### Example 5.4.1. Pin and Roller.

The L-shaped body is supported by a roller at \(B\) and a frictionless pin at \(A\text{.}\) The body supports a \(\lb{250}\) vertical force at \(C\) and a \(\ftlb{500}\) couple-moment at \(D\text{.}\) Determine the reactions at \(A\) and \(B\text{.}\)

This problem will be solved three different ways to demonstrate the advantages and disadvantages of different approaches.

Solution 1.
Solution 2.
Solution 3.

Solutions always start with a free-body diagram, showing all forces and moments acting on the object. Here, the known loads \(C =
\lb{250}\) (down) and \(D= \ftlb{500}\) (CCW) are red, and the unknown reactions \(A_x\text{,}\) \(A_y\) and \(B\) are blue.

The force at \(B\) is drawn along its known line-of-action perpendicular to the roller surface, and drawn pointing up and right because that will oppose the rotation of the frame about A caused by load C and moment D. The force at \(A\) is represented by unknown components \(A_x\) and \(A_y\text{.}\) The sense of these components is unknown, so we have arbitrarily assigned the arrowheads pointing left and up.

We have chosen the standard coordinate system with positive \(x\) to the right and positive \(y\) pointing up, and resolved force \(A\) into components in the \(x\) and \(y\) directions.

The magnitude of force \(B\) is unknown but its direction is known, so the \(x\) and \(y\) components of B can be expressed as

\begin{align*}
B_x \amp= B \sin \ang{60} \amp B_y \amp = B \cos \ang{60}
\end{align*}

.

We choose to solve equation set \(\{A\}\text{,}\) and choose to take moments about point \(A,\) because unknowns \(A_x\) and \(A_y\) intersect there. Substituting the variables into the equation and solving for the unknowns gives

\begin{align*}
\sum F_x \amp = 0 \\
B_x - A_x \amp = 0\\
A_x \amp = B\ \sin \ang{60} \amp \amp (1)\\
\\
\sum F_y \amp = 0\\
B_y - C + A_y \amp= 0\\
A_y \amp= C - B\ \cos \ang{60}\amp \amp (2)\\
\\
\sum M_A \amp = 0 \\
-B_x( 3) -B_y(7)+C(4) + D \amp = 0\\
3 B \cos \ang{60} + 7 B \sin \ang{60} \amp = 4 C + D \\
B (3 \sin \ang{60} + 7 \cos \ang{60})\amp = 4 C + D \\
B \amp = \frac{4 C + D}{6.098} \amp \amp (3)
\end{align*}

Of these three equations only the third can be evaluated immediately, because we know \(C\) and \(D\text{.}\) In equations \((1)\) and \((2)\) unknowns \(A_x\) and \(A_y\) can't be found until \(B\) is known. Inserting the known values into \((3)\) and solving for \(B\) gives

\begin{align*}
B \amp = \frac{4 (250) + 500 }{6.098} \\
\amp = \frac{\ftlb{1500}}{\ft{6.098}} \\
\amp = \lb{246.0}
\end{align*}

Now with the magnitude of \(B\) known, \(A_x\) and \(A_y\) can be found with \((1)\) and \((2)\text{.}\)

\begin{align*}
A_x \amp = B \sin \ang{60} \\
\amp = 246.0 \sin \ang{60} \\
\amp = \lb{213.0}\\
\\
A_y \amp= C - B \cos \ang{60}\\
\amp= 250 - 246.0 \cos \ang{60}\\
\amp= \lb{127.0}
\end{align*}

The positive signs on these values indicate that the directions assumed on the free-body diagram were correct.

The magnitude and direction of force \(\vec{A}\) can be found from the scalar components \(A_x\) and \(A_y\) using a rectangular to polar conversion.

\begin{gather*}
A = \sqrt{A_x^2 + A_y^2} = \lb{248.0}\\
\\
\theta = \tan^{-1} \left | \frac{ A_y}{A_x} \right | =
\ang{30.8}
\end{gather*}

The final values for \(\vec{A}\) and \(\vec{B}\text{,}\) with angles measured counter-clockwise from the positive \(x\) axis are

\begin{equation*}
\vec{A} = \lb{248.0}\ \measuredangle\ \ang{149.2},
\end{equation*}

\begin{equation*}
\vec{B} = \lb{246.0}\ \measuredangle \ 30°
\end{equation*}

.

This solution demonstrates a fairly standard approach appropriate for many statics problems which should be considered whenever the free-body diagram contains a frictionless pin. Start by taking moments there.

In this solution, we have rotated the coordinate system \(\ang{30}\) to align it with force \(\vec{B}\) and also chosen the components of force \(\vec{A}\) to align with the new coordinate system.

