48610 Chap.10 Supports, Reactions and Free Body Diagrams
Supports, Reactions and Free Body Diagrams
Week 8 makes the rest of statics usable. Chapters 8 and 9 dealt with forces that were given; here you work out the forces the world exerts back.
The method is printed as a boxed three step procedure: draw the outlined shape, picturing the body lifted clear of everything it touches and being sure to take every support away; show all known and unknown external forces and moments, which come from applied loadings, the reactions where it was held or where it touched another body, and its own weight; then label each one with its magnitude and direction so the labels can be used unchanged in the equations.
Each support is read by asking what motion it prevents.
A roller stops movement perpendicular to its surface and contributes one unknown; a pin stops movement in both directions but allows rotation and contributes two; a built in support stops everything including rotation and contributes two forces and a moment; a smooth contact and a cable each contribute one. Counting unknowns this way, before any algebra, says whether a single free body diagram can solve the problem.
Two shortcuts then reduce the count: a two force member carries force along the line joining its two loaded points, and a body under exactly three non parallel forces must have all three lines of action passing through one point. The same three equations also answer questions about tipping, by setting the reaction at the support about to lift to zero.
What this chapter covers
- 01
The published three step free body diagram procedure
- 02
Removing supports, and why beginners leave them in
- 03
Roller, pin, fixed, smooth contact and cable, and their unknown counts
- 04
Weight as a force at the centre of mass
- 05
Stating assumptions and relating them to the diagram
- 06
Smooth contact, weightless members, frictionless pins and rigidity
- 07
Solving a beam: horizontal, then moments, then vertical, then check
- 08
The sum check: reactions must add to the total load
- 09
Two force members and three force members
- 10
Tipping as the limiting case where a reaction reaches zero
- 11
Why the tipping load is a ratio of distances
Contact forces on a loader, and the load that tips it
- +1Free the machine. Remove the ground and replace each wheel contact with a single upward force, then add the machine's weight at its centre of mass and the load's weight at the bucket. Two unknowns, so the moment equation plus vertical equilibrium will do.
- +1Take moments about the front axle, because that is where tipping will occur. Measuring horizontally from it, the rear reaction acts at 2.40 m behind, the machine weight at 0.800 m behind and the load at 1.60 m ahead.
- +1Substitute the 700 kg case. The machine term is 0.800 times 4200 times 9.81, which is 32 962 newton metres, and the load term is 1.60 times 700 times 9.81, which is 10 987. The rear reaction is their difference divided by 2.40, that is 9156 N or 9.16 kN.
- +1Vertical equilibrium gives the front axle. The total weight is 4900 times 9.81, which is 48 069 N, so the front reaction is 48 069 minus 9156, that is 38 913 N or 38.9 kN, about four times the rear.
- +1Set the rear reaction to zero for the tipping load. The same moment equation becomes 0.800 times 4200 times g equals 1.60 times the load times g, so g cancels and the load is 0.800 times 4200 divided by 1.60, that is 2100 kg, exactly half the machine mass.
Key terms
- Reaction
- The force or moment a support exerts on a body in place of the motion it prevents. Every support removed from a free body diagram must be replaced by its reaction.
- Roller support
- A support that prevents movement perpendicular to its surface only, contributing a single unknown force normal to that surface and allowing both rotation and sliding.
- Pin support
- A support that prevents movement in two directions but allows rotation, contributing two unknown force components and no moment.
- Built in support
- A support that prevents translation in both directions and rotation as well, contributing two force components and a reaction moment, so three unknowns in total.
- Smooth contact
- An idealised contact carrying no friction, so the reaction acts along the common normal to the two surfaces. It is unsafe to assume wherever a joint grips or clamps.
- Three force member
- A body in equilibrium under exactly three non parallel forces, whose lines of action must all pass through one point, which fixes the direction of an unknown force.
- Tipping
- The condition reached when the reaction at one support falls to zero and the body begins to pivot about another. It is solved as an ordinary statics problem with that reaction set to zero.
Supports, Reactions and Free Body Diagrams FAQ
Why must the supports be removed rather than drawn?
Because a free body diagram is a statement about one body, and the wall or the ground is not part of it. Its entire contribution is the force it exerts, so you replace it with that force and delete it. Leaving the support drawn is not merely untidy: it invites you to forget to write its reaction as an unknown, which is the commonest way a diagram ends up with fewer forces than the body actually feels.
How do I know how many unknowns a support contributes?
Ask what motion it prevents, and count one unknown for each prevented translation plus one for a prevented rotation. A roller prevents motion in one direction, so one. A pin prevents two translations but allows rotation, so two. A built in support prevents everything, so three. A cable prevents motion along its own line only, so one, and it can only pull.
Why do I need to state my assumptions?
Because every support symbol is an idealisation of something messier, and the idealisation you choose decides the answer. Assuming smooth contact deletes friction from the diagram; assuming a weightless member deletes a force; assuming rigidity keeps the lever arms as drawn.
The Week 8 exercises ask you to state any assumptions and say how they relate to how the diagram was drawn, and writing them as three short lines beside the diagram makes it possible for a reviewer to disagree with an assumption rather than with your answer.
What order should I solve the equations in?
Look for the equation containing exactly one unknown and start there. On a typical beam that means the horizontal equation first because it is often trivially zero, then moments about the support carrying the most unknowns, then vertical equilibrium for whatever is left, then a second moment equation about a different point purely as a check.
Taking the equations in the order they appear on a formula sheet is how simple problems turn into simultaneous ones.
Assessment move
Draw more diagrams than you solve. Take every mechanism you can see, a bench clamp, a bicycle brake lever, a door closer, and draw the free body diagram of one part of it, removing the supports and labelling every force, without computing anything.
The skill being trained is recognising what each connection can resist, and it transfers directly to the project, whose report asks for free body diagrams based on pre defined scenarios. When you do solve, use the sum check every time: the reactions must add to the total applied load, and a mistake in the moment equation almost never leaves that sum intact.
Working through Supports, Reactions and Free Body Diagrams in 48610? Sia is AskSia’s AI Engineering tutor — ask any 48610 Supports, Reactions and Free Body Diagrams question and get a clear, step-by-step explanation grounded in how 48610 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
