Equilibrium of a rigid body

Why a handle is far from the hinge

• Push near a door hinge and you need a large force. Push at the handle and the same force has a larger turning effect 转动效应. To predict balance we must know both the forces and where they act.
• Model the object as a rigid body 刚体: its shape and the distances between its parts stay fixed. We consider coplanar forces 共面力, all in one plane. A particle model would lose the information needed to predict turning.

Measure a moment from the line of action

• The moment 力矩 of a force about a point is force multiplied by the perpendicular distance 垂直距离 from that point to the force's line of action 作用线. The unit is $\text{N m}$. Choose a sign, such as anticlockwise positive.
• A $20\ \text{N}$ upward force acting $0.6\ \text{m}$ to the right of a pivot 支点 has moment $+12\ \text{N m}$. If the line of action passes through the pivot, its moment is zero even when the force itself is not zero.

An angled force needs its perpendicular arm

• A force of $30\ \text{N}$ acts at $30^\circ$ above a horizontal rod, $4\ \text{m}$ from the pivot. Its perpendicular 垂直 component is $30\sin30^\circ=15\ \text{N}$, giving moment $15(4)=60\ \text{N m}$ anticlockwise.
• Equivalently use the full force and arm $4\sin30^\circ=2\ \text{m}$. These are two ways to calculate the same moment; do not insert the sine factor twice.

Equilibrium needs two separate balances

Replace gravity by one force at the centre of mass

• In a uniform gravitational field, the weight $F_W=mg$ acts through the centre of mass 质心. A uniform 均匀 body has mass distributed evenly over its length, area or volume, as appropriate.
• Use symmetry 对称: a uniform rod has its centre of mass at its midpoint; a uniform rectangle has it where its diagonals meet. The centre of mass lies on every symmetry axis of the mass distribution. Shape symmetry alone is not enough if the density 密度 varies, and a centre of mass may lie in a hole outside the material.

Locate the centre of a triangular lamina

• A lamina 薄片 is a thin plate. For a uniform triangular lamina, the centre of mass lies two-thirds of the way along each median 中线 from its vertex, or one-third from the opposite side. The MF19 formula is given; a proof is not required here.
• For vertices $(0,0)$, $(6,0)$ and $(0,3)$ in metres, the midpoint opposite the origin is $(3,1.5)$. Two-thirds of this displacement gives $(\bar x,\bar y)=(2,1)\ \text{m}$. Equivalently, average the three vertex coordinates for a uniform triangle.

Read the reference point in a solid-body formula

• The formula list places the centre of mass of a uniform solid cone or pyramid $3h/4$ from its vertex, hence $h/4$ from its base. For a solid hemisphere of radius $r$, it is $3r/8$ from the base-circle centre along the axis towards the curved surface.
• A thin hemispherical shell 半球壳 instead has distance $r/2$. A shell is not a solid hemisphere. Convert every position to the same origin and signed axis before combining bodies.

Distinguish an arc from a sector

• For a uniform circular arc 圆弧 of radius $r$ and angle $2\alpha$, MF19 gives centre-of-mass distance $r\sin\alpha/\alpha$ from the circle centre on the symmetry axis. For a uniform circular sector 扇形 lamina, the distance is $2r\sin\alpha/(3\alpha)$. Here $\alpha$ is the half-angle in radians.
• A semicircle has $\alpha=\pi/2$: the wire arc gives $2r/\pi$, while the sector plate gives $4r/(3\pi)$. Use length for the mass of a uniform wire and area for a uniform plate of fixed thickness; these objects do not have the same mass model.

Replace each component by one particle

• A composite body 复合物体 consists of simpler parts. Replace each part by a particle of mass $m_i$ at its own centre of mass $(x_i,y_i)$. Taking moments of the weights gives

• Mass is proportional to length for a uniform wire, area for a uniform lamina of fixed thickness, or volume for a uniform solid. Use an area or volume ratio only when the relevant densities are equal. A plain average of component positions works only for equal component masses.

