Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: static equilibrium (203-SN1-RE)

Q: Why is a zero sum of forces not enough for equilibrium?

Because it concerns translation only. Two equal and opposite forces applied at two different points form a couple: their sum is zero, but they make the object rotate indefinitely. The condition on the torques must therefore be added, which rules out rotation. The two conditions describe two distinct motions.

Q: How do I choose the pivot in a statics problem?

You place it on an unknown force, preferably the one whose direction you also do not know, such as a hinge. Its torque then becomes zero and the equation contains only one unknown. Since the sum of the torques is zero about any point, this choice is entirely free, and it is the main technique of the chapter.

Q: Why does the statement say the wall is smooth in the ladder problem?

Because a smooth surface can exert only a force perpendicular to itself, so here strictly horizontal. Without that statement, there would be a fourth unknown, the friction of the wall, for only three equations, and the problem would be unsolvable at the CEGEP level.

Q: When does an object tip rather than slide?

It tips when the vertical through its centre of gravity leaves its base of support, and it slides when the horizontal force exceeds the maximum static friction. These are two independent thresholds, which must be calculated separately: the one reached first decides. Pushing low favours sliding, pushing high favours tipping.

Q: What does a negative support reaction mean?

It means the support would have to pull downward instead of pushing upward. A simple support cannot do that, you need an anchor or a bolt. It is therefore not a calculation error but information about the structure, and it is often the point the reasoning question is looking for.

Q: Where does a distributed load on a beam act?

At the midpoint of the zone it covers, if it is uniform. You replace it by a single force equal to the load per unit length multiplied by the loaded length, placed at the centre of that length. Placing it at the end of the beam doubles its torque and throws off every reaction.

In 203-SN1-RE (formerly 203-NYA), statics is won before the calculation begins. A well-chosen pivot eliminates two unknowns at once and turns a system of three equations into a single line. A badly chosen pivot leaves three unknowns and gives the false impression that the statement lacks data.

This sheet gives the method in the order in which it is written up, and gathers the errors that really cost marks: the forgotten condition, the misplaced centre of gravity, the smooth surface credited with friction, and the two ways of losing equilibrium that must be tested separately. It is written for CEGEP students in the Science program (Sciences de la nature).

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The thread of the chapter

Two conditions, not one: $\Sigma\vec{F}=\vec{0}$ AND $\Sigma\tau=0$ . The first alone lets an object spin in place, and that is exactly what the marker checks first on a statics paper.

This chapter is part of 203-SN1-RE, Mechanics

The essentials

Two conditions, and why one alone is not enough

• $\Sigma F_{x}=0$ , $\Sigma F_{y}=0$ and $\Sigma\tau=0$ : three equations in two dimensions, so at most three unknowns.
• Two equal and opposite forces applied at different points form a COUPLE: the resultant is zero, the torque is not, and the object rotates.
• At equilibrium, $\Sigma\tau=0$ about ANY point, including a point outside the body. This is what allows you to choose the most convenient pivot.
• Sign convention: positive counterclockwise. It must be the same for every term of the equation.
• A force applied AT the pivot has zero torque: that is the whole point of placing the pivot on an unknown.

The two forces are equal and opposite, so

\Sigma\vec{F}=\vec{0}

: the first condition is met. Applied at two different points, their torques add, and the bar rotates.

The number of equations is fixed, the number of unknowns is not: beyond three, the problem is statically indeterminate and falls outside the course. If you count four, an assumption has been forgotten, often “the surface is smooth”.

Concurrent forces: the lamp hung from two cables

• When all the forces pass through the same point, the torque condition is automatic: $\Sigma F_{x}=0$ and $\Sigma F_{y}=0$ are enough.
• Project each tension: $T_{1}\cos\theta_{1}=T_{2}\cos\theta_{2}$ horizontally, $T_{1}\sin\theta_{1}+T_{2}\sin\theta_{2}=W$ vertically.
• The cable closest to the vertical carries the most; a nearly horizontal cable demands an enormous tension.

Cables at

30°

and

60°

from the horizontal:

T_{1}=W\sin 30°=50

T_{2}=W\sin 60°=87

N. Placed head to tail, the three forces close into a triangle.

Visual check: placed head to tail, the forces on a body in equilibrium form a CLOSED polygon.

