Question

find the reactions at contact points a, b, and c, given:
\\(f = 95\text{ lbs}\\), \\(l_1 = 0.9\text{ ft}\\), \\(l_2 = 1.6\text{ ft}\\), \\(l_3 = 3.5\text{ ft}\\), \\(\theta = 28^\circ\\), \\(\phi = 36^\circ\\)

\\(n_a =\\) number \\(\text{lbs}\\) (enter a positive number)
\\(n_b =\\) number \\(\text{lbs}\\) (enter a positive number)
\\(n_c =\\) number \\(\text{lbs}\\) (enter a positive number)

Explanation:

🆕 New Concept Discovered: Static Equilibrium of Rigid Bodies
Balancing forces and moments on a member in contact with smooth surfaces.

Step 1: Identify the contact forces (reactions)

The member is in contact with smooth surfaces at points \(A\), \(B\), and \(C\). Since the surfaces are smooth, the reaction forces are normal (perpendicular) to the contacting surfaces:
- At \(A\), the contact is horizontal, so the reaction force \(N_A\) is vertical (pointing downwards on the member to prevent it from moving up).
- At \(B\), the contact is horizontal, so the reaction force \(N_B\) is vertical (pointing upwards on the member).
- At \(C\), the corner contact acts perpendicular to the inclined member \(BC\). Thus, the reaction force \(N_C\) is perpendicular to the bar, pointing upwards and to the left at an angle of \(\theta\) with the vertical (or \(90^\circ - \theta\) with the horizontal).

Let's set up a coordinate system where the \(x\)-axis is horizontal (positive to the right) and the \(y\)-axis is vertical (positive upwards):
- \(N_A\) acts vertically downwards: \(\vec{N}_A = -N_A \hat{j}\)
- \(N_B\) acts vertically upwards: \(\vec{N}_B = N_B \hat{j}\)
- \(N_C\) is perpendicular to the bar. The bar is inclined at angle \(\theta = 28^\circ\) above the horizontal. The normal force \(N_C\) points perpendicular to this incline (upwards and leftwards):
\[ \vec{N}_C = -N_C \sin\theta \hat{i} + N_C \cos\theta \hat{j} \]

Step 2: Resolve the applied force \(F\)

The force \(F = 95\text{ lbs}\) is applied at the free end. The angle \(\phi = 36^\circ\) is measured relative to the dashed line, which is perpendicular to the bar.
- The bar is inclined at \(\theta = 28^\circ\) above the horizontal.
- The dashed line perpendicular to the bar points downwards and to the right, making an angle of \(\theta = 28^\circ\) with the vertical (or \(90^\circ - \theta = 62^\circ\) below the horizontal).
- The force \(F\) is rotated by \(\phi = 36^\circ\) clockwise from this perpendicular line.
- Therefore, the angle of \(F\) below the horizontal is:
\[ \alpha = (90^\circ - \theta) - \phi = (90^\circ - 28^\circ) - 36^\circ = 62^\circ - 36^\circ = 26^\circ \]
- The components of \(\vec{F}\) are:
\[ F_x = F \cos\alpha = 95 \cos(26^\circ) \approx 85.386\text{ lbs} \]
\[ F_y = -F \sin\alpha = -95 \sin(26^\circ) \approx -41.645\text{ lbs} \]

Step 3: Set up the equations of equilibrium

For the system to be in static equilibrium, the sum of forces in both directions and the sum of moments about any point must equal zero.

1. Force balance in the \(x\)-direction:

\[ \sum F_x = 0 \implies -N_C \sin\theta + F_x = 0 \]
\[ N_C \sin(28^\circ) = 95 \cos(26^\circ) \]
\[ N_C = \frac{95 \cos(26^\circ)}{\sin(28^\circ)} \approx \frac{85.386}{0.46947} \approx 181.877\text{ lbs} \]

