Engineering Statics: Open and Interactive

1. Assumptions.

   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 which is normal to the surface since it is frictionless and can't prevent motion along the surface.

2. Givens.

   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 \(N\text{.}\)

3. Free Body Diagram.

   You should always begin a statics problem by drawing a free-body diagram. It gives you an opportunity 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 the applied force respectively. The angles could be given symbols too, but since we know their values it isn't necessary.

   

   The free-body can be a quick sketch or an accurate drawing but it must show all the forces acting on the particle and define the symbols. In most cases you won't know the magnitudes of all the forces, so the lengths of the vectors are just approximate.

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

4. Force Triangle.

   Use the known information to carefully and accurately construct the force triangle.

   1. Start by placing point \(A\) at the origin.

   2. Draw force \(\vec{W}\) straight down from \(A\) with a length of 1, and place point \(B\) at its tip. The length of this vector represents the weight.

   3. We know the direction of force \(\vec{P}\) but not its magnitude. For now, just draw line \(BC\) passing through point \(B\) with an angle of 10° from the horizontal.

   4. Similarly we know force \(\vec{N}\) acts at 25° from vertical because it is perpendicular to the inclined surface, and it will close the triangle. So draw line \(CA\) passing through point \(A\) and at a 25° angle from the \(y\) axis.

   5. Call the point where lines \(BC\) and \(CA\) intersect point \(C\text{.}\) Points \(A\text{,}\) \(B\text{,}\) and \(C\) define the force triangle.

   6. Now draw force \(\vec{P}\) from point \(B\) to point \(C\text{,}\) and

   7. Draw force \(\vec{N}\) from point \(C\) back to point \(A\text{.}\)

5. Results.

   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 counterclockwise from the positive \(x\) axis.

   These are not the answers we are looking for, but we’re close. Remember that for this diagram, our scale is

   \begin{equation*} 1 \text{ unit} = 100 \text{ lbs}\text{,} \end{equation*}

   so scaling the lengths of p and n by this factor gives

   \begin{align*} P \amp = (\unit{0.438}) (\lb{100}/\unit{})\\ \amp = \lb{43.8} \text{ at } 10° \measuredangle\\ N \amp = ( \unit{1.02}) ({\lb{100}}/\unit{})\\ \amp = \lb{102} \text{ at } 115° \measuredangle\text{.} \end{align*}

   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 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 an 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.

1. Draw diagrams.

   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.

   

   
2. Find angles.

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

   \begin{equation*} \alpha = \tan^{-1}\left(\frac{1}{4}\right) = 14.0°\text{.} \end{equation*}

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

   \begin{equation*} \beta = 90° - 40° = 50° \end{equation*}

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

   \begin{align*} \theta_1 \amp = \alpha + \beta = 64.0°\\ \theta_2 \amp = 90° - \alpha = 76.0°\\ \theta_3 \amp = 90° - \beta = 40.0° \end{align*}
3. Solve force triangle.

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

   \begin{equation*} \frac{\sin \theta_1}{W} = \frac{\sin \theta_2}{C} = \frac{\sin \theta_3}{T} \end{equation*}
   \begin{align*} T \amp = W \left(\frac{\sin \theta_3}{\sin \theta_1}\right) \amp C \amp = W \left(\frac{\sin \theta_2}{\sin \theta_1}\right)\\ T \amp= \kN{24} \left(\frac{\sin 40.0°}{\sin 64.0°}\right) \amp C \amp =\kN{24} \left(\frac{\sin 76.0°}{\sin 64.0°}\right)\\ T\amp= \kN{17.16}\amp C \amp= \kN{25.9} \end{align*}

1. Assumptions.

   A utility pole isn't two-dimensional, but we will consider the top view and forces in the horizontal plane only.

   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 cables 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 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.

2. Givens.

   \(\displaystyle T = \kN{10.0}\)

3. Procedure.

   Begin by drawing a neat, labeled, free-body diagram of the pole, establishing a coordinate system and indicating the directions of the forces. Although it is not necessary, it simplifies this problem considerably to note the symmetry and establish the \(x\) axis along the axis of symmetry. Let \(T\) be the tension in one wire, and \(G\) be the tension of the guy wire.

   

   To solve apply the equations of equilibrium. The symmetry of this problem means that the \(\Sigma F_x\) equation is sufficient.

   \begin{align*} \Sigma F_x \amp = 0\\ G - 6\,T_x \amp= 0\\ G \amp= 6\,(T \cos 76°)\\ \amp= \kN{14.5} \end{align*}

This problem could have also been solved using the force triangle method. See Subsection 3.4.3.

