John Abbott 203-SNB-AB tutoring
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Optional course listed in the catalogue. It may not run every term. Ask us whether we can tutor you in it.
Check whether both supports can keep pushing
If statics is on your 203-SNB-AB outline, use this original example to connect equilibrium equations to the kind of support in the diagram. John Abbott lists statics and structures analysis within this optional course; its options rotate, so the listing does not confirm a Fall 2026 section.
A uniform rigid beam is 3.0 m long and weighs 90 N. It rests horizontally on two smooth supports underneath it: A at the left end, and B 2.0 m to the right of A. The beam overhangs B by 1.0 m. A downward point load P = 30 N acts at the right end. Each support can push upward but cannot pull downward; there are no clamps or ties. Ignore deformation. This is a classroom model, not a test of a real structure’s safety.
Find the vertical reactions and , then decide whether both contacts are possible. Before calculating, predict which support carries more. Take upward forces and counterclockwise moments as positive.
Show a hint
The beam’s own weight acts at its midpoint, 1.5 m from A. Take moments about A first so that contributes no moment. Both reactions must then be nonnegative for these push-only supports.
Show the worked solution
1. Draw the beam’s external forces
Draw upward at x = 0, upward at x = 2.0 m, the 90 N weight downward at x = 1.5 m, and P downward at x = 3.0 m. Here x is distance to the right of A. All forces are vertical, so the horizontal force equation is already satisfied.
2. Balance moments, then forces
Every lever arm below is the perpendicular distance from A to that force’s line of action. The moment equation is
The clockwise load moment is 225 N·m, so = 112.5 N. Vertical force balance gives + = 90 N + 30 N, hence = 7.5 N. Support B carries more because the right-end load has a long moment arm about A.
3. Check the other pivot and the contacts
About B, A is 2.0 m to the left, the beam’s weight is 0.5 m to the left, and P is 1.0 m to the right. The symbol means the sum of the moments about B:
Also, 7.5 + 112.5 − 90 − 30 = 0 N. Both reactions are positive, so neither support is being asked to pull. Equal 60 N reactions would balance the forces but would fail the moment equation. Force balance alone is not enough.
Retry: increase the overhanging load
Keep the beam and supports unchanged, but set P = 60 N. Find the reactions predicted by the same two-contact equations. Can the assumed arrangement remain at rest? Then find the value of P at which first reaches zero. Attempt both questions before opening the answer.
Check the retry and the model
Moment balance about A gives = [90(1.5) + 60(3.0)]/2.0 = 157.5 N. Force balance then gives = 150 − 157.5 = −7.5 N. A negative sign would require support A to pull downward, which this contact cannot do. The assumed two-contact static state is impossible; A loses contact rather than providing a downward reaction.
For a variable load P, the same equations give
Thus = 0 at P = 45 N. For nonnegative loads up to 45 N the reactions satisfy the push-only sign condition; 45 N is the limiting zero-reaction case. This condition does not establish resistance to disturbances, material strength or safe working loads. Beyond it, a new motion/contact model is needed.
Choose the next step from your error
Bring your force diagram and both balance equations. A tutor can check the midpoint weight, ask you to measure every lever arm from the same pivot, and compare your reaction signs with the actual support type. Then change a load position and predict which contact becomes lighter before calculating independently.
For a simpler pivot check, use the balanced-bar example. Then try question 4 on changing the pivot. Those exercises check force and moment balance; the push-only contact test above adds a separate condition.
Consult John Abbott’s official science course list for the catalogue context. Use your current section outline for its other chapters and assessments. Send your chapter and attempt to confirm tutoring coverage before booking.
Related subject areas
Choose the area that matches the chapter you’re studying.
Common difficulties
- Turning the situation into a model
- Using vectors, signs, and units consistently
- Selecting the relevant physical principle
How tutoring can help
Lessons follow your course materials and focus on what you want help with. Your tutor adapts to your needs and pace, explaining difficult ideas patiently, step by step. There’s time to ask questions, revisit unclear steps and practise together to build understanding and confidence.
Exam preparation for 203-SNB-AB
Choose a permitted practice question from your current review list and try it without its solution. Mark the first step you cannot explain. After reviewing that step, redo the question and explain why your method applies.
Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.
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