Mechanics: Six Questions to Check Your Setup

Nablio · 6 questionsPublished October 7, 2026

Use these six questions to find the first step you need to revisit in a force problem. Write your setup before doing the arithmetic, then compare it with the worked solution.

The questions cover forces on one object, elevator motion, static and kinetic friction, contacting objects and a fixed pulley. Keep the hints and solutions closed until you have tried the question.

Measure motion relative to the room or building and treat it as nonaccelerating. Use g = 9.8 m/s² throughout. Ignore air resistance, tipping and rotation of the objects; question 6 states the pulley assumptions.

Notes for your working

For each question, name the object or system, draw its external forces and choose positive directions. Write the force equations in symbols before substituting numbers.
Include units and directions in your answers. Keep unrounded values during calculation. The solutions round decimal answers to two decimal places where useful for comparison.

1. Which object receives the force?

An inspection light of mass $m=1.3\,\mathrm{kg}$ hangs motionless from one taut vertical cable of negligible mass, attached to a fixed support. Nothing else touches the light.

A student lists four forces: Earth pulling on the light; the cable pulling on the light; the light pulling on the cable; and the light pulling on Earth.

Before calculating

Choose the light as the system. From the list, select the forces that belong on its free-body diagram and name the source of each.
Choose a positive vertical direction and write Newton’s second law for the light before calculating.

Then answer

Find the cable’s tension.
For each force on the light, identify its Newton’s-third-law partner. Explain whether the two forces on the light form such a pair, even though they balance.

Hint

A force’s name has a source and a recipient. Keep only forces whose recipient is the light. Reverse source and recipient to find a third-law partner.

Worked solution

Choose the object and axis

The system is the light alone. Earth exerts a downward gravitational force on the light; the cable exerts tension upward. The light’s forces on the cable and on Earth act on other objects and are absent from this diagram.

Take upward as positive. The light remains at rest, so its vertical acceleration is $a_y=0$ .

Write the equation, then calculate

T denotes the cable’s tension magnitude, measured in newtons (N). The light’s weight has magnitude mg. The signed vertical sum is:

\begin{aligned}T-mg&=ma_y\\&=0\end{aligned}

\begin{aligned}T&=mg\\&=(1.3\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\\&=12.74\,\mathrm N\end{aligned}

Pair the interactions

The partner of the cable’s force on the light is the light’s force on the cable. The partner of Earth’s gravitational force on the light is the light’s gravitational force on Earth.

Weight and tension both act on the light. Their equality here follows from zero acceleration. A third-law pair instead acts on two different objects.

Common mistake: Putting a force exerted by the light on the light’s own diagram.

Review this step

2. Read motion from an elevator force measurement

A $7.0\,\mathrm{kg}$ instrument case stays on a force plate on the level floor of an elevator. The plate measures the upward normal force on the case and reads a constant $75.6\,\mathrm N$ during a $0.90\,\mathrm s$ interval. The case stays at rest relative to the elevator.

At the start of the interval, the elevator is descending at $1.6\,\mathrm{m/s}$ . No other contact force acts on the case.

Before calculating

Isolate the case, draw its forces and choose a positive vertical direction relative to the building. Call the normal-force magnitude N and the signed acceleration $a_y$ .
Write the signed force equation and a velocity-change equation valid during this interval. Explain why the acceleration is constant.

Then answer

Find the case’s acceleration and its velocity at the end of the interval. Is the elevator still descending, and is it speeding up or slowing down?
Later, the elevator descends at constant velocity. What normal-force reading would the same case produce then?

Hint

Compare the measured normal force with mg first. Its difference from the weight fixes acceleration. Use the initial direction of motion separately when finding the final velocity.

Worked solution

Use the case as the system

The case receives normal force N upward and weight mg downward. The plate and elevator are outside the system. Take upward as positive in the building frame.

The initial velocity is $v_i=-1.6\,\mathrm{m/s}$ . Let $v_f$ be the final velocity and $\Delta t=0.90\,\mathrm s$ . N, m and g stay constant, so the net force and acceleration stay constant.

Connect the force measurement to motion

N-mg=ma_y

\begin{aligned}a_y&=\frac{N-mg}{m}\\&=\frac{75.6-7.0(9.8)}{7.0}\,\mathrm{m/s^2}\\&=+1.00\,\mathrm{m/s^2}\end{aligned}

\begin{aligned}v_f&=v_i+a_y\Delta t\\&=-1.6+(1.00)(0.90)\\&=-0.70\,\mathrm{m/s}\end{aligned}

Interpret the signs

The elevator is still descending because its final velocity is negative. Its speed has fallen from 1.6 m/s to 0.70 m/s: the upward acceleration slows the downward motion. The 75.6 N support force exceeds the 68.6 N weight.