Where each force acts

• The weight of a body acts at its CENTRE OF GRAVITY, which is at the midpoint only if the body is uniform and symmetric.
• Composite body: $x_{G}=\dfrac{\sum m_{i}x_{i}}{\sum m_{i}}$ . For a uniform plate, the masses are proportional to the AREAS.
• Plate with a hole: treat the hole as a NEGATIVE mass, which avoids any complicated cutting up.
• A load spread uniformly over a length $\ell$ is equivalent to its resultant applied at the MIDPOINT of $\ell$ , not at the end of the beam.
• A SMOOTH surface can exert only a force NORMAL to itself: this assumption is what makes the ladder problem solvable.

Each weight starts from its own centre of gravity, the beam at its midpoint, the load at

2.00

m; each reaction starts from its support. Forces to scale.

A body at rest stays in equilibrium as long as the vertical through its centre of gravity falls INSIDE the base of support. As soon as it leaves it, the body tips, whatever its mass.

The lever and the human arm

• Lever: a pivot, a driving force, a load. At equilibrium, the torques are equal, $F_{E}d_{E}=F_{L}d_{L}$ : the longer arm needs the smaller force.
• The forearm is a lever that is unfavourable in force: the biceps is attached $5$ cm from the elbow, the load is in the hand at $35$ cm. The muscle therefore pulls about eight times harder than the load.
• Pivot at the elbow: the joint force drops out of the torque equation; it is then found from $\Sigma F_{y}=0$ .

387

N at the biceps for

49.0

N in the hand, and the elbow compressed by

323

N. To scale, except the weight of the forearm, lengthened to remain visible.

A beam balance is a lever with equal arms: it compares two weights, which is why it gives the same reading on the Moon.

The cantilevered beam held by a cable

• Beam hinged on a wall and held by an oblique cable: pivot at the hinge, which eliminates its TWO unknown components at once.
• Only the component of the cable perpendicular to the beam has a torque, $T\sin\theta\times d$ . The flatter the cable, the greater the tension must be.
• The hinge components then follow from $\Sigma F_{x}=0$ and $\Sigma F_{y}=0$ ; a negative sign says it pulls instead of pushing.

Crane boom isolated, forces to scale: a cable at

40°

pulls

4675

N for a load of

1960

N, and the hinge pushes almost horizontally.

A sign at the end of a bracket is exactly this problem: the tension in the cable almost always exceeds the weight of the sign.

The two ways of losing equilibrium

• SLIDING: the horizontal driving force exceeds $\mu_{s}N$ . The condition involves only forces.
• TIPPING: the vertical through the centre of gravity leaves the base of support, or equivalently the normal force would have to act outside the contact surface.
• The two conditions are INDEPENDENT: you must test both and keep the one that happens first.
• A force applied high favours tipping; a force applied low favours sliding. That is why you push furniture low to move it.
• Stress and strain: $\sigma=\dfrac{F}{A}$ , $\varepsilon=\dfrac{\Delta L}{L_{0}}$ , and $\sigma=E\varepsilon$ with $E$ Young's modulus, in pascals.

As long as the vertical through

G

falls between the two edges of the base, the body rights itself. On the right, it passes exactly through the edge: one more degree and the weight tips the body instead of righting it.

A question that asks “what happens first” explicitly expects both calculations and a comparison. One of the two alone, even if correct, does not answer the question asked.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Concluding equilibrium from the condition on the forces alone

the whole reasoning question, 4 marks

What not to write

“The sum of the forces is zero, so the beam is in equilibrium.”

What to write

“ $\Sigma\vec{F}=\vec{0}$ only guarantees that the centre of mass does not accelerate. $\Sigma\tau=0$ is still needed to rule out rotation.”

Why: A couple of forces has a zero resultant and yet makes the object rotate indefinitely. The two conditions describe two different motions, and neither replaces the other.

2. Changing pivot in the middle of a single torque equation

the whole question, 6 marks

What not to write

“The torque of the weight about the left support, plus that of the load about the right support, equals zero.”

What to write

“All the torques in a single equation are calculated about the SAME pivot, stated at the start of the line: ‘torques about the left support’.”

Why: $\Sigma\tau=0$ holds about each point taken separately, never about two points mixed together. The error is invisible in the calculation and gives a plausible but wrong result.

