Md15, Md16 - Equilibrium, sliding or toppling

Md15, Md16 - Equilibrium, sliding or toppling

Md15 combines zero resultant force with zero resultant moment to decide whether a rigid body can remain in equilibrium. Md16 checks the physical limits of that equilibrium: friction may run out, or the contact force may reach the edge of the supporting base, producing impending sliding or toppling.

Balance both forces and moments

Can making a surface rougher change a block's first failure from sliding to toppling? Extra friction can prevent slipping while the applied force still creates a turning effect. We need to check both limits, not assume that more friction guarantees equilibrium.

You already know

A particle requires zero resultant force for equilibrium. A rigid body also retains turning effects: a force has moment FdFd, with perpendicular distance dd, and a couple adds a signed moment without adding a resultant force.

Force balance prevents translational acceleration; moment balance prevents an unbalanced turning effect. A stationary rigid body needs both conditions at once.

Equilibrium of a planar rigid body

A rigid body is in equilibrium when its resultant force is zero and the sum of moments about any axis is zero. For coplanar forces, use

∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_O=0

where the moment axis is perpendicular to the plane through a stated point OO. Given force balance, zero moment about one such point implies zero moment about every point.

Not in the formula booklet: learn it.

Choose a moment point that removes unknown forces passing through it. Then use force balance for the remaining unknowns. A support underneath a body can supply an upward normal reaction, but not a negative one; an algebraic negative answer means that contact assumption fails.

Worked example 1: two supports and a point load

Question. A uniform horizontal beam ABAB, length 4 m4\,\mathrm m and weight 80 N80\,\mathrm N, rests on supports at its ends. A downward load of 40 N40\,\mathrm N is placed 1 m1\,\mathrm m from AA. Find the upward support reactions.

Plan. Take moments about AA to eliminate its reaction, then balance vertical forces.

4RB=80(2)+40(1)moments about ARB=50 NRA+RB=80+40vertical equilibriumRA=70 N\begin{aligned} 4R_B&=80(2)+40(1) && \text{moments about }A \\ R_B&=50\,\mathrm N \\ R_A+R_B&=80+40 && \text{vertical equilibrium} \\ R_A&=70\,\mathrm N \end{aligned}

Why this step? The reaction at AA has zero moment about AA, leaving one unknown in the moments equation.

Answer. RA=70 NR_A=70\,\mathrm N, RB=50 NR_B=50\,\mathrm N.

Check. Their sum is 120 N120\,\mathrm N. About BB, 70(4)=80(2)+40(3)=280 N m70(4)=80(2)+40(3)=280\,\mathrm{N\,m}.

Change only the point load's position, from 11 to 3 m3\,\mathrm m from AA. The total load stays the same, while its turning effect changes.

Worked example 2: moving the point load

Question. Find the reactions when the 40 N40\,\mathrm N load is instead 3 m3\,\mathrm m from AA.

Plan. Use moments about AA again, then the unchanged total vertical load.

4RB=80(2)+40(3)new load lever armRB=70 NRA=120−70=50 Nvertical balance\begin{aligned} 4R_B&=80(2)+40(3) && \text{new load lever arm} \\ R_B&=70\,\mathrm N \\ R_A&=120-70=50\,\mathrm N && \text{vertical balance} \end{aligned}

Why this step? A force's contribution to force balance does not change when moved along the beam, but its contribution to moment balance does.

Answer. RA=50 NR_A=50\,\mathrm N, RB=70 NR_B=70\,\mathrm N.

Check. Reflecting the arrangement in the beam's midpoint swaps the support reactions, exactly as the values show.

Force and moment equations have to describe the same body. Include the beam's own weight as well as any added load, at their separate positions.

Watch out

Wrong: use only RA+RB=120R_A+R_B=120 and conclude that both reactions are 60 N60\,\mathrm N.

Right: the moments equation decides how the load is shared. Equal reactions require the appropriate symmetry, not merely equal support heights.

The next load has a new magnitude at the first position.

What equilibrium reveals about the force system

Three non-zero coplanar forces in equilibrium must have concurrent or parallel lines of action. To see why, suppose two non-parallel lines meet at OO. Those two forces have zero moment about OO, so the third force's line must also pass through OO. If the forces are parallel, they may balance without sharing one point.

Concurrency or parallelism is necessary, not sufficient: the force magnitudes must still give zero resultant and zero moment. For two forces alone, equilibrium requires equal opposite forces on the same line.

If all applied forces have zero resultant but a non-zero total moment, an additional couple can supply the missing opposite moment. A couple cannot correct a non-zero resultant force.

