Want to keep learning?

This content is taken from the UNSW Sydney's online course, Through Engineers' Eyes - Introducing the Vision: Engineering Mechanics by Experiment, Analysis and Design. Join the course to learn more.
4.5

UNSW Sydney

Skip to 0 minutes and 9 seconds In week two, we could find the unknown forces by adding two forces that meet at a point. But what if we have a rigid body rather than a point? In the design task for week two, we saw how to use the sliding vector property to convert forces that don’t act at the same point into equivalent forces that do. But what if we have a rigid body with several unknown forces and several known ones too? Or what if we have parallel unknown forces that will never meet at a point, like in experiment 3.2? The answer is an extension of equilibrium. This is a most important concept to grasp. Be prepared for longer than usual presentation.

Skip to 1 minute and 0 seconds Take it in chunks, if you like but it will need your full attention. First you need an FBD. It’s good to include axes, they don’t have to be vertical and horizontal. Sometimes other directions work better. Let’s assume that only the weight W is known and we want to find the other three forces. Then this story is force components in each of two directions that are not parallel must add up to zero. In this case, we have chosen directions x and y, shown by the dotted axes. That gives us two equations but we need three equations to solve for three unknowns. Our third equation comes from considering moments. Moments about a point must add to zero. We can choose any point.

Skip to 1 minute and 59 seconds Here we use point A because it eliminates one of the unknowns. Can you see why? Other sets of equations are available. There are some restrictions on where you can take moments but we’ll stick to the main story for now. If you do choose an invalid point, you generally just find nought equals nought, which is true but not new. You can then try somewhere else. Now we’ll try this out on the results from the experiments on the suspended cardboard. We’ll assume that we don’t know the weight of the cardboard or the loaded weight pin. We’ll calculate them from the force measurements. We’ll guide you to find the weight of the cardboard shape.

Skip to 2 minutes and 46 seconds You’ve probably worked out that it is the sum of the two forces but write it out formally and use equilibrium. First, pause the video and draw an FBD. Call the forces in the force transducers F A and F B, and the weight of the cardboard W. The cardboard is symmetrical so the centre of gravity is on the centre line. Include the dimensions shown.

Skip to 3 minutes and 23 seconds You should get something like this. Next write an expression for the sum of the forces in the vertical direction and re-arrange it to find W, that is the weight of the cardboard. You should get something like this F a plus F b minus W equals naught. So W equals F a plus F b. We’ve proven that our intuitive ideas were correct. The weight of the shape is the sum of the forces in the transducers but we’ve shown it in a formal fashion. If we substitute the test data, we’ll get W, the weight of the cardboard. These data came from an actual experiment with the cardboard shape.

Skip to 4 minutes and 8 seconds So we got F A equals 3 washers, F B equals 3 washers, and the distance a is 70 millimetres. So far, so good. Next we’ll repeat the process for the test with the load in the middle and find the added load. We’ll call the weight of the pin plus washers W P, we’ll be formal, no shortcuts. Here’s the updated FBD and equilibrium equation. It’s much as before but with an extra weight force. Substituting our calculated value for W and the test data gives W P equals 53 washers. We know that the weight of the pan is two washers and we loaded it with 50 washers, so we can compare the calculated value with the actual value.

Skip to 5 minutes and 3 seconds So the calculated value is 53 washers and the actual value is 52 washers. I think that’s reasonable agreement. Now for something new. We’ll find the offset in the offset test. Here’s the FBD. Summing forces in the vertical direction should confirm the value of the load W P, let’s check it. Here’s the analysis. Sum of the forces in the y-direction equals nought, F a plus F b minus W minus W P equals naught. So W P equals F a plus F b minus W as before. Let’s check it with the result of the experiment. The sum of the two forces in the force transducer should be the same as for the test with the weight pan in the middle.

Skip to 5 minutes and 59 seconds The two values we got were 53 washers and seven washers, a total of 60 washers, compared with 59 washers before. I think that agreement is reasonable. But how are we going to get the offset B? We’ll need to take moments and apply equilibrium. It’s probably simplest to take moments about one of the suspension points because it eliminates one of the suspension forces, but many points will work. Pause the video and try it yourself.

Skip to 6 minutes and 46 seconds Here we go. Some of the moments about A was not positive clock wise. a times W plus a minus b times W P minus 2 a F b equals zero. Rearranging we get b equals a W plus a W P minus 2 a F b and the whole lot is divided by W P. Substituting the measured data, that gives b equals 59 millimetres, measuring on the cardboard gives 60 millimetres. Not too bad. Next is a short presentation on couples.

Analysis: Finding forces in rigid objects

So, you’ve got the vocabulary and you know how to calculate the magnitude of a moment.

Now you will use your new capabilities to analyse the data you obtained from your experiments (or use the data we’ve supplied in the Downloads section below).

Considering moments opens up new ways for understanding the effects of forces on rigid bodies.

Talking points

• In the balancing experiment the cardboard was pinned in its centre. Explain what would happen if it was off centre; use the concept of moment arm.