3. Placing the weight at the geometric midpoint of a non-uniform body

4 marks

What not to write

“The beam is $6.0$ m long, so its weight acts at $3.0$ m from the end, even though it carries a block welded at $5.0$ m.”

What to write

“The weight of the beam alone does act at $3.0$ m, but the block has its own weight applied at $5.0$ m. Two forces, two points of application, or else a recalculated centre of gravity.”

Why: The centre of gravity is an average weighted by the masses. Adding two bodies without recalculating amounts to assuming that they have the same mass and the same position.

4. Giving a friction force to a wall declared smooth

the whole question, 8 marks

What not to write

“The ladder leans on the smooth wall: the wall exerts a normal force and a friction force that stops it from sliding down.”

What to write

“A SMOOTH surface exerts only a force NORMAL to itself: here, a strictly HORIZONTAL force. That is what leaves three unknowns for three equations.”

Why: Without this assumption, the problem would have four unknowns and would be unsolvable at the CEGEP level. The word “smooth” in the statement is not a detail of the scenery, it is the given that closes the system.

5. Writing the floor friction equal to $\mu_{s}N$ when the ladder is not yet slipping

3 marks, and a wrong limiting position

What not to write

“The ladder is resting and motionless, so $f=\mu_{s}N=0.40\times 882=353$ N.”

What to write

“As long as it is not slipping, $f$ equals what horizontal equilibrium requires, that is $N_{wall}$ . We write $f=\mu_{s}N$ ONLY to find the limiting position.”

Why: It is the same trap as the motionless block in the dynamics chapter, in a different setting. The equality holds only at the instant of breakaway, never before.

6. Testing only one of the two ways equilibrium can fail

the whole question, 5 marks

What not to write

“The crate starts to slide at $F=180$ N, so that is the answer.”

What to write

“The tipping force must also be calculated, here $F=140$ N. Since $140<180$ , the crate TIPS before it slides.”

Why: The two conditions are independent and occur at different thresholds. The question “what happens first” requires both numbers, then their comparison.

7. Replacing a distributed load by a force applied at the wrong point

4 marks, and a doubled torque

What not to write

“The load of $200$ N/m is spread over the left $3.0$ m: I replace it by $600$ N applied at the end of the beam.”

What to write

“ $600$ N applied at the MIDPOINT of the loaded zone, that is at $1.5$ m from the left end. The resultant of a uniform load passes through the centre of the zone.”

The resultant of a uniform load starts from the MIDPOINT of the loaded zone, at

1.5

m: placing it at the end, at

3.0

m, doubles its torque,

1800

N·m instead of

900

Why: The resultant of a uniform distribution acts at its centre of gravity, exactly like the weight of a uniform beam. Placing it at the end multiplies its torque by two.

8. Choosing a pivot that simplifies nothing

no marks lost, but ten minutes and a pointless system

What not to write

“I take torques about the centre of the beam, and I get an equation with both support reactions unknown.”

What to write

“Place the pivot ON one of the two reactions: its torque becomes zero and the equation contains only one unknown, which can be calculated immediately.”

Why: The pivot is a free choice, therefore a tool. Taking it at the centre out of reflex means giving up its only virtue: cancelling the torque of what you do not know.

9. Forgetting to check the sign of a support reaction

2 marks, and the physical conclusion missed

What not to write

“The reaction of the second support is $-120$ N, so I made a calculation error somewhere.”

What to write

“A negative reaction means the support must PULL downward: an anchor is needed, a simple support is not enough. It is a physical answer, not an error.”

Why: A simple support can only push. A negative reaction is therefore valuable information about the structure, and it is often the reasoning question of the problem.

Which method to choose

Where to place the pivot

The list of unknown forces, before writing a single equation

If two support reactions are unknown → pivot on one of the two, the equation gives the other directly

Example: beam on two supports, right reaction obtained in one line
If a hinge or an axle exerts a force of unknown direction → pivot on that hinge, which eliminates TWO unknowns at once

Example: crane boom hinged to the mast, ladder resting on the floor
If you are looking precisely for the force applied at the obvious pivot → pivot elsewhere, then return to $\Sigma F=0$ to get the force sought

Example: floor reaction on a ladder, obtained from the vertical sum
If all the forces are vertical → any point of the beam will do, choose the one that gives the simplest lever arms

Example: horizontal beam with suspended loads
If three non-parallel forces maintain equilibrium → their lines of action are CONCURRENT, and the meeting point is the best pivot

this theorem gives the direction of the third force with no calculation

Once the pivot is chosen, write it out in full at the start of the line. A marker who does not know where the pivot is cannot validate any sign, and an equation whose signs cannot be validated earns nothing.

