Taking moments about an object that is hanging over an edge.

Question

Question: A 5000 kg bus hangs 12 m over the edge of a cliff and has 1000 kg of gold at the the front. The gold sits on a wheeled cart. A group of n people, each weighing 70 kg, stands at the other end. The bus is 20 m long.

Write down the total clockwise moment about the cliff edge in terms of n.

A diagram is also supplied with the question

The official answer to the question is Total moment = $(22 000 − 560n)g$ [Nm] (sorry, I cannot figure out how to format this in maths text)

To make it clear what stage I am at with the question, I have labelled the forces so far as follows. I believe you are supposed to interpret the weight of the group of people, and the weight of gold to be distributed at the very end of the bus.

My confusion arises when considering where the reaction between the bus and the cliff will lie. In the answer to the question, the reaction force is considered to be at the edge of the cliff. As a result, moments can then be taken around the edge of the cliff.

My question: Why is the reaction force acting at the edge of the cliff? Surely if the bus is not on the point of tilting then the reaction force would act further within the cliff.

Answer 1

You are summing up the moments about the edge of the cliff. The gold provides a clockwise torque of $(1000)(g)(12)$. The people provide a counterclockwise torque of $(70n)(g)(8)$.

Then for the bus, for problems like this, we assume the mass is evenly distributed across its full length. And when that is the case, you can calculate the torque as if the mass was all concentrated at its midpoint. Since the midpoint of the bus is 2 meters to the right of the point about which we are calculating the moment, we calculate this as a clockwise torque. $(5000)(g)(2)$

This gives a total clockwise torque of $(12000)g-(560n)g+(10000)g=(22000-560n)g$

Is that what you were asking?

Answer 2

Moment from the bus is $5000\times2\times g=10000g(NM)$

Moment from the gold is $1000\times12\times g=12000g(NM)$

Moment from the people is $70\times8\times n\times g=560ng(Nm)$

Now, the total moment is $(22000-560n)g(NM)$