Applications of Vectors - Forces

Find the resultant force pulling the toboggan forward from a stop by adding the vectors.

\(\vert\vec{f_{R}}\vert^2=65^2+60^2-2(65)(60)\cos{(140°)}\)

\(\vert\vec{f_{R}}\vert=117.5 \; N\)

\( \dfrac{\sin{θ}}{65}=\dfrac{\sin{140°}}{117.5} \implies θ=21°\)

\(\vec{f_{R}}=117.5 \; N \) (21° off of 60 N force)

The equilibrant force is equal in magnitude and opposite in direction to the resultant force. This force would cause the net force to equal zero on the toboggan.

\(\vec{f_{E}}=117.5N \) 159° off of 60N force

Each force is resolved into its x and y components. Summing the x components, we have:

\(\vec{f_{R_{x}}} = \sum \vec{f_{x}}\)

\( \vec{f_{R_{x}}} = -400 \; N + 250\sin{(45°)} \; N - 200 \frac{4}{5} \; N = -383.2 \; N\)

The negative sign indicates that \( \vec{f_{R_{x}}} \) acts to the left, i.e., in the negative x direction.

\(\vec{f_{R_{y}}} = \sum \vec{f_{y}}\)

\( \vec{f_{R_{y}}} = 250\cos{(45°)} \; N + 200 \frac{3}{5} \; N = 296.8 \; N\)

The resulting force has a magnutude of

\( |\vec{f_{R}}| = \sqrt{(-383.2N)^2 + (296.8N)^2} = 485N\)

From the vector addition, the direction angle θ is

\( θ = tan^{-1}\left(\dfrac{296.8}{383.2}\right) = 37.8°\)

Note how convenient this method is compared to applications of the parallelogram law for each vector.