Inferring an additional couple

If the known forces satisfy ∑F=0\sum\mathbf F=\mathbf0 and have total signed moment MM, equilibrium requires an added couple of signed moment

C=−MC=-M

Because the force resultant is already zero, this value is independent of the reference point.

Not in the formula booklet: learn it.

Keep a chosen moment sign throughout, and use the couple's units N m\mathrm{N\,m}. It is a turning effect, not another vertical force.

Worked example 3: forces balance but moments do not

Question. A rigid body is acted on by upward forces 30 N30\,\mathrm N at x=0x=0 and 50 N50\,\mathrm N at x=4 mx=4\,\mathrm m, and a downward force 80 N80\,\mathrm N at x=2 mx=2\,\mathrm m. Determine the additional couple required for equilibrium.

Plan. Check force balance, then calculate the signed moment about x=0x=0, taking anticlockwise as positive.

∑Fy=30+50−80=0force balance already holdsM0=50(4)−80(2)=40 N manticlockwise positiveC=−40 N mcancel the moment\begin{aligned} \sum F_y&=30+50-80=0 && \text{force balance already holds} \\ M_0&=50(4)-80(2)=40\,\mathrm{N\,m} && \text{anticlockwise positive} \\ C&=-40\,\mathrm{N\,m} && \text{cancel the moment} \end{aligned}

Why this step? Adding another force would disturb force balance. A couple changes the moment without changing the zero resultant.

Answer. A 40 N m40\,\mathrm{N\,m} clockwise couple is required.

Check. About x=4 mx=4\,\mathrm m, the forces give −30(4)+80(2)=40 N m-30(4)+80(2)=40\,\mathrm{N\,m} anticlockwise, confirming the same required couple.

Change only the downward force's position to x=3 mx=3\,\mathrm m. The force resultant stays zero, but the residual moment reverses.

Worked example 4: the required couple reverses

Question. Determine the required couple when the same 80 N80\,\mathrm N downward force acts at x=3 mx=3\,\mathrm m.

Plan. Retain force balance and recalculate the moment about x=0x=0.

∑Fy=30+50−80=0unchanged forcesM0=50(4)−80(3)=−40 N mnew lever armC=40 N mopposite signed moment\begin{aligned} \sum F_y&=30+50-80=0 && \text{unchanged forces} \\ M_0&=50(4)-80(3)=-40\,\mathrm{N\,m} && \text{new lever arm} \\ C&=40\,\mathrm{N\,m} && \text{opposite signed moment} \end{aligned}

Why this step? The required couple opposes the calculated moment, rather than automatically being clockwise.

Answer. A 40 N m40\,\mathrm{N\,m} anticlockwise couple.

Check. Adding it gives total moment −40+40=0-40+40=0, without changing any force component.

The added couple must cancel the residual moment, including its sign.

Watch out

Wrong: C=M=−40 N mC=M=-40\,\mathrm{N\,m} in the second example. That reinforces the existing clockwise moment instead of cancelling it. Correct: C=−M=40 N mC=-M=40\,\mathrm{N\,m}, anticlockwise with the chosen sign convention.

These three forces are parallel in both examples, yet still need an added couple. That is a concrete reason not to treat the three-force geometry condition as a complete equilibrium test.

Exam tip

State the reference point and the positive turning sense before a moments equation. A clockwise couple has the same clockwise effect about every point; its sign does not change when your pivot moves.

The next system also has zero resultant force, leaving its moment as the deciding calculation.

A pushed block has two different thresholds

Model a uniform upright block on rough horizontal ground. Its weight WW acts through the geometric centre. A horizontal push PP acts at height hh, and its base has width bb in the push direction. Increase PP slowly from zero, so the question is which static equilibrium condition fails first.

Vertical and horizontal balance give R=WR=W and required friction F=PF=P. Sliding is impending at P=μWP=\mu W. Meanwhile the resultant normal contact can move towards the leading edge of the base, but cannot move beyond it.

[DIAGRAM: asset_name: Sliding and toppling thresholds for a pushed block; asset_slug: h645_b059_push; caption: Sliding exhausts friction while toppling moves the contact resultant to the leading edge; description: A uniform upright block on horizontal rough ground has base width b, downward weight W through its centre, and rightward push P at height h. Leftward friction and upward normal reaction act at the ground. A separate limiting diagram places both contact components at the lower right pivot. Dimensions mark the width b and push height h. Equations show P slide equals mu W and P topple times h equals W b over two.]
Sliding exhausts friction while toppling moves the contact resultant to the leading edge

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