2. Moment balance about point \(B\):

Let's define the coordinates of the points relative to \(B(0,0)\):
- The segment \(BA\) of length \(L_1 = 0.9\text{ ft}\) is perpendicular to the horizontal, so it is vertical. Thus, point \(A\) is at:
\[ (x_A, y_A) = (0, L_1) = (0, 0.9) \]
- Point \(C\) is along the inclined bar of length \(L_2 = 1.6\text{ ft}\) at angle \(\theta = 28^\circ\):
\[ (x_C, y_C) = (L_2 \cos\theta, L_2 \sin\theta) = (1.6 \cos(28^\circ), 1.6 \sin(28^\circ)) \approx (1.4127, 0.7512) \]
- The end of the bar where \(F\) is applied is at a distance of \(L_2 + L_3 = 1.6 + 3.5 = 5.1\text{ ft}\) from \(B\):
\[ (x_F, y_F) = (5.1 \cos(28^\circ), 5.1 \sin(28^\circ)) \approx (4.5030, 2.3943) \]

Now, calculate the sum of moments about \(B\) (counter-clockwise is positive):
- Moment due to \(N_A\):
The force \(\vec{N}_A = -N_A \hat{j}\) acts at \((0, L_1)\).
\[ M_{A/B} = x_A (F_y) - y_A (F_x) = 0 \cdot (-N_A) - L_1 \cdot (0) = 0 \]
*Note:* Since \(N_A\) is vertical and acts along the vertical line passing through \(B\), its line of action passes directly through \(B\). Thus, its moment about \(B\) is zero.

- Moment due to \(N_C\):
The force \(N_C\) acts perpendicular to the bar at a distance \(L_2 = 1.6\text{ ft}\) from \(B\). Since it pushes counter-clockwise around \(B\):
\[ M_{C/B} = N_C \cdot L_2 = 181.877 \cdot 1.6 \approx 291.003\text{ lb}\cdot\text{ft} \]

- Moment due to \(F\):
The force \(F\) acts at a distance \(L_2 + L_3 = 5.1\text{ ft}\) from \(B\). The component of \(F\) perpendicular to the bar is \(F \cos\phi\) acting clockwise:
\[ M_{F/B} = -F \cos\phi \cdot (L_2 + L_3) = -95 \cos(36^\circ) \cdot 5.1 \approx -76.857 \cdot 5.1 \approx -391.972\text{ lb}\cdot\text{ft} \]

- Let's re-verify if there is any horizontal reaction at \(A\) or \(B\). Since the contacts at \(A\) and \(B\) are smooth horizontal surfaces, they can only exert vertical normal forces. Thus, the only horizontal forces are the horizontal component of \(N_C\) and the horizontal component of \(F\), which we balanced in Step 1.

Since \(\sum M_B = 0\):
\[ M_{C/B} + M_{F/B} + M_{A/B} = 0 \implies N_C \cdot L_2 - F \cos\phi \cdot (L_2 + L_3) = 0 \]
But wait, let's check if this is consistent with our calculated \(N_C\):
\[ N_C = \frac{F \cos\phi \cdot (L_2 + L_3)}{L_2} = \frac{95 \cos(36^\circ) \cdot 5.1}{1.6} \approx \frac{391.972}{1.6} \approx 244.983\text{ lbs} \]
However, from the horizontal force balance, we got \(N_C \approx 181.877\text{ lbs}\).

Why is there a discrepancy?
Let's look at contact \(A\). The contact at \(A\) is a vertical surface or a horizontal surface?
Looking at the diagram, the blue structure has a horizontal surface at the top contacting the vertical-ish member at \(A\). Since the member \(L_1\) is perpendicular to \(L_2\), and the angle of \(L_2\) is \(\theta = 28^\circ\), the member \(L_1\) is inclined at \(90^\circ + \theta = 118^\circ\) from the horizontal.
The contact at \(A\) is with a horizontal ceiling. Thus, the normal force \(N_A\) is vertical.
The contact at \(B\) is with a horizontal floor. Thus, the normal force \(N_B\) is vertical.
The contact at \(C\) is a corner contact. A corner contact on an inclined beam exerts a force perpendicular to the beam.

