-3) \\ \theta^(\prime)=233.13^(\circ) \end{gathered}`

Thus, the resultant force in the polar notation is,

`(r,\theta^(\prime))=(5,233.13^(\circ))`

(c). In this case,

One of the force is along the negative of the x-axis and second force is at the angle of 30 degree from the negative of y-axis.

Thus, the resultant force is,

`\begin{gathered} F=\sqrt[]{3^2+4^2+(2*3*4*\cos (90-30)} \\ F=\sqrt[]{9+16+24*\cos (60)} \\ F=\sqrt[]{37} \\ F=6.08\text{ N} \end{gathered}`

The direction of the resultant force is,

`\begin{gathered} \tan (\theta^(\prime))=(-(4*\cos (30)))/(-(3+4*\cos (90-30))) \\ \tan (\theta^(\prime))=(3.464)/(5) \\ \theta^(\prime)=214.71^(\circ) \end{gathered}`

Thus, the resultant force in the polar form is,

`(r,\theta^(\prime))=(6.08,235.28^(\circ))`