Question

in (figure 1), take that w = 6.0 kn/m. replace the distributed loading with an equivalent resultant force.

problem 6

part a
determine the magnitude of the equivalent resultant force.
express your answer to three significant figures and include the appropriate units.

part b
specify its location on the beam measured from point o.
express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Calculate area of triangular - distributed load

The distributed - load is triangular. The area of a triangle is $A=\frac{1}{2}bh$. Here, the base $b=(3 + 1.5)\text{ m}$ and the height $h = w_0=6.0\text{ kN/m}$. So, $F_R=\frac{1}{2}\times(3 + 1.5)\times6.0$.

Step2: Compute the value of the resultant force

$F_R=\frac{1}{2}\times4.5\times6.0 = 13.5\text{ kN}$.

Step3: Determine the location of the resultant force for a triangular load

For a triangular distributed load, the resultant force acts at a distance $\bar{x}$ from the vertex of the triangle. For a triangle, $\bar{x}=\frac{1}{3}$ of the base length from the vertex. The total length from point $O$ is $d=\frac{1}{3}\times(3 + 1.5)=1.5\text{ m}$ from point $O$.

Answer:

Part A:
$F_R = 13.5\text{ kN}$
Part B:
The location of the resultant force from point $O$ is $1.5\text{ m}$