In the later constant-velocity descent, acceleration is zero. The same force equation then gives the reading below. Velocity’s downward direction does not change this balance.

\begin{aligned}N&=mg\\&=68.60\,\mathrm N\end{aligned}

Common mistake: Choosing the acceleration sign from the direction of velocity alone.

Review this step

3. Find the friction needed by two opposing pulls

A $3.6\,\mathrm{kg}$ display stand starts at rest on a level, stationary floor. Two cords pull horizontally: one exerts $F_L=17\,\mathrm N$ to the left, and the other exerts $F_R=25\,\mathrm N$ to the right. The coefficient of static friction between stand and floor is $\mu_s=0.25$ .

In a separate trial, again starting from rest, the leftward pull is $26\,\mathrm N$ and the rightward pull remains $25\,\mathrm N$ . The stand does not tip.

Before calculating

Choose the stand alone as the system. Draw all its forces and choose horizontal and vertical axes.
For each trial, write the equations that would permit rest. Let N be the normal-force magnitude and $f_x$ the signed horizontal friction component. State the static-friction condition you must test.

Then answer

For each trial, determine whether the stand can remain at rest. Give the actual friction magnitude and direction, and the maximum available static friction.
Explain why the larger leftward pull in the second trial does not require a larger friction magnitude.

Hint

For rest, friction must cancel the sum of the two cord forces. Compare the magnitude required with $\mu_sN$ ; the maximum is a limit, not an automatic force value.

Worked solution

Test the model of rest

The system is the stand. Its forces are weight, floor normal, the two cord pulls and floor friction. Take right as positive x and up as positive y. Let $f_x$ be friction’s signed horizontal component.

If the stand can stay at rest, both acceleration components vanish. The two horizontal cord pulls have no vertical component.

N-mg=0

F_R-F_L+f_x=0

|f_x|\leq\mu_sN

Find the available friction

\begin{aligned}N&=(3.6)(9.8)\,\mathrm N\\&=35.28\,\mathrm N\end{aligned}

\begin{aligned}f_{s,\max}&=\mu_sN\\&=0.25(35.28)\,\mathrm N\\&=8.82\,\mathrm N\end{aligned}

Find the required friction in each trial

In trial 1, the cords alone would pull right. Equilibrium needs 8 N of friction left. This is below the 8.82 N limit, so rest is possible.

In trial 2, the cords alone would pull left with a net force of only 1 N. Equilibrium needs 1 N of friction right, also below the limit. Starting from rest, the stand stays at rest in both trials under this model.

\begin{aligned}\text{Trial 1: }f_x&=F_L-F_R\\&=17-25\\&=-8.00\,\mathrm N\end{aligned}

\begin{aligned}\text{Trial 2: }f_x&=26-25\\&=+1.00\,\mathrm N\end{aligned}

Check the force that friction must oppose

The larger leftward pull brings the two cord forces closer to balancing each other. That reduces the friction required and reverses its direction. The maximum available friction stays 8.82 N because neither the normal force nor the coefficient changed.

Common mistake: Setting actual static friction equal to $\mu_sN$ before checking what equilibrium requires.

Review this step

4. An angled pull while the sled slides the other way

An equipment sled of mass $m=7.5\,\mathrm{kg}$ is already sliding left across a level, stationary floor. At the instant considered, a tow line pulls with magnitude $F=28\,\mathrm N$ at angle $\theta=40^\circ$ above the horizontal toward the right. The coefficient of kinetic friction is $\mu_k=0.25$ .

The sled stays flat. Test the assumption that it remains in contact with the floor. Consider the interval while it is still sliding left.

Before calculating

Isolate the sled, choose horizontal and vertical axes and draw the four forces. Determine friction’s direction from the sliding at the floor.
Let N be the normal-force magnitude, $f_k$ the kinetic-friction magnitude and $a_x$ the signed horizontal acceleration. Write both force equations and the kinetic-friction law before substituting values.

Then answer

Find the normal force, kinetic-friction magnitude and horizontal acceleration.
Does the sled speed up or slow down at this instant? Explain why the same friction direction cannot automatically be carried through the instant when sliding stops.

Hint

The sled slides left relative to the floor, so friction acts right. Use the vertical component of the tow force to find N before calculating friction.

Worked solution

Choose the sled and the axes

Take right as positive x and up as positive y. The four forces are weight down, normal up, the angled tow force up and right, and kinetic friction right. Friction points right because the sled slides left relative to the floor.

The vertical component of the tow force is $F\sin\theta$ ; its horizontal component is $F\cos\theta$ . These components describe the same tow force. While contact with the level floor holds, $a_y=0$ .

