Vanier College 203-SN1-RE Mechanics Tutoring

About 203-SN1-RE

Vanier lists 203-SN1-RE as Mechanics in Pure and Applied Science. Its Physics Wiki brings together material on motion, forces and conservation laws. The example below connects a change in velocity to the force a scale measures.

Topics to review

Units, measurement and vectors
Motion in one and two dimensions; rotational kinematics
Newton’s laws and force diagrams
Static equilibrium and torque
Work, energy and power
Linear and angular momentum

For work–energy practice, read a force–position graph and check which positions can be reached. The worked example and four optional questions connect signed area, initial kinetic energy and speed.

Start here

Connect braking force and stopping distance
Balance forces and torques on a beam

Before you start

You will use signed numbers, the slope of a straight velocity–time graph and a one-step force equation. Acceleration describes how velocity changes; its direction can differ from the direction of travel.

What to watch for

Inferring acceleration from a change in velocity
Distinguishing weight from the scale’s normal force

Mechanics: stopping distance and torque balance

Find a stopping time and distance, then balance a beam. Choose positive directions before writing the force and torque equations.

Before you start: Allow 20–25 minutes to try both questions, then 15 minutes to check and retry any missed steps.

1. A force gives a stopping distance

A $2.0\,\mathrm{kg}$ cart moves right at $4.0\,\mathrm{m/s}$ . A constant net horizontal force of $4.0\,\mathrm N$ acts left until it stops. Find the stopping time and distance.

Show the explained solution

With right positive, $a=F_{\mathrm{net}}/m=-2.0\,\mathrm{m/s^2}$ .
From $0=4.0-2.0t$ , the stopping time is $t=2.0\,\mathrm s$ .
Using $0^2=4.0^2+2(-2.0)\Delta x$ gives $\Delta x=4.0\,\mathrm m$ . Check with work: the initial $16\,\mathrm J$ is removed by $(-4.0)(4.0)=-16\,\mathrm J$ .

Check the stopping time, distance and braking interval

2. Balance rotation and translation

A massless horizontal beam is hinged at its left end. A $60\,\mathrm N$ downward load acts $0.80\,\mathrm m$ from the hinge. An upward vertical cable force acts $1.20\,\mathrm m$ from the hinge. Find the cable tension and the vertical hinge force in static equilibrium.

Show the explained solution

Let $T$ be the upward cable tension and $H_y$ the hinge-force component, positive upward. Take torques about the hinge, so its force contributes no torque. With counterclockwise positive, $1.20T-0.80(60)=0.$
Thus $T=40\,\mathrm N$ . Next impose vertical force balance: $H_y+40-60=0$ .
The hinge force is $H_y=20\,\mathrm N$ upward. Zero net torque alone does not establish zero net force; both conditions are needed.

Review a hinged beam and cable

Use your answers to plan your revision

Stopping problems
Use one sign convention from the force equation through the motion calculation. Check the distance against the loss of kinetic energy.
Equilibrium problems
Choose the torque pivot deliberately, then write both rotational and translational equilibrium equations.

When requesting mechanics support, include your free-body diagram, axis choice and your section’s permitted review questions. That makes it easier to target the setup you need.

Request 203-SN1-RE exam preparation help

Original Nablio practice, not an official examination.

Course topic sources

A rising elevator: what does the scale read?

A passenger of mass $m=60\,\mathrm{kg}$ stands on a force scale in an elevator. The elevator is rising, but its upward velocity decreases uniformly from $3.0\,\mathrm{m/s}$ to $1.0\,\mathrm{m/s}$ over $2.0\,\mathrm s$ . The passenger stays at rest relative to the elevator. Use $g=9.8\,\mathrm{m/s^2}$ . Model the passenger as a single point, ignoring body size. Neglect air resistance. Find the scale reading in newtons during this interval. First predict: is it greater than, equal to or less than the passenger’s weight?

Show a hint

Take upward as positive. Find acceleration from the change in velocity before writing the force equation. The scale reads the normal force it exerts on the passenger.