Which condition to write depending on the question

The verb of the question

If “determine the reaction”, “the tension in the cable”, “the force of the hinge” → the three equilibrium equations, with a well-chosen pivot

Example: beam, crane, hanging sign
If “how far can it go”, “what maximum distance”, “at what limiting position” → set $f=\mu_{s}N$ at the limit, then solve for the position

Example: ladder, person walking out on an overhanging plank
If “does the object slide or tip” → calculate BOTH thresholds and compare, never just one

Example: crate pushed at varying height
If “where is the centre of gravity” → average weighted by the masses, or by the areas if the body is uniform, with holes counted negatively

Example: L-shaped plate, disc with a hole
If “by how much does the cable stretch” → $\Delta L=\dfrac{FL_{0}}{AE}$ , after obtaining $F$ from equilibrium

statics gives the force, Young's modulus gives the deformation

Statics questions are almost always posed in this order: first the reactions, then the limit, finally the deformation. Handling the second before the first leads to writing $\mu_{s}N$ with an $N$ that has not been calculated yet.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a ladder-against-a-wall problem

When to use it: An elongated object leans on rough ground on one side and on a wall on the other, and a force or a limiting position is asked for.

1 Draw the diagram and place ALL the forces on it with their points of application: normal force of the floor, friction of the floor, normal force of the wall, weight of the ladder at its midpoint, weight of the load at its position.
2 State explicitly that the wall is smooth, so that its force is horizontal. This is the sentence that makes the system solvable.
3 Write $\Sigma F_{y}=0$ to get the normal force of the floor, then $\Sigma F_{x}=0$ to relate the friction to the force of the wall.
4 Write $\Sigma\tau=0$ about the FOOT of the ladder, which cancels the normal force of the floor and the friction in one stroke.
5 Set $f=\mu_{s}N$ for the limiting condition, then solve for the quantity asked and conclude by comparing it with the length of the ladder.

Concluding sentence

“Since the wall is smooth, its force is horizontal. Torques about the foot of the ladder give $N_{wall}L\sin\theta=mg\dfrac{L}{2}\cos\theta+Mgd\cos\theta$ ; at the limit of slipping $N_{wall}=\mu_{s}N=353$ N, hence $d=3.74$ m, that is $75\%$ of the length of the ladder.”

The trap: Taking the pivot at the top of the ladder. The equation then contains three unknowns instead of one, and you think a given is missing.

Marking: Typically 2 marks for the complete diagram, 1 for the smooth-wall assumption, 3 for the three equations, 2 for the limiting condition, 1 for the conclusion compared with the length of the ladder.

Locating the centre of gravity of a composite body

When to use it: The statement describes a plate in several pieces, a part with a hole, or a beam carrying masses.

1 Choose an origin and an axis, and draw them: without a reference frame, no coordinate means anything.
2 Split the body into simple shapes whose centre of gravity is known, and draw up a table with three columns: mass or area, x-coordinate, product.
3 Count holes as NEGATIVE masses, with their own x-coordinate.
4 Apply $x_{G}=\dfrac{\sum m_{i}x_{i}}{\sum m_{i}}$ , then check that the result falls inside the body, or explain why it does not.

Concluding sentence

“With the origin at the lower left corner, the table gives $\sum m_{i}=100$ kg and $\sum m_{i}x_{i}=420$ kg·m, so $x_{G}=4.20$ m, a value comfortably between the two pieces.”

The trap: A centre of gravity can lie OUTSIDE the material, for example at the centre of a ring or in the notch of an L-shaped piece. This is not an error, and you should say so rather than correct the result.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

A second pivot, taken at random

Recalculate $\Sigma\tau$ about a different point, with the values found. The sum must be zero there too, otherwise a reaction is wrong.

A sum of torques of $+180$ N·m about the right support when it is zero about the left: the left reaction is wrong.
The sum of the reactions against the total load

For a horizontal beam, the sum of the vertical reactions must equal exactly the total weight supported. The check takes five seconds.