There is no particular advantage to this approach over the first one, but with two unknown forces aligned with the \(x'\) direction, \(A_{y'}\) can be found directly after breaking force \(C\) into components.

\begin{align*}
\sum F_{x'} \amp = 0 \\
B - C_{x'} + A_{x'} \amp = 0\\
A_{x'} \amp = -B + C \sin \ang{30} \amp \amp (1)\\
\\
\sum F_{y'} \amp = 0\\
-C_{y'} + A_{y'} \amp= 0\\
A_{y'} \amp= C \cos \ang{30} \amp \amp (2)\\
\\
\sum M_A \amp = 0 \\
-B_x( 3) -B_y(7)+C(4) + D \amp = 0\\
3 B \cos \ang{60} + 7 B \sin \ang{60} \amp = 4 C + D \\
B (3 \cos \ang{60} + 7 \sin \ang{60})\amp = 4 C + D \\
B \amp = \frac{4 C + D}{7.56} \amp \amp (3)
\end{align*}

Solving equation (2) yields

\begin{equation*}
A_{y'} = \lb{216.5}
\end{equation*}

.

Solving equation (3) yields the same result as previously

\begin{equation*}
B = \lb{246.0}
\end{equation*}

.

Substituting \(B\) and \(C\) into equation (1) yields

\begin{align*}
A_{x'} \amp = -B + C \sin \ang{30} \\
\amp = - 246.0 + 250 \sin \ang{30} \\
\amp = -\lb{121.0}
\end{align*}

The negative sign on this result indicates that our assumed direction for \(A_{x'}\) was incorrect, and that force actually points \(\ang{180}\) to the assumed direction.

Resolving the \(A_{x'}\) and \(A_{y'}\) gives the magnitude and direction of force \(\vec{A}\text{.}\)

\begin{gather*}
A = \sqrt{A_{x'}^2 + A_{y'}^2} = \lb{248.0}\\
\\
\theta = \tan^{-1} \left | \frac{ A_y}{A_x} \right | =
\ang{60.8}\\
\alpha = \ang{180} - (\theta - \ang{30}) = \ang{149.2}
\end{gather*}

Again, the final values for \(\vec{A}\) and \(\vec{B}\text{,}\) with angles measured counter-clockwise from the positive \(x\) axis are

\begin{equation*}
\vec{A} = \lb{248.0}\ \measuredangle\ \ang{149.2},
\end{equation*}

\begin{equation*}
\vec{B} = \lb{246.0}\ \measuredangle \ 30°
\end{equation*}

This approach was slightly more difficult than solution one because of the additional trigonometry involved to find components in the rotated coordinate system.

For this solution, we will use the same free-body diagram as solution one, but will use three moment equations, about points \(B\text{,}\) \(C\) and \(D\text{.}\)

\begin{align*}
\sum M_B \amp = 0 \\
- A_x(3) + A_y(7) - C(3) + D \amp = 0 \\
- 3 A_x + 7 A_y \amp = 250 \amp \amp (1)\\
\\
\sum M_C \amp = 0\\
-A_x(3) + A_y(4) - B_y(3) + D \amp= 0\\
-3A_x + 4 A_y - 3 B \cos \ang{60} \amp= -D \\
3A_x - 4 A_y + 1.5 B \amp = 500 \amp \amp (2)\\
\\
\sum M_D \amp = 0 \\
- A_x(1.5) -B_x( 1.5 ) -B_y(7) + C(4) + D \amp = 0\\
1.5 A_x + 1.5 B \sin \ang{60} + 7 B \cos \ang{60} \amp = 4 C +
D \\
1.5 A_x + 4.799 B \amp = 1500 \amp \amp (3)
\end{align*}

This set of three equations and three unknowns can be solved with some algebra.

Adding (1) and (2) gives

\begin{align*}
3 A_y + 1.5 B \amp = 750 \amp \amp (4)
\end{align*}

Dividing equation (2) by 2 and subtracting it from (3) gives

\begin{align*}
2 A_y + 4.049 B = 1250 \amp \amp (5)
\end{align*}

Multiplying (4) by 2/3 and subtracting from (5) eliminates \(A_y\) and gives

\begin{gather*}
3.049 B = 750
\end{gather*}

\begin{gather*}
B = \lb{246.0}
\end{gather*}

, the same result as before.

Substituting \(B\) into (3) gives \(A_x = \lb{213.0}\text{,}\) and substituting this into (1) gives \(A_y = \lb{127.0}\text{,}\) again the same result as before.

An alternate approach is to set these three equations up for a matrix solution and use technology to do the algebra, as done here with Sage.

\begin{equation*}
\begin{bmatrix} -3 \amp 7 \amp 0 \\ 3 \amp -4 \amp 1.5 \\ 1.5
\amp 0 \amp 4.799 \end{bmatrix} \begin{bmatrix} A_x \\ A_y \\ B
\end{bmatrix} = \begin{bmatrix} 250 \\ 500 \\ 1500 \end{bmatrix}
\end{equation*}

This is a good example of an inefficient solution because of all the algebra involved. The issue here was the poor choice of \(B\text{,}\) \(C\) and \(D\) as moment centers. Whenever possible you should take moments about a point where the line of action of two unknowns intersect as was done in solution one. This gives a moment equation which can be solved immediately for the third unknown.