How Do Beam Forces and Support Reactions Relate in Equilibrium Analysis?

In summary: After you have solved of Ax and Ay, you can find the values for FB and FA using the equations given at the beginning of the problem. In summary, the problem involves finding the reaction forces FA and FB at points A and B, respectively, using the equations of static equilibrium. The solution involves solving for the moments and forces in both the x and y directions, and using the properties of a 3-4-5 triangle to find the x and y components of the reactions at point A.
  • #1
sami23
76
1

Homework Statement


As shown, a roller at point A and a pin at point B support a uniform beam that weighs 42.0 lb. The beam is subjected to the forces F1 = 21.0 lb and F2 = 22.0 lb. The dimensions are L1 = 1.30 ft and L2 = 8.00 ft. What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible.

Homework Equations


[tex]\Sigma[/tex]Fx = 0
[tex]\Sigma[/tex]Fy = 0
[tex]\Sigma[/tex]MA = 0

FA = sqrt{Ax^2 + Ay^2}
FB = sqrt{Bx^2 + By^2}

The Attempt at a Solution


[tex]\Sigma[/tex]MA = 0
F1L1 + F2cos(15)*(L1+L2) = By(L1+L2)
21(1.3) + 22cos(15)(9.3) = 9.3By
By = 24.186 lb (upward)

[tex]\Sigma[/tex]Fy = 0
Ay + By = -F1 -F2cos(15) - 42 = 0
Ay + 24.186 = -21 - 22cos(15) - 42
Ay = 108.47 lb (downward)

[tex]\Sigma[/tex]Fx = 0
Ax + Bx = F2sin(15)= 0
Ax + Bx = 5.694 lb
How do I find the x components for A and B?

[tex]\theta[/tex] = tan-1(3/4) = 36.87[tex]\circ[/tex]

I'm not sure about my A and B components.
 

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  • #2
sami23 said:

Homework Statement


As shown, a roller at point A and a pin at point B support a uniform beam that weighs 42.0 lb. The beam is subjected to the forces F1 = 21.0 lb and F2 = 22.0 lb. The dimensions are L1 = 1.30 ft and L2 = 8.00 ft. What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible.


Homework Equations


[tex]\Sigma[/tex]Fx = 0
[tex]\Sigma[/tex]Fy = 0
[tex]\Sigma[/tex]MA = 0

FA = sqrt{Ax^2 + Ay^2}
FB = sqrt{Bx^2 + By^2}

The Attempt at a Solution


[tex]\Sigma[/tex]MA = 0
F1L1 + F2cos(15)*(L1+L2) = By(L1+L2)
21(1.3) + 22cos(15)(9.3) = 9.3By
By = 24.186 lb (upward)
You forgot to include the moment from the beams weight
[tex]\Sigma[/tex]Fy = 0
Ay + By = -F1 -F2cos(15) - 42 = 0
Ay + 24.186 = -21 - 22cos(15) - 42
Ay = 108.47 lb (downward)
you mean upward. The terms on the right should all be plus terms ( that is , if Q - P = 0, then +Q = +P) But first correct the error in the moment equation
[tex]\Sigma[/tex]Fx = 0
Ax + Bx = F2sin(15)= 0
Ax + Bx = 5.694 lb
How do I find the x components for A and B?

[tex]\theta[/tex] = tan-1(3/4) = 36.87[tex]\circ[/tex]

I'm not sure about my A and B components.
Once you correctly solve for Ay, then Ax is related to Ay by the properties of the 3-4-5 triangle (the reaction at A must be perprendicular to the diagonal, since it is a roller support)
 

Related to How Do Beam Forces and Support Reactions Relate in Equilibrium Analysis?

1. What is the definition of "equations of equilibrium" in physics?

The equations of equilibrium in physics refer to a set of mathematical equations that describe the state of a system in which all forces acting on the system are balanced, resulting in a state of either rest or constant motion.

2. What are the three types of equations of equilibrium?

The three types of equations of equilibrium are the translational equilibrium equation, the rotational equilibrium equation, and the static equilibrium equation. These equations are used to describe the forces acting on a system in different states of equilibrium.

3. How are equations of equilibrium used in real-world applications?

Equations of equilibrium are used in various fields of physics and engineering, such as mechanics, structural analysis, and fluid dynamics. They are applied to design and analyze structures, machines, and systems to ensure stability and safety.

4. What are the key principles behind equations of equilibrium?

The key principles behind equations of equilibrium are Newton's Laws of Motion, which state that an object at rest will remain at rest, and an object in motion will continue moving in a straight line at a constant speed unless acted upon by an external force. These principles are crucial in understanding how forces interact in a system to achieve equilibrium.

5. How do equations of equilibrium relate to the concept of torque?

Equations of equilibrium and torque are closely related. Torque is a measure of the turning or twisting force applied to an object, and it is essential in rotational equilibrium equations. In systems with rotational motion, torque must be balanced to achieve equilibrium, as described by the equation Στ = 0, where Στ represents the sum of all torques acting on the system.

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