If all reactions at \(A\) and \(B\) are purely vertical, then the horizontal force balance must be:
\[ \sum F_x = 0 \implies -N_C \sin\theta + F_x = 0 \implies N_C = \frac{F_x}{\sin\theta} \]
But if we also must satisfy \(\sum M_B = 0\), we have:
\[ \sum M_B = -N_A \cdot d_A + N_C \cdot L_2 - F \cos\phi \cdot (L_2 + L_3) = 0 \]
where \(d_A\) is the horizontal distance from \(B\) to \(A\).
Since \(L_1\) is perpendicular to the bar \(BC\), the vector from \(B\) to \(A\) is perpendicular to the bar:
\[ x_A = -L_1 \sin\theta = -0.9 \sin(28^\circ) \approx -0.4225\text{ ft} \]
Since \(N_A\) acts vertically downwards at \(A\), its moment about \(B\) is:
\[ M_{A/B} = -N_A \cdot |x_A| = -N_A \cdot L_1 \sin\theta \]
This resolves the discrepancy! \(N_A\) does produce a moment about \(B\) because \(A\) is horizontally offset from \(B\).

Let's write down the complete system of 3 equations with 3 unknowns (\(N_A\), \(N_B\), \(N_C\)):

1. \(\sum F_x = 0\):
\[ -N_C \sin\theta + F \cos\alpha = 0 \implies N_C = \frac{F \cos\alpha}{\sin\theta} \]

2. \(\sum M_B = 0\):
\[ -N_A (L_1 \sin\theta) + N_C L_2 - F \cos\phi (L_2 + L_3) = 0 \]

3. \(\sum F_y = 0\):
\[ -N_A + N_B + N_C \cos\theta - F \sin\alpha = 0 \]

Step 4: Calculate the values

- **Calculate \(N_C\):**
\[ N_C = \frac{95 \cos(26^\circ)}{\sin(28^\circ)} \approx \frac{85.3863}{0.46947} \approx 181.877\text{ lbs} \]

- **Calculate \(N_A\):**
Using the moment equation:
\[ N_A (0.9 \sin(28^\circ)) = N_C (1.6) - 95 \cos(36^\circ)(5.1) \]
\[ N_A (0.42252) = 181.877(1.6) - 76.8573(5.1) \]
\[ N_A (0.42252) = 291.003 - 391.972 = -100.969 \]
Since \(N_A\) must be positive (contact force pushing away from the ceiling), let's re-verify the direction of \(N_A\).
If the bar wants to rotate clockwise due to \(F\), the end \(A\) will push *upwards* against the ceiling, so the ceiling pushes *downwards* on the bar with force \(N_A\).
Let's check the sign of the moment:
- \(N_C\) creates a counter-clockwise (positive) moment: \(+N_C L_2\)
- \(F\) creates a clockwise (negative) moment: \(-F \cos\phi (L_2 + L_3)\)
- Since the clockwise moment from \(F\) (\(391.97\)) is larger than the counter-clockwise moment from \(C\) (\(291.00\)), we need a counter-clockwise moment to balance it.
- A downward force \(N_A\) at \(A\) (which is to the left of \(B\), \(x_A < 0\)) produces a counter-clockwise moment:
\[ \vec{r}_A \times \vec{N}_A = (-x_A \hat{i} + y_A \hat{j}) \times (-N_A \hat{j}) = x_A N_A \hat{k} \]
Since \(x_A = -L_1 \sin\theta\), the moment is indeed counter-clockwise:
\[ M_{A/B} = + N_A L_1 \sin\theta \]
Therefore, the moment equation is:
\[ N_A (L_1 \sin\theta) + N_C L_2 - F \cos\phi (L_2 + L_3) = 0 \]
\[ N_A (0.42252) + 291.003 - 391.972 = 0 \]
\[ N_A (0.42252) = 100.969 \implies N_A \approx 238.97\text{ lbs} \]

- **Calculate \(N_B\):**
Using the vertical force balance:
\[ N_B = N_A - N_C \cos\theta + F \sin\alpha \]
\[ N_B = 238.97 - 181.877 \cos(28^\circ) + 95 \sin(26^\circ) \]
\[ N_B = 238.97 - 160.59 + 41.64 \approx 120.02\text{ lbs} \]

Answer:

• \(N_A = 239.0\) lbs
• \(N_B = 120.0\) lbs
• \(N_C = 181.9\) lbs