1. Givens.

   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 \(R\text{.}\)

2. Procedure.

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

   \begin{equation*} R_y = \Sigma F_y = A_y + B_y + C_y = 0 \end{equation*}

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

   \begin{align*} R_y = A_y + B_y + C_y \amp = 0\\ A + B \sin \alpha - C \cos \alpha \amp = 0\\ 20 + 20 \sin \alpha - 30 \cos \alpha \amp = 0\\ 2 + 2 \sin \alpha - 3 \cos \alpha \amp= 0 \end{align*}

   This is a single equation with a single unknown, although is is not particularly easy to solve with algebra. One approach is described here. 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 \(C\text{.}\)

   \begin{align*} R_x \amp = A_x + B_x + C_x\\ \amp = A + B \cos \alpha + C \sin \alpha\\ \amp = 0 + 10 \cos(22.62°) + 20 \sin(22.62°)\\ \amp = \N{26.15} \end{align*}

   By inspection the sum of \(A\text{,}\) \(B\text{,}\) and \(C\) acts to the right, and the resultant force \(\vec{R}\) is the vector sum of \(R_x\) and \(R_y\text{.}\) In this case, because \(R_y\) = 0,

   \begin{align*} \vec{R} \amp = R_x \rightarrow \\ \amp = \langle \N{26.15}, 0 \rangle \end{align*}

1. Strategy.

   1. Select a coordinate system, in this case horizontal and vertical.

   2. Draw a free-body digram

   3. Solve the equations of equilibrium using the scalar approach.

   
2. Procedure.
   \begin{align*} \Sigma F_x \amp = 0 \amp \Sigma F_y \amp = 0\\ -P_x + N_x \amp = 0 \amp P_y + N_y \amp = 0\\ N \cos \ang{80} \amp = P \cos \ang{40} \amp P \sin \ang{40} + N \sin \ang{80} \amp = W\\ N \amp = P \left(\frac{0.766}{0.174} \right) \amp 0.643 P + 0.985 N \amp = \lb{160} \end{align*}

   Solving simultaneously for \(P\)

   \begin{align*} 0.643 P + 0.985 ( 4.40 P) \amp = \lb{160}\\ 4.98 P \amp = \lb{160}\\ P \amp = \lb{32.1} \end{align*}

1. Strategy.

   1. Rotate the standard coordinate system \(\ang{10}\) clockwise to align the new \(y'\) axis with with force \(N\text{.}\)

   2. Draw a free-body digram and calculate the angles between the forces and the rotated coordinate system.

   3. Solve for force \(P\) directly.

   
2. Procedure.
   \begin{align*} \Sigma F_{x'} \amp = 0 \\ -P_{x'} + W_{x'} \amp = 0 \\ P \cos \ang{30} \amp = W \sin \ang{10} \\ P \amp = \lb{160} \left( \frac{0.1736}{0.866} \right)\\ P \amp = \lb{32.1} \end{align*}

1. Strategy.

   Following the General Procedure we identify the particles as points A and B, and draw free-body digrams 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:

   \begin{align*} W \amp = \N{100}\\ \alpha \amp = \tan^{-1} \left( \frac{40}{20} \right) = 63.4°\\ \beta \amp = \tan^{-1} \left( \frac{10}{80} \right) = 7.13°\\ \phi \amp = \tan^{-1} \left( \frac{50}{50} \right) = 45° \end{align*}

   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.

2. Solve Particle C.
   \begin{align*} \Sigma F_x \amp = 0 \amp \Sigma F_y \amp = 0\\ - C_x + D_x \amp = 0 \amp C_y + D_y - W \amp = 0\\ C \cos \beta \amp = D \cos \phi \amp C \sin \beta + D \sin \phi \amp= W\\ C \amp = D \left(\frac{\cos \ang{45}}{\cos \ang{7.13}}\right) \amp C \sin \ang{7.13} + D \sin \ang{45} \amp = \N{100}\\ C \amp = 0.713 D \amp 0.124 C + 0.707 D \amp = \N{100} \end{align*}

   Solving these two equations simultaneously gives

   \begin{align*} C \amp =\N{89.5} \amp D \amp = \N{125.7}. \end{align*}

   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.

3. Solve Particle B.

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

   \begin{align*} \Sigma F_x \amp = 0 \amp \Sigma F_y \amp = 0\\ - A_x - P_x + C_x \amp = 0 \amp A_y - P_y - C_y \amp = 0\\ P \cos \theta \amp = C \cos\beta - A \cos\alpha \amp P \sin \theta \amp = A \sin\alpha -C \sin\beta \end{align*}

   Since \(A = C = \N{89.5}\text{,}\) substituting and solving simultaneously gives

   \begin{align*} P \cos \theta \amp = \N{48.7} \amp P \sin \theta \amp = \N{68.9}\\ P\amp = \N{84.3} \amp \theta \amp = \ang{54.7}. \end{align*}

   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.

   \begin{align*} \vec{P} \amp= \langle - P \cos\theta, - P \sin\theta \rangle\\ \amp = \langle \N{-48.7}, \N{-68.8} \rangle \end{align*}