# Calculating torque when the lever arm has mass

by threecorners
Tags: force, kinematics, torque
 P: 2 1. The problem statement, all variables and given/known data A board is set on top of a scaffolding, with dimensions as shown in the diagram. The board weighs 15 kg. A man weighing 70 kg stands on the board as shown. How far out can he stand before the board falls? 2. Relevant equations Torque: $\tau = rFsin(\theta)$ Force due to gravity: $F = mg$ 3. The attempt at a solution I decided to use calculus, and then combine with the torque equation for the torque the man applies. I am assuming the leftmost point where the scaffolding touches the board is the fulcrum, and the rightmost point only serves to keep the board from rotating clockwise. First, since gravity is our only force and it points downwards, we have $\tau=rmg$. Now we use calculus: $d\tau=rg\cdot dm$, and $dm=\frac{15}{5.5}\cdot dr$, so $d\tau=\frac{15g}{5.5}r\cdot dr$. Now I assume the first 1.5m of board on each side of the fulcrum cancel out, so the torque on the right of the fulcrum is $\tau=\int_{1.5}^{4} \frac{15g}{5.5}r\cdot dr=...=18.75g$. I leave the g because I am about to cancel it out. Now for the lefthand side of my equation, i just need the torque from the man, which will be at a distance of $x$ out, which we must solve for. So $x\cdot 70\cdot g = 18g$, and so $x\cong 0.27m$. Okay, that's it for my solution. The book says the answer should be $1.2m$. I haven't done physics in a while but I have done a lot of math, which is why my solution is overly mathy. I assume there's an easier thing to do here that doesn't require calculus like I did...