The boom AB is supported at A by a ball-and-socket joint and by two cables BD and BC. Neglect self-weights of the boom and a cylinder has a weight of 490N. A. Determine the tension... The boom AB is supported at A by a ball-and-socket joint and by two cables BD and BC. Neglect self-weights of the boom and a cylinder has a weight of 490N. A. Determine the tension in cable BD and BC? B. Calculate the x, y, z components of reaction at joint A?
Understand the Problem
The question involves analyzing a mechanical system where a boom is supported by cables and needs to determine specific tensions and reaction forces at various joints based on the provided forces and geometry.
Answer
$T_{BD}, T_{BC}; R_{Ax}, R_{Ay}, R_{Az}$ (exact numeric values require more calculations).
Answer for screen readers
- Tension in cable BD: $T_{BD} = 490N \times \frac{12m}{d_{BD}}$ (exact value requires further calculations).
- Tension in cable BC: $T_{BC} = ...$ (similar calculation).
- Reaction at joint A: $R_{Ax}$, $R_{Ay}$, and $R_{Az}$ (specific values depend on system of equations).
Steps to Solve
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Identify Forces Acting on the Boom
The weight of the cylinder is given as 490 N acting down at point B. Additionally, there are applied forces of 12 kN, 6 kN, 6 kN, and 10 kN on the boom at specified angles. We must determine the tensions in cables BD and BC while considering these forces.
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Free Body Diagram of the Boom
Draw a free-body diagram (FBD) of the boom. Include the forces acting on the boom: the weight (490 N) acting down at B, the reaction at joint A (let's call it $R_A$), and the tensions in cables BD (let’s denote it as $T_{BD}$) and BC (denote it as $T_{BC}$).
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Apply Equilibrium Equations
Use the conditions of static equilibrium:
- Sum of forces in x-direction: $\sum F_x = 0$
- Sum of forces in y-direction: $\sum F_y = 0$
- Sum of moments about point A: $\sum M_A = 0$
Setting up the equations based on these conditions will help isolate $T_{BD}$ and $T_{BC}$.
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Moment Calculation on Point A
Calculate moments about point A for forces causing rotation about this point. For example, the moment due to the weight at B, $490 N \times 12 m$, and moments due to tensions in cables BD and BC adjusted for their angles will help set up the moment equation.
$$ \sum M_A = T_{BD} \cdot d_{BD} - T_{BC} \cdot d_{BC} - 490N \cdot 12 m = 0 $$
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Solve for Tensions
Substitute values from the previous steps into the equations and solve for $T_{BD}$ and $T_{BC}$.
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Determine Reaction Components at Joint A
Once tensions are found, substitute back into the equilibrium equations from step 3 to solve for the reaction components at joint A ($R_{Ax}$, $R_{Ay}$, $R_{Az}$).
- Sum of forces in x-direction: $R_{Ax} + T_{BC} \cdot \cos(60^\circ) - T_{BD} \cdot \sin(60^\circ) = 0$
- Sum of forces in y-direction: $R_{Ay} + T_{BD} \cdot \cos(60^\circ) + T_{BC} \cdot \sin(60^\circ) - 490 N = 0$
- Sum of forces in z-direction: Just include the vertical components and solve accordingly.
- Tension in cable BD: $T_{BD} = 490N \times \frac{12m}{d_{BD}}$ (exact value requires further calculations).
- Tension in cable BC: $T_{BC} = ...$ (similar calculation).
- Reaction at joint A: $R_{Ax}$, $R_{Ay}$, and $R_{Az}$ (specific values depend on system of equations).
More Information
The tensions must be calculated based on equilibrium equations involving the angles and distances provided. The final values depend on solving the simultaneous equations obtained from static equilibrium conditions.
Tips
- Neglecting components of tension from cables; always resolve forces into their components.
- Incorrectly applying the moment equation; ensure correct distances and force directions.
- Missing any external forces acting on the boom, so make sure to include all applied loads.
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