Question

Construction cranes are able to balance heavy loads by using a counterweight. The crane operator can slide the entire torque arm horizontally in order to balance the load. If the torque arm has a total length of 56m and a mass of 750 kg, and the counterweight has a mass of 2500kg, where should the operator position the pivot point to balance a 860 kg object?

Answer 1

We need to make a couple of assumptions here. First, let's say that the counterweight is located at one extreme of the beam and the load is located at the other. Second, let's fix the origin at the end of torque arm where the load is located. Additionally, the positive \(x\) direction will be directed towards the counterweight and consequently, the pivot point. Let's draw a free body diagram. \(x_{CG}\) is the location of the center of mass of the torque arm, \(x_P\) is the location of the pivot point, R is the resultant force of the torque arm, \(F_p\) is the force exerted by the pivot point, L is the load, and CW is the counterweight.

Now, let's find the location of the center of mass of the torque arm, assuming an equal mass distribution. It will be located at the middle of beam. Therefore, \[x_{CG} = 28 m\]

Now we can balance the torques about the origin. These, of course, must equal zero for the crane to be in a static state. We will take clockwise to be the positive torque direction. \[\sum \tau = 0 = R \cdot x_{CG} + m_{CW} \cdot g \cdot L_{TA} - F_P \cdot x_P \]
Now because the torque arm doesn't plument towards earth, the forces in the vertical direction must also balance. \[F_P = R + m_{CW} \cdot g\]

Note that \(R = m_{TA} \cdot g\)

You should be able to solve both equations for \(F_P\) and \(x_P\)