Torque — Question 9

Question 9

A distributed load has a torque-weighted application point

A horizontal beam from $x=0$ to $L$ carries a downward load density $w(x)=w_0x/L$ in N/m. Find moments about its left end.

1. Find the total downward force and torque.

2. Find the application point of an equivalent single downward force and explain why the midpoint is wrong.

Question 9 - Solution

A distributed load has a torque-weighted application point

(a) Derivation and calculation

Integrate the force density over the beam: $$\boxed{F=\int_0^L\frac{w_0x}{L}\,dx=\frac{w_0L}{2}}\quad\text{downward}.$$ The signed moment weights each contribution by its distance: $$\boxed{\tau_z=-\int_0^Lx\frac{w_0x}{L}\,dx=-\frac{w_0L^2}{3}}.$$

(b) Result and interpretation

A single equivalent downward force must reproduce both results: $$-Fx_{\mathrm{eq}}=\tau_z\Rightarrow \boxed{x_{\mathrm{eq}}=\frac{w_0L^2/3}{w_0L/2}=\frac{2L}{3}}.$$ The midpoint would give only $-w_0L^2/4$, so it would not reproduce the load’s moment.

Physical check

The equivalent point lies toward the more heavily loaded right end.