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Concordia University · PHYS 204

Concordia PHYS 204: mechanics tutoring

Mechanics: one-to-one online tutoring and exam preparation for Concordia PHYS 204. Start with a question about forces or motion. Your tutor can help you draw the diagram, choose the physical principle and build the equations step by step.

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Course overview

Concordia PHYS 204 introduces mechanics through kinematics, Newton’s laws, conservation principles, rotation, and periodic motion.

Start here

Practise the reasoning between steps

Choose a workshop when the topic is included in your instructor’s announced material. These original exercises are not an exam reproduction or a complete course review.

Connect equilibrium and static friction

When rigid-body equilibrium is in your assigned material, practise force diagrams, torques and the friction limit in one model. This supplement is not a complete course review or an exam prediction.

Ladder equilibrium: maximum distance and minimum angle

PHYS 204 topics to review

KinematicsNewton’s lawsStatics and dynamicsMomentum and energyRotational motionPeriodic motion

Common difficulties

  • Drawing only forces acting on the selected system
  • Choosing between forces, energy and momentum methods

Forces on an incline: what changes when the rope is released?

A 4.0 kg4.0\text{ kg} block rests on a fixed, smooth ramp inclined at 30∘30^\circ above the horizontal. A light rope parallel to the ramp holds the block at rest. Use g=9.8 m/s2g=9.8\text{ m/s}^2. Ignore air resistance.

Before calculating, draw separate force diagrams for the block while the rope holds it and just after release. Include only forces acting on the block. Does the normal force equal its full weight? Find the rope tension and normal force before release, then the acceleration just after release.

Show a hint

Choose one axis along the ramp and one perpendicular to it. Weight points vertically downward; it does not point along either axis. After release, remove the rope force from the diagram.

Show the worked solution

1. Identify the forces

While the rope holds the block, three forces act: weight mgmg vertically downward, the normal force NN perpendicular to the ramp, and tension TT uphill. The smooth surface exerts no friction. The weight is 39.2 N39.2\text{ N}.

Forces before and after releasing the rope: Rope holds the blockOn a ramp rising to the right, the held block has weight downward, a normal force perpendicular away from the ramp, and tension uphill. After release, only weight and the normal force remain.Rope holds the blockmgNT30°+y+xDashed lines: axes+x: downhillForces before and after releasing the rope: After releaseOn a ramp rising to the right, the held block has weight downward, a normal force perpendicular away from the ramp, and tension uphill. After release, only weight and the normal force remain.After releasemgN30°+y+xDashed lines: axes+x: downhill
The rope’s tension disappears on release. Weight and the normal force remain; downhill is positive for the release calculation. Force arrows are schematic.

2. Resolve weight and use equilibrium

The weight component down the ramp is mgsin⁡30∘mg\sin30^\circ; the component into the ramp is mgcos⁡30∘mg\cos30^\circ. Choose downhill and away from the ramp as the positive directions. The block is held at rest, so the equations are mgsin⁡30∘−T=0mg\sin30^\circ-T=0 along the ramp and N−mgcos⁡30∘=0N-mg\cos30^\circ=0 perpendicular to it. Solving gives T=mgsin⁡30∘=19.6 N,N=mgcos⁡30∘=19.63 N.\begin{aligned}T&=mg\sin30^\circ=19.6\text{ N},\\N&=mg\cos30^\circ=19.6\sqrt3\text{ N}.\end{aligned} The normal force is about 34 N34\text{ N}. It balances only the perpendicular component of weight.

3. Remove the force that disappears

After release, tension is zero. Taking downhill as positive gives ma=mgsin⁡30∘ma=mg\sin30^\circ, so a=4.9 m/s2a=4.9\text{ m/s}^2 downhill. The perpendicular acceleration remains zero, so the normal force is unchanged while contact with this straight ramp is maintained.

4. Check the model

If the ramp angle approached zero, the downhill acceleration would approach zero and the normal force would approach mgmg. A diagram with both the full weight and its components as separate forces would count gravity twice.

Retry without the solution

The same block is now given an initial speed of 2.45 m/s2.45\text{ m/s} uphill on the same smooth ramp, with no rope attached. The brief push has ended; only gravity and contact with the ramp act afterward. Choose uphill as positive. What is its acceleration while it moves uphill, and how long does it take to stop momentarily?

Check the retry

The forces are still weight and the normal force. With uphill positive, a=−gsin⁡30∘=−4.9 m/s2a=-g\sin30^\circ=-4.9\text{ m/s}^2. The block moves uphill while accelerating downhill. From 0=2.45−4.9t0=2.45-4.9t, the stopping time is 0.50 s0.50\text{ s}. Zero velocity at that instant does not mean zero acceleration; the block then starts sliding downhill.

Choose the next step

For help with this example, your tutor can draw the before-and-after force diagrams with you, explain the choice of axes and build the signed equations. You can then change the ramp angle and try the same method with guidance.

Preparing for this course

Review vector components, trigonometry, units and position–velocity–acceleration relationships.

Registration requirements

MATH 203 or equivalent must be completed previously or taken concurrently.

Check the official course catalogue
Vector refresher

Try a course problem

A displacement has components 6 metres east and 8 metres north. Find its magnitude and its direction, measured north of east.

Show the reasoning

Let dd denote the displacement magnitude. East and north are perpendicular, so the components form the legs of a right triangle. Pythagoras gives d=62+82=100=10 md=\sqrt{6^2+8^2}=\sqrt{100}=10\text{ m}. Both components are positive, so the direction is in quadrant I. With the calculator in degree mode, θ=arctan⁡(8/6)≈53.13∘\theta=\arctan(8/6)\approx53.13^\circ north of east. Adding 6 and 8 would measure a two-part route, not the straight displacement.

Add displacement components and find the resultant direction

Use force components to find the resultant and acceleration

Optional spring-oscillation practice

If spring oscillations are in your assigned periodic-motion work, use this optional example to connect starting position and velocity with phase and a first future crossing. Check your current outline before adding it to exam review.

How tutoring can help

If you are unsure how to begin a PHYS 204 problem, your tutor can help you choose the object or system to study and draw the forces acting on it. You can then decide together whether to use Newton’s laws, energy or momentum, and write the equations one step at a time. Your tutor can explain the signs and components, check the result with you and guide you through a similar problem.

Exam preparation for PHYS 204

Practise motion, force, energy and momentum problems in mixed order. State conservation assumptions and finish with units and a limiting-case check.

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.

Useful links for preparation or further study—not transfer equivalencies or a required registration sequence.

MATH 203 tutoringDifferential and Integral Calculus IMATH 204 tutoringVectors and MatricesPHYS 205 tutoringElectricity and Magnetism
Independent practice

Free practice and study guides for PHYS 204

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PHYS 204 · official course reference