N+F\sin\theta-mg=0

f_k=\mu_kN

F\cos\theta+f_k=ma_x

Find the normal force and friction first

\begin{aligned}N&=mg-F\sin\theta\\&=7.5(9.8)-28\sin40^\circ\approx55.50\,\mathrm N\end{aligned}

f_k=\mu_kN\approx13.88\,\mathrm N

Calculate and interpret the acceleration

The positive normal force is consistent with continued contact. Calculate $a_x$ using unrounded values.

The acceleration is rightward while the velocity is leftward, so the sled slows down. Both the tow force’s horizontal component and friction act right in this interval.

\begin{aligned}a_x=\frac{F\cos\theta+\mu_k(mg-F\sin\theta)}{m}\\&\approx+4.71\,\mathrm{m/s^2}\end{aligned}

Stop at the boundary of the model

Once the sled reaches rest, leftward sliding has ended. Test whether static friction can hold it before deciding what happens next. No static-friction coefficient was supplied, so these data do not determine that next phase. If it later slides right, kinetic friction points left.

Common mistake: Using $N=mg$ even though the pull has an upward component.

Review this step

5. Two carriers in contact, one with a brake

Carrier A has mass $m_A=4.5\,\mathrm{kg}$ and is immediately to the left of carrier B, with mass $m_B=1.5\,\mathrm{kg}$ . They move right together on a level track and remain pressed together. A horizontal drive force $F=18\,\mathrm N$ pushes A right. A controlled brake exerts a horizontal force $R=6.0\,\mathrm N$ left on B.

All other horizontal resistance is negligible. The drive and brake forces stay constant during the interval. Each carrier remains on the track.

Before calculating

Draw a free-body diagram for A and another for B. Choose one common positive horizontal direction and define C as the magnitude of their contact force.
Write one horizontal force equation for each carrier. Then identify the external horizontal forces on the combined system A + B and write its equation. State why both carriers have the same horizontal acceleration.

Then answer

Find the shared acceleration and the contact force on each carrier, including its direction.
A student uses $C=m_Ba$ for B. What force is missing? Check whether your contact-force result is consistent with the carriers remaining pressed together.

Hint

The combined system gives the acceleration without C. When you return to B alone, include both the contact force and the brake force.

Worked solution

Set up the separate systems

Take right as positive x and up as positive y for both carriers. Maintained contact along the straight track keeps their separation constant, so they share the horizontal acceleration a.

A receives drive F right, contact C left, track normal $N_A$ up and weight $m_Ag$ down. B receives contact C right, brake R left, track normal $N_B$ up and weight $m_Bg$ down.

F-C=m_Aa

C-R=m_Ba

\begin{gathered}N_A-m_Ag=0,\\N_B-m_Bg=0\end{gathered}

Use A + B for the acceleration

The contact interaction is internal to A + B. Adding the horizontal equations cancels its equal-and-opposite components. The drive and brake forces remain external.

F-R=(m_A+m_B)a

\begin{aligned}a&=\frac{18-6.0}{4.5+1.5}\,\mathrm{m/s^2}\\&=+2.00\,\mathrm{m/s^2}\end{aligned}

Return to B for the contact force

\begin{aligned}C&=m_Ba+R\\&=(1.5)(2.00)+6.0\\&=9.00\,\mathrm N\end{aligned}

Check both carriers

A exerts 9.00 N right on B; B exerts 9.00 N left on A. On B the net force is 9.00 − 6.0 = 3.00 N right, matching $m_Ba$ . On A it is 18 − 9.00 = 9.00 N right, matching $m_Aa$ .

$C=m_Ba$ omits the brake. The contact force must both overcome that brake and produce B’s acceleration. C is positive, so a compressive contact can supply the force; the maintained-contact model is consistent.

Common mistake: Treating the contact force as the entire net force on an object that also receives another horizontal force.

Review this step

6. A fixed pulley with an extra downward pull

Two masses, $m_L=1.6\,\mathrm{kg}$ on the left and $m_R=2.4\,\mathrm{kg}$ on the right, hang from the ends of one rope over a fixed pulley. A separate device below L pulls L vertically downward with constant force $P=12.8\,\mathrm N$ . The device follows L freely; it imposes no additional motion constraint.

The connecting rope stays taut, does not stretch and has negligible mass. The pulley has negligible rotational inertia and a frictionless axle. Its centre is fixed. Release the masses from rest, with room for both to move.