Add a rectangle and a triangle

• Join a rectangle $0\le x\le4$, $0\le y\le2$ to a triangular plate with vertices $(4,0)$, $(4,2)$ and $(7,0)$, with lengths in metres. Both have the same thickness and density. Their areas are $8$ and $3$, and their centres are $(2,1)$ and $(5,2/3)$.
• Therefore $\bar x=(8\cdot2+3\cdot5)/11=31/11\ \text{m}$ and $\bar y=(8\cdot1+3\cdot2/3)/11=10/11\ \text{m}$. The larger part contributes more to each moment.

Subtract a removed part from both mass and moments

• Start with a $6\ \text{m}$ by $4\ \text{m}$ uniform rectangular plate. Remove the upper-right $4\ \text{m}$ by $2\ \text{m}$ rectangle, leaving an L-shape. Using the lower-left corner as origin, the original area is $24$ at $(3,2)$ and the removed area is $8$ at $(4,3)$.
• Subtract the hole's area and both its moments:

Change the mass weights when the material changes

• Two uniform thin rods of the same material and cross-section form an L: a $6\ \text{m}$ horizontal rod from the origin, and a $4\ \text{m}$ vertical rod. Use lengths $6$ and $4$ as mass weights. Their centres are $(3,0)$ and $(0,2)$, giving $(\bar x,\bar y)=(1.8,0.8)\ \text{m}$.
• If the vertical rod has twice the mass per unit length, the weights become $6$ and $8$, giving $(9/7,8/7)\ \text{m}$. Do not use the area of the empty region between the rods.

Join solids with a signed common axis

• Join a uniform solid cone of radius $r$, height $4r$, to a uniform solid hemisphere of radius $r$ at their circular bases. They have the same density. Put the common base at $z=0$ and take positive $z$ towards the cone's vertex.
• The cone has volume $4\pi r^3/3$ and centre $z=r$. The hemisphere has volume $2\pi r^3/3$ and centre $z=-3r/8$. Thus

Axial sections of the solids; use their volumes as mass weights, rather than the drawn triangle and semicircle areas.

Remove a cavity without changing its original centre

• A uniform cone has base radius $2r$ and height $4r$. Remove a solid hemisphere of radius $r$ whose base lies in the cone's base plane, with the same axis and its curved surface inside the cone. The original cone's volume and centre are $16\pi r^3/3$ and $z=r$; the removed hemisphere's are $2\pi r^3/3$ and $z=3r/8$.
• With $z$ measured upwards from that base, the remaining centre is

Removing material near the base moves the centre upwards. A hole has a negative mass contribution in the calculation; it does not have an unknown new centre.

A freely suspended body puts its centre below the support

• When a body hangs freely from a smooth pivot, the support force and weight must share a vertical line. For stable rest, the centre of mass is vertically below the support. A vertical line with the centre above the support is the unstable orientation.
• Suspend our L-shaped plate from its upper-left corner $D=(0,4)$. In the plate's original coordinates, $G-D=(2.5,-2.5)$. The angle between the downward edge direction and $DG$ is $\tan^{-1}(2.5/2.5)=45^\circ$. The plate rotates until $DG$ points vertically down.

Check whether a parameter root leaves a real support point

• A $4$ by $6$ uniform plate has an upper-right rectangle of width $h$ and height $2$ removed, where $0<h<4$. Distances are in metres; the support is its surviving upper-left corner $D=(0,6)$. The area method gives

• If the original downward edge makes angle $\theta$ with the hanging vertical and $\tan\theta=1/2$, use $\bar x/(6-\bar y)=1/2$. This gives $(h-3)(h-4)=0$. Only $h=3$ is allowed: at $h=4$ the entire top strip, including $D$, has been removed. Algebraic roots must obey the body's geometry.

