Section 3.4 2D Particle Equilibrium

We must assume that the block is in equilibrium, that is, either motionless or moving at a constant velocity in order to use the equilibrium equations. We will represent the block’s weight and the force between the incline and the block as concentrated forces. The force of the inclined surface on the block must act in a direction that is normal to the surface since it is frictionless and can’t prevent motion along the surface.

The knowns here are the weight of the block, the direction of the applied force, and the slope of the incline. The slope of the incline provides the direction of the normal force.

The unknown values are the magnitudes of forces \(P\) and \(N\text{.}\)

You should always begin a statics problem by drawing a free-body diagram. It allows you to think about the situation, identify knowns and unknowns, and define symbols.

We define three symbols, \(W\text{,}\) \(N\text{,}\) and \(P\text{,}\) representing the weight, normal force, and applied force respectively. The angles could be given symbols too, but since we know their values it isn’t necessary.

Notice that the force \(N\) is represented as acting 25° from the \(y\) axis, which is 90° away from the direction of the surface.

In steps 6 and 7, Geogebra tells us that p = (0.438;10.0°) which means force \(P\) is 0.438 units long with a direction of 10°, similarly n = (1.02;115°) means \(N\) is 1.02 units long at 115°. These angles are measured counter-clockwise from the positive \(x\) axis.

If you use technology such as Geogebra, as we did here, or CAD software to draw the force triangle, it will accurately produce the solution.

If technology isn’t available to you, such as during an exam, you can still use a ruler and protractor to draw the force triangle, but your results will only be as accurate as your diagram. In the best case, using a sharp pencil and carefully measuring lengths and angles, you can only expect about two significant digits of accuracy from a hand-drawn triangle. Nevertheless, even a roughly drawn triangle can give you an idea of the correct answers and be used to check your work after you use another method to solve the problem.

Start by identifying the particle and drawing a free-body diagram. The particle in this case is point \(B\) at the end of the boom because it is the point where all three forces intersect. Let \(T\) be the tension of the topping lift, \(C\) be the force in the boom, and \(W\) be the weight of the load. Let \(\alpha\) and \(\beta\) be the angles that forces \(T\) and \(C\) make with the horizontal.

Rearrange the forces acting on point \(B\) to form a force triangle as was done in the previous example.

Angle \(\alpha\) can be found from the slope of the topping lift.

Angle \(\beta\) is the complement of the 40° angle the boom makes with the vertical kingpost.

Use these values to find the three angles in the force triangle.

With the angles and one side of the force triangle known, apply the Law of Sines to find the two unknown sides.

A utility pole isn’t two-dimensional, but we can solve this problem as if it was by first considering the force components acting in a horizontal plane, and then considering the components in a vertical plane.

It also isn’t a concurrent force problem because the lines of action of the forces don’t all intersect at a single point. However, we can make it into one by replacing the forces of the three power lines in each direction with a single force three times larger. This is an example of an equivalent transformation, a trick engineers use frequently to turn complex situations into simpler ones. It works here because all the tensions are equal, and the outside wires are equidistant from the center wire. You must be careful to justify all equivalent transformations because they will lead to errors if they are not applied correctly. Equivalent transformations will be discussed in greater detail in Section 4.6 later.

\(T = \kN{10.0}\) and 38° and 152° angles.

Begin by drawing a neat, labeled, free-body diagram of the top view of the pole, establishing a coordinate system and indicating the directions of the forces.

Call the tension in one power line \(T\) and the tension in the guy wire \(G\text{.}\) Resolve the the tension of the guy wire into a horizontal component \(G_h\text{,}\) and a vertical component \(G_v\text{.}\) Only the horizontal component of \(G\) is visible in the top view.

Although it is not necessary, it simplifies this problem considerably to note the symmetry and establish the \(x\) axis along the axis of symmetry.

Solve for \(G_h\) by applying the equations of equilibrium. The symmetry of this problem means that the \(\Sigma F_x\) equation is sufficient.

Once \(G_h\) is determined, the tension of the guy wire \(G\) is easily found by considering the components of \(G\) in the side view. Note that the vertical component \(G_v\) tends to compress the pole.

We are given magnitudes of forces \(A= \N{20}\text{,}\) \(B = \N{20}\text{,}\) and \(C = \N{30}\text{.}\) The unknowns are angle \(\alpha\) and resultant force \(R\text{.}\)

Since the rod is frictionless, it cannot prevent the slider from moving vertically. Consequently, the slider will only be in equilibrium if the resultant of the three load forces is horizontal. Since a horizontal force has no \(y\) component, we can establish this equilibrium condition:

Inserting the known values into the equilibrium relation and simplifying gives an equation in terms of unknown angle \(\alpha\text{.}\)

This is a single equation with a single unknown, although it is not particularly easy to solve with algebra. One approach is described at socratic.org. An alternate approach is to use technology to graph the function \(y(x) = 2 + 2 \sin x - 3 \cos x\text{.}\) The roots of this equation correspond to values of \(\alpha\) which satisfy the equilibrium condition above. The root occurring closest to \(x=0\) will be the answer corresponding to our problem, in this case \(\alpha = 22.62°\) which you can verify by plugging it back into the equilibrium equation. Note that -90° also satisfies this equation, but it is not the solution we are looking for.

Once \(\alpha\) is known, we can find the reaction force by adding the \(x\) components of \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

Solving simultaneously for \(P\)

Following the General Procedure we identify the particles as points A and B, and draw free-body diagrams of each. We label the rope tensions \(A\text{,}\) \(C\text{,}\) and \(D\) for the endpoints of the rope segments, and label the angles of the forces \(\alpha\text{,}\) \(\beta\text{,}\) and \(\phi\text{.}\) We will use the standard Cartesian coordinate system and use the scalar components method.

Weight \(W\) was given, and we can easily find angles \(\alpha\text{,}\) \(\beta\text{,}\) and \(\phi\) so the knowns are:

Counting unknowns we find that there are two on the free-body diagram of particle \(C\) (\(C\) and \(D)\text{,}\) but four on particle \(B\text{,}\) (\(A\)\(C\text{,}\) \(P\) and \(\theta\)).

Two unknowns on particle \(C\) means it is solvable since there are two equilibrium equations available, so we begin there.

Solving these two equations simultaneously gives

With particle \(C\) solved, we can use the results to solve particle \(B\text{.}\) There are three unknowns remaining, tension \(A\text{,}\) magnitude \(P\text{,}\) and direction \(\theta\text{.}\) Unfortunately, we still only have two available equilibrium equations. When you find yourself in this situation with more unknowns than equations, it generally means that you are missing something. In this case, it is the pulley. When a cable wraps around a frictionless pulley the tension doesn’t change. The missing information is that \(A = C\text{.}\) Knowing this, the magnitude and direction of force \(\vec{P}\) can be determined.

Referring to the FBD for particle \(B\) we can write these equations.

Since \(A = C = \N{89.6}\text{,}\) substituting and solving simultaneously gives

These are the magnitude and direction of vector \(\vec{P}\text{.}\) If you wish, you can express \(\vec{P}\) in terms of its scalar components. The negative signs on the components have been applied by hand since \(\vec{P}\) points down and to the left.