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

[sami23]

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 F_A and F_B of the reaction forces F_A and F_B at points A and B, respectively? The beam's height and width are negligible.

Homework Equations

$$\Sigma$$F_x = 0
$$\Sigma$$F_y = 0
$$\Sigma$$M_A = 0

F_A = sqrt{A_x^2 + A_y^2}
F_B = sqrt{B_x^2 + B_y^2}

The Attempt at a Solution

$$\Sigma$$M_A = 0
F₁L₁ + F₂cos(15)*(L₁+L₂) = B_y(L₁+L₂)
21(1.3) + 22cos(15)(9.3) = 9.3B_y
B_y = 24.186 lb (upward)

$$\Sigma$$F_y = 0
A_y + B_y = -F₁ -F₂cos(15) - 42 = 0
A_y + 24.186 = -21 - 22cos(15) - 42
A_y = 108.47 lb (downward)

$$\Sigma$$F_x = 0
A_x + B_x = F₂sin(15)= 0
A_x + B_x = 5.694 lb
How do I find the x components for A and B?

$$\theta$$ = tan^-1(3/4) = 36.87$$\circ$$

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

 

[PhanthomJay]

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 F_A and F_B of the reaction forces F_A and F_B at points A and B, respectively? The beam's height and width are negligible.

Homework Equations

$$\Sigma$$F_x = 0
$$\Sigma$$F_y = 0
$$\Sigma$$M_A = 0

F_A = sqrt{A_x^2 + A_y^2}
F_B = sqrt{B_x^2 + B_y^2}

The Attempt at a Solution

$$\Sigma$$M_A = 0
F₁L₁ + F₂cos(15)*(L₁+L₂) = B_y(L₁+L₂)
21(1.3) + 22cos(15)(9.3) = 9.3B_y
B_y = 24.186 lb (upward)

You forgot to include the moment from the beams weight

$$\Sigma$$F_y = 0
A_y + B_y = -F₁ -F₂cos(15) - 42 = 0
A_y + 24.186 = -21 - 22cos(15) - 42
A_y = 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

$$\Sigma$$F_x = 0
A_x + B_x = F₂sin(15)= 0
A_x + B_x = 5.694 lb
How do I find the x components for A and B?

$$\theta$$ = tan^-1(3/4) = 36.87$$\circ$$

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)