Before calculating

Choose each hanging mass as a separate system and draw its forces. Use upward as the positive vertical direction for both masses; call their signed accelerations $a_L$ and $a_R$ .
Write a signed force equation for each mass. Then write the acceleration constraint imposed by the connecting rope. State separately why the tension magnitude T is the same on both sides.

Then answer

Find both signed accelerations and the tension. Which mass starts moving downward?
Explain why “the heavier mass must descend” is insufficient here, and why the two upward tension forces do not cancel when the two common-axis force equations are simply added.

Hint

If one hanging rope segment lengthens, the other shortens equally. With up positive for both masses, their acceleration components have opposite signs. Include P only in L’s equation.

Worked solution

Choose each mass and a common upward axis

L receives tension T up, weight $m_Lg$ down and the device’s force P down. R receives tension T up and weight $m_Rg$ down. Use upward as positive for both $a_L$ and $a_R$ .

Negligible pulley inertia and axle friction require the two rope tensions to produce equal and opposite torques about the axle; their equal lever arms therefore give equal tension magnitudes. Negligible rope mass makes tension uniform along each free segment. Separately, fixed rope length requires one vertical segment to shorten as much as the other lengthens.

T-m_Lg-P=m_La_L

T-m_Rg=m_Ra_R

a_L+a_R=0

Use the constraint to eliminate tension

Substitute $a_R=-a_L$ into R’s equation, giving $T=m_R(g-a_L)$ . Use that in L’s equation.

(m_R-m_L)g-P=(m_L+m_R)a_L

\begin{aligned}a_L&=\frac{(2.4-1.6)(9.8)-12.8}{1.6+2.4}\,\mathrm{m/s^2}\\&=-1.24\,\mathrm{m/s^2}\end{aligned}

\begin{aligned}a_R&=-a_L\\&=+1.24\,\mathrm{m/s^2}\end{aligned}

Calculate tension and check the motion

Released from rest, L starts down and R starts up. The extra downward pull on L is larger than the difference in the two weights. Mass comparison alone therefore gives the wrong prediction.

The positive tension is consistent with a taut rope. Check L using the unrounded tension: 26.496 − 15.68 − 12.8 = −1.984 N, which equals (1.6 kg)(−1.24 m/s²).

\begin{aligned}T&=m_R(g-a_L)\\&=2.4[9.8-(-1.24)]\,\mathrm N\\&=26.496\,\mathrm N\approx26.50\,\mathrm N\end{aligned}

Keep force signs and motion signs consistent

Both tension forces point up, so both enter these common-axis equations as +T. Simply adding the equations gives +2T, not zero tension. They are not a third-law pair: each rope-on-mass force is paired with that mass’s force on the rope.

You may instead choose opposite local positive directions and use one common acceleration magnitude, but you must change both the force signs and the acceleration relation consistently. With the axes specified here, $a_L$ and $a_R$ have opposite signs.

Common mistake: Giving both hanging masses the same signed acceleration when upward is positive for both.

Review this step

Choose what to revisit

After accounting for your chosen axes, identify the first step in your setup that is inconsistent with the model. Use that solution’s review link, then close it and rebuild the diagram and symbolic equations. A correct number is most useful when you can explain which object receives each force and why each sign belongs.

Choose the questions that match the topics in your current course outline.

Useful for

Work on the step you’re missing.

If you still cannot justify the chosen object, force direction or sign, see CEGEP mechanics tutoring and find your college’s course page. Bring your diagram, first equation and current outline when asking for individual help.

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Mechanics: Six Questions to Check Your Setup

1. Which object receives the force?

Before calculating

Then answer

Choose the object and axis

Write the equation, then calculate

Pair the interactions

2. Read motion from an elevator force measurement

Before calculating

Then answer

Use the case as the system

Connect the force measurement to motion

Interpret the signs

3. Find the friction needed by two opposing pulls

Before calculating

Then answer

Test the model of rest

Find the available friction

Find the required friction in each trial

Check the force that friction must oppose

4. An angled pull while the sled slides the other way

Before calculating

Then answer

Choose the sled and the axes

Find the normal force and friction first

Calculate and interpret the acceleration

Stop at the boundary of the model

5. Two carriers in contact, one with a brake

Before calculating

Then answer

Set up the separate systems

Use A + B for the acceleration

Return to B for the contact force

Check both carriers

6. A fixed pulley with an extra downward pull

Before calculating

Then answer

Choose each mass and a common upward axis

Use the constraint to eliminate tension

Calculate tension and check the motion

Keep force signs and motion signs consistent

Choose what to revisit

Useful for

Work on the step you’re missing.

Free study guides and practice

Free-Body Diagrams and Friction: From Forces to Equations

Motion Graphs: From the Story to the Equation

Connected Objects: Contact Forces, Tension and Pulleys