Question

A uniform rigid beam AB has mass 20 kg and is 4m long. It rests on the fulcrum F and FB is 1 m. A 5 kg mass hangs from A and a vertical wire attached to the beam midway between F and B prevents the beam from rotating. Calculate the tension in the wire in Newtons.

Answer 1

Take the balance of torque around F, letting clockwise be the positive direction and A be to the left of F. First, we should draw a free body diagram. Let's note some lengths. \[\left(\begin{matrix}L_{AF} = 1 \\ L_{AC} = 0.5 \\ L_{FB} = 3 \\ L_{FD} = 1.5\end{matrix}\right)\]Second, we need to find the mass distribution of the beam on a per unit length basis. \[w = {m \over L} = {20 \over 4} = 5 ~\left[ \rm kg \over m \right]\]Now, we need to find the center of gravity of the portion of the beam to the left and right of the fulcrum. \[CG_{AF} = 0.5 [{\rm m}]\]\[CG_{FB} = 1.5 ~ [\rm m]\]Now, let's find the resultant force of the beam's mass distribution, which acts at the center of gravity. \[R_{AF} = L_{AF} \cdot w \cdot g \approx 50 ~ [\rm N]\]\[R_{FB} = L_{FB} \cdot w \cdot g \approx 150 ~[\rm N]\]Now, we can balance the torque. \[\sum \tau_F = 0 = -(m_A \cdot g \cdot L_{AF}) - (R_{AF} \cdot CG_{AF}) + (R_{FB} \cdot CG_{FB}) - \left(T \cdot {L_{FB} \over 2} \right)\]This, of course, assumes the tension in the wire, \(T\) acts upwards. If we get a negative value for \(T\), we know it acts downwards.