# Calculating the moments due to reaction forces on a bent beam

I am trying to solve the following problem:

## EXAMPLE 2.10

Determine the reactions on the beam in Fig. 2–30a. Assume $B$ is a pin and the support at $B$ is a roller (smooth surface).

### SOLUTION

Free-Body Diagram. As shown in Fig. 2–30b, the support ("roller") at $B$ exerts a normal force on the beam at its point of contact. The line of action of this force is defined by the 3-4-5 triangle.

Equations of Equilibrium. Resolving $\mathbf{N}_B$ into x and y components and summing moments around $A$ yields a direct solution for $N_B$. Why? Using this result, we can then obtain $A_x$ and $A_y$.

\begin {align} \\ \Sigma M_A = 0; \quad & -3500 (3.5) + (\frac{4}{5}) N_B (4) + (\frac{3}{5}) N_B (10) = 0 \\ & \qquad N_B = 1331.5\ \mathrm{lb} = 1.33\ \mathrm{k} \\ \Sigma F_x = 0; \quad & A_x - \frac{4}{5}(1331.5) = 0 & A_x = 1.07\ \mathrm{k}\\ \Sigma F_y = 0; \quad & A_y - 3500 + \frac{3}{5}(1331.5) = 0 & A_x = 2.70\ \mathrm{k}\\ \end {align}

For the moment about $A$, why does the reaction at $B$ produce a moment of $(\frac{4}{5})N_B(4)$? I think it's not necessary to include that moment because the moment arm $d = 0.4\ \mathrm{m}$ is not measured directly from point A, but from the bend in the beam, which I circled red in figure (b).

• I'm not sure why you have a problem with the second term in the sum of moments but not the third term. How would you propose writing that equation instead?
– Air
Jan 11 '17 at 19:42

The reason is that both the horizontal and the vertical component of $N_B$ generate a moment around $A$. Since the moment due to a force is equal to the product of the force and the perpendicular lever-arm between the force's line of action and the point being considered, this therefore means that the total moment around $A$ is equal to:

\begin{align} N_{B,x} &= \dfrac{4}{5}N_B \text{ (horizontal component)} \\ N_{B,y} &= \dfrac{3}{5}N_B \text{ (vertical component)} \\ M_A &= -3500 \cdot 3.5 + 4N_{B,x} + 10N_{B,y} = 0 \end{align}

After all, the horizontal distance from $A$ to $B$ is 10 ft, so that's the lever arm for the vertical component, while the vertical distance is 4 ft, so that's used for the horizontal component.

Hoping to make this as clear as possible, here's a diagram of the forces, their lines of action and the respective perpendicular distances. They are color-coded: everything relevant for the horizontal component is green, everything for the vertical component is red.

• I can't see how to answer this question beyond saying that's what it is. The moment is the force times the perpendicular distance, but the question seems to be saying why is the moment the force times the perpendicular distance. Jan 11 '17 at 16:08
• referrring to the last figure uploaded in the original post , we can see that the d (prepedicular distance) for NBy is measured directly from A , but for the NBx ( blue part) , it's measured directly from a point which is 10 m from A . So , for the moment 4NBx , it's not moment about A , am i right ? Why the author consider it in the calculation ? Jan 12 '17 at 0:21