Two reactions of $400$ N and $300$ N for a total load of $800$ N: $100$ N is missing somewhere.
The limiting position compared with the geometry

A limiting distance must fall within the dimensions of the problem. A $d$ greater than the length of the ladder means the ladder never slips, which must then be stated.

A $d=6.2$ m on a $5.0$ m ladder: either an error, or the conclusion that the person can climb all the way to the top.
The limiting case of the angle

Let the angle of the ladder tend to $90°$ in the final expression: the friction needed must tend to zero, since a vertical ladder does not slip.

An expression that demands more friction when the ladder is stood up has swapped sine and cosine.
The direction of each reaction reread on the diagram

Check that a positive reaction really pushes in the direction drawn. A negative value must be interpreted, not silently corrected.

A negative support reaction on an overhang: the support must hold down, an anchor is needed.

The typical problem, taken apart

The ladder against a smooth wall, and how far up you can climb

A uniform ladder of $5.00$ m and $20.0$ kg leans against a SMOOTH vertical wall, making an angle of $60.0°$ with the horizontal floor. The coefficient of static friction between the foot of the ladder and the floor is $\mu_{s}=0.400$ .

A person of $70.0$ kg starts to climb. How far $d$ , measured along the ladder from the foot, can they climb before the ladder starts to slip? Take $g=9.80$ m/s $^{2}$ .

Since the wall is smooth, its force is strictly horizontal: that is what leaves three unknowns for three equations. The foot of the ladder is the best pivot, since two of the unknown forces act there.

Step 1

Forces present: $\vec{N}$ vertical and $\vec{f}$ horizontal at the foot, $\vec{N}_{wall}$ horizontal at the top because the wall is SMOOTH, the weight of the ladder $mg$ at $2.50$ m from the foot, that of the person $Mg$ at the distance $d$ .

Why

Listing the forces with their points of application is half the work. The mention of the smooth wall is not a formality: it removes the fourth unknown and makes the system solvable.

Step 2

$\Sigma F_{y}=0$ : $N=(m+M)g=(20.0+70.0)(9.80)=882$ N.

Why

The vertical equation contains only one unknown, since the wall force is horizontal. It is therefore handled first, and its result will be used for the limiting condition.

Step 3

$\Sigma F_{x}=0$ : $f=N_{wall}$ . At the limit of slipping, $f=\mu_{s}N=0.400\times 882=353$ N, so $N_{wall}=353$ N.

Why

The equality $f=\mu_{s}N$ is written only here, because the question is explicitly about the LIMITING position. As long as the person is lower, $f$ is less than that.

Step 4

Torques about the FOOT of the ladder: $N_{wall}(L\sin 60°)=mg\left(\dfrac{L}{2}\cos 60°\right)+Mg\,(d\cos 60°)$ , that is $353\times 4.33=20.0(9.80)(2.50)(0.500)+70.0(9.80)\,d\,(0.500)$ .

Why

The pivot at the foot cancels the torques of $N$ and $f$ in one stroke, the two forces applied there. That is exactly why the pivot is never taken at the top here.

Step 5

$1528=245+343\,d$ , so $d=\dfrac{1283}{343}=3.74$ m.

Why

The calculation reduces to a first-degree equation in one unknown, solely thanks to the choice of pivot. With another pivot, you would have to solve a system of three equations.

Step 6

Comparison: $3.74$ m out of $5.00$ m, that is $74.8\%$ of the ladder, at a height of $3.74\sin 60°=3.24$ m above the ground.

Why

A distance along the ladder is not a height. The fact that the result is less than the total length also confirms that a limiting position does exist, which was not guaranteed in advance.

Step 7

Check using the limiting case: with $\theta=90°$ , the term $\cos\theta$ vanishes and $d$ would become infinite, which correctly expresses that a vertical ladder never slips.

Why

This test validates the structure of the formula, not just the arithmetic. If the limiting case gave a zero $d$ , sine and cosine would have been swapped.

The conclusion, written out

“The person can climb up to $3.74$ m along the ladder, about three quarters of its length and a height of $3.24$ m; beyond that, the force required of the wall exceeds what the friction of the floor can balance and the ladder slips.”

The classic mistake on this problem: Taking the pivot at the top of the ladder. The equation then contains $N$ , $f$ AND $d$ , and you conclude that a given is missing. The problem is perfectly determined, it was the pivot that was badly placed.