Show the reasoning

1. Separate motion from change in motion
Both velocities are positive because the elevator is moving upward. However, velocity is decreasing, so acceleration points downward. A positive velocity does not require a positive acceleration.
2. Find the acceleration
Let $v_i$ and $v_f$ be the initial and final vertical velocities, $\Delta t$ the elapsed time, and $a$ the vertical acceleration. Because the velocity changes uniformly, $a$ is constant.
$a=\frac{v_f-v_i}{\Delta t}$
$a=\frac{(1.0-3.0)\,\mathrm{m/s}}{2.0\,\mathrm s}$
$a=-1.0\,\mathrm{m/s^2}$
3. Draw forces on the passenger
There are two vertical forces: the normal force $N$ from the scale upward, and the gravitational force $mg$ downward. Here $m$ is mass in kilograms and $g$ is gravitational acceleration in $\mathrm{m/s^2}$ .
The magnitude of the weight is:
$mg=(60)(9.8)\,\mathrm N$
$mg=588\,\mathrm N$
Draw one dot for the passenger, an upward arrow labelled $N$ and a downward arrow labelled $mg$ . The elevator’s upward velocity is not an extra force.
4. Use the net force
With upward positive, Newton’s second law is:
$N-mg=ma$
Rearranging gives:
$N=m(g+a)$
Substitute the values in SI units:
$N=60(9.8-1.0)\,\mathrm N$
$N=528\,\mathrm N$
The scale reads $528\,\mathrm N$ , less than the $588\,\mathrm N$ weight.
5. Check the result
$N-mg=(528-588)\,\mathrm N$
$N-mg=-60\,\mathrm N$
This agrees with:
$ma=(60)(-1.0)\,\mathrm N$
$ma=-60\,\mathrm N$
Also, $\mathrm{kg\cdot m/s^2}$ is a newton. The passenger is still moving upward while the net force slows that upward motion. The mass and the gravitational force have not changed.

Upward velocity and downward acceleration are separate from the passenger’s two forces: N = 528 N upward and mg = 588 N downward. — Upward is positive. The motion arrows indicate only the directions of velocity v and acceleration a. The two force arrows start at the point representing the passenger; their lengths are proportional to 528 N and 588 N.

Try the opposite direction

The same passenger now travels downward. With upward still positive, the elevator’s velocity changes uniformly from $-3.0\,\mathrm{m/s}$ to $-1.0\,\mathrm{m/s}$ in $2.0\,\mathrm s$ . Predict whether the scale reading is above or below $588\,\mathrm N$ , then calculate it.

Check your answer

Find the sign from the velocity change
$\begin{gathered}a=\\\frac{[-1.0-(-3.0)]\,\mathrm{m/s}}{2.0\,\mathrm s}\end{gathered}$
$a=+1.0\,\mathrm{m/s^2}$
The elevator is moving downward and slowing down, so acceleration points upward.
Reuse the force equation
$N=m(g+a)$
$N=60(9.8+1.0)\,\mathrm N$
$N=648\,\mathrm N$
The reading is above $588\,\mathrm N$ . The sign of acceleration controls this comparison. At constant vertical velocity, $a=0$ and the scale would read $588\,\mathrm N$ in either direction.

If the acceleration sign was the difficult part, review motion graphs. If the force equation was the difficult part, review free-body diagrams.

How a tutor can help

A tutor can ask you to predict the scale reading before calculating, check your chosen positive direction and help you separate a velocity arrow from a force arrow. You can then fix the first incorrect equation and try the downward-motion check without copying the worked answer.

Prepare for an assessment

Bring the topics announced for your assessment and a practice problem. Include any work you have started, or ask your tutor to help you begin. Practise explaining a sign or force choice before doing the calculation, then redo a missed question with the direction of motion changed.

Independent practice

Choose a lesson to review

40 min

Motion graphs and kinematics

Use a velocity–time graph’s slope to determine acceleration and keep signs consistent.

20 min

Free-body diagrams and friction

Choose the object and turn its interactions into a force equation.

75 min

Torque and static equilibrium

For a separate equilibrium question: choose a pivot and a perpendicular lever arm. This lesson focuses on static equilibrium.

6 questions

Check your force setup in six questions

Try mixed force problems, then use the optional hints and worked solutions to find the step to revisit.

45 min

Rolling without slipping: friction, forces and energy

Optional extension when assigned: rolling bodies, the static-friction threshold and translational plus rotational energy. Legacy enrichment coverage does not establish current assessment scope.

Angular momentum: choose the system and axis

For a rotation problem, extend your force diagram by naming the system and axis. The six questions ask when angular momentum is conserved, how a changed inertia affects rotation and what happens when an object sticks to a disk. Use them when this topic is in your current chapter.

Try six angular-momentum questions

Bring a question to a tutor. You can build the diagram and first angular-momentum equation together, or review what you have started. You can check the external-torque assumption, the signs and the energy comparison before retrying a missed question.

Official course information

Before a lesson

How do I request tutoring?

Send the topic you want help with and your preferred times. Start with a problem you’re stuck on or a concept you’d like to understand better. Your tutor can help you build confidence from the basics. You can include a practice question or any work you have started. We’ll confirm whether we can help and agree on a time before payment.

Does this guide match my section?

Use your current course outline and your instructor’s announced topics to decide which parts of this guide to practise. Follow your section’s assessment instructions.

Need help with 203-SN1-RE?

Choose a topic or send the step where you get stuck.

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

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

Vanier 203-SN1-RE: mechanics tutoring