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Applications of Vectors - Forces

When objects are at rest or move at a constant volocity in a straight line, then the forces that act upon the object cancel each other out. In other words, there is no net force. The net, or resultant force, can be found by adding the forces which are vectors. The counteracting force is called the equilibrant force and is the opposite vector to the resultant force.

Observe the resultant force and the corresponding equilibrant force for the following forces:

Collinear forces at Equilibrium Pair of collinear forces at equilibrium with resulting force and equilibrant force vectors.

Coplanar Forces at Equilibrium Pair of coplanar forces at equilibrium with a resultant and equilibrant force.

Describe the forces that act on the object to keep in a state of equilibrium.
An aircraft flying at constant velocity
Aircraft force diagram with lift, thrust, weight, and drag forces acting on it.
An object hanging on two wires.
Force diagram of object hanging on two wires with tension and weight forces acting on it.
A box standing on a ramp.
Force diagram of box standing on ramp with normal, weight, and gravity forces acting on it.
Jake and Maria are towing their freinds on toboggan. Jake is exerting a force of \(65\; [\text{N}]\) and Maria a force of \(60 \; [\text{N}]\). Since they are walking side by side, the ropes pull to either side of the toboggan at \(40°\) to each other. Determine the resultant and equilibrant forces.

Diagram of resultant and equilibrant forces acting on tobaggon with individual force vectors.


Step 1

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)


Step 2

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

The end of the boom is subject to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force.

Diagram of 3 coplanar and concurrent forces acting on boom end.


Step 1

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.


Step 2

Summing the \(y\) components, we have:

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

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

Diagram of horizontal, vertical, and resultant forces acting on boom end.

Step 3

The resulting force has a magnutude of:

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


Step 4

From the vector addition, the direction angle \(\theta\) is:

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

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



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