Learn by heart

• Equilibrium: $\Sigma F_{x}=0$ , $\Sigma F_{y}=0$ AND $\Sigma\tau=0$ . Three equations, at most three unknowns.
• $\Sigma\tau=0$ about ANY point: choose the one that cancels the most unknowns.
• A force applied at the pivot has zero torque; so does a force whose line of action passes through the pivot.
• A SMOOTH surface exerts only a NORMAL force. That is the assumption that closes the ladder system.
• The weight acts at the centre of gravity: $x_{G}=\dfrac{\sum m_{i}x_{i}}{\sum m_{i}}$ , holes counted negatively.
• Uniform distributed load: resultant at the MIDPOINT of the loaded zone.
• Sliding and tipping are two independent thresholds: calculate both and compare.
• $\sigma=\dfrac{F}{A}$ , $\varepsilon=\dfrac{\Delta L}{L_{0}}$ , $\sigma=E\varepsilon$ , so $\Delta L=\dfrac{FL_{0}}{AE}$ .

Frequently asked questions

Why is a zero sum of forces not enough for equilibrium?

Because it concerns translation only. Two equal and opposite forces applied at two different points form a couple: their sum is zero, but they make the object rotate indefinitely. The condition on the torques must therefore be added, which rules out rotation. The two conditions describe two distinct motions.

How do I choose the pivot in a statics problem?

You place it on an unknown force, preferably the one whose direction you also do not know, such as a hinge. Its torque then becomes zero and the equation contains only one unknown. Since the sum of the torques is zero about any point, this choice is entirely free, and it is the main technique of the chapter.

Why does the statement say the wall is smooth in the ladder problem?

Because a smooth surface can exert only a force perpendicular to itself, so here strictly horizontal. Without that statement, there would be a fourth unknown, the friction of the wall, for only three equations, and the problem would be unsolvable at the CEGEP level.

When does an object tip rather than slide?

It tips when the vertical through its centre of gravity leaves its base of support, and it slides when the horizontal force exceeds the maximum static friction. These are two independent thresholds, which must be calculated separately: the one reached first decides. Pushing low favours sliding, pushing high favours tipping.

What does a negative support reaction mean?

It means the support would have to pull downward instead of pushing upward. A simple support cannot do that, you need an anchor or a bolt. It is therefore not a calculation error but information about the structure, and it is often the point the reasoning question is looking for.

Where does a distributed load on a beam act?

At the midpoint of the zone it covers, if it is uniform. You replace it by a single force equal to the load per unit length multiplied by the loaded length, placed at the centre of that length. Placing it at the end of the beam doubles its torque and throws off every reaction.

Practise it

Corrected exercises: Static equilibrium, 203-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: static equilibrium (203-SN1-RE)

The essentials

Two conditions, and why one alone is not enough

Concurrent forces: the lamp hung from two cables

Where each force acts

The lever and the human arm

The cantilevered beam held by a cable

The two ways of losing equilibrium

The mistakes that cost marks

1. Concluding equilibrium from the condition on the forces alone

2. Changing pivot in the middle of a single torque equation

3. Placing the weight at the geometric midpoint of a non-uniform body

4. Giving a friction force to a wall declared smooth

5. Writing the floor friction equal to μsN\mu_{s}Nμs​N when the ladder is not yet slipping

6. Testing only one of the two ways equilibrium can fail

7. Replacing a distributed load by a force applied at the wrong point

8. Choosing a pivot that simplifies nothing

9. Forgetting to check the sign of a support reaction

Which method to choose

Where to place the pivot

Which condition to write depending on the question

How the answer is expected to be written

Solving a ladder-against-a-wall problem

Locating the centre of gravity of a composite body

Check before you hand in

The typical problem, taken apart

The ladder against a smooth wall, and how far up you can climb

Learn by heart

Frequently asked questions

Why is a zero sum of forces not enough for equilibrium?

How do I choose the pivot in a statics problem?

Why does the statement say the wall is smooth in the ladder problem?

When does an object tip rather than slide?

What does a negative support reaction mean?

Where does a distributed load on a beam act?

Corrected exercises: Static equilibrium, 203-SN1-RE

See also

Looking for a 203-SN1-RE tutor in Montreal?

5. Writing the floor friction equal to $\mu_{s}N$ when the ladder is not yet slipping