Draw every external force before choosing equations

• A free-body diagram 受力图 contains the body and the forces acting on it. Weight acts at the centre of mass; a light string pulls along itself; a normal reaction 法向反力 is perpendicular to a contact surface. A smooth contact has no friction. A hinge can exert horizontal and vertical reaction components.
• At a rough contact, friction 摩擦力 opposes the direction of impending relative sliding 滑动. Static friction obeys $|F_f|\le\mu F_N$, where $\mu$ is the coefficient of friction 摩擦系数. The limiting friction 最大静摩擦 is $\mu F_N$. Equality holds at limiting equilibrium 极限平衡, not at every resting contact. A support that can only push must also have $F_N\ge0$.

Balance a beam and then find its reaction

A loaded beam needs the load positions

• A uniform horizontal beam $AB$ of length $6\ \text{m}$ and weight $120\ \text{N}$ rests on supports at its ends. A $60\ \text{N}$ load is $5\ \text{m}$ from $A$. The upward reactions satisfy $R_A+R_B=180$.
• Moments about $A$: $6R_B=120(3)+60(5)=660$, giving $R_B=110\ \text{N}$ and $R_A=70\ \text{N}$. Both reactions are positive. If a calculated reaction at a simple support is negative, the assumed contact cannot supply it; check whether the body loses contact or topples 倾倒.

Resolve an angled rod and its hinge reaction

• A uniform rod of length $4\ \text{m}$ and weight $100\ \text{N}$ is hinged at $A$ and held at $30^\circ$ above horizontal by a horizontal light string attached at its other end. The string pulls towards the wall through $A$.
• Moments about $A$ use the vertical arm for the horizontal tension and horizontal arm for the weight:

The hinge reaction components are $50\sqrt3\ \text{N}$ rightwards and $100\ \text{N}$ upwards. Its magnitude is $50\sqrt7\ \text{N}$, not their sum.

Set up the geometry of a rough-wall rod

• A uniform rod $AB$ has length $4a$ and weight $F_W$, with its lower end $A$ against a rough vertical wall. Its midpoint is $C$. A light string of length $2a$ joins $C$ to a point $D$ higher on the wall, so $AC=CD$. The rod makes angle $\theta$ with the wall, where $\tan\theta=4/3$.
• A particle of weight $2F_W$ is attached at $AE=ka$, between $C$ and $B$. The coefficient of friction at $A$ is $1/4$, and the rod is about to slip down. The unknowns are the string tension, contact forces and position parameter $k$.

Resolve the rough-wall forces first

• The isosceles triangle $ACD$ makes the string's angle with the vertical equal to $\theta$. Friction acts upwards against impending downward slipping. Using $\sin\theta=4/5$, $\cos\theta=3/5$ and the limiting relation $F_f=F_N/4$ gives

• Substitute the first equation into the second: $4F_T/5=3F_W$. Hence $F_T=15F_W/4$, $F_N=3F_W$ and $F_f=3F_W/4$. The extra particle doubles its own weight contribution; the total downward weight is $3F_W$.

Finish the rough-wall problem with one moment equation

• Taking moments about $A$ removes both contact forces. Resolve the tension at $C$, or use its perpendicular arm. The moment equation is

• Substitute $F_T=15F_W/4$, $\sin2\theta=24/25$ and $\sin\theta=4/5$. This gives $36/5=(8/5)(1+k)$, hence $k=7/2$. Check $2<k<4$: the particle is between $C$ and $B$, as the model requires.

A ring can need friction even when its centre is stationary

• A ring of radius $r$ and weight $F_W$ rests on a rough plane inclined at $\alpha$, with $\tan\alpha=3/4$. A horizontal string pulls rightwards at a point $P$ whose radius $OP$ points uphill, parallel to the plane. Let $F_f$ act uphill and $F_N$ normal to the plane.
• Moments about the centre give $F_fr=F_Tr\sin\alpha$. Resolving along the plane gives $F_f+F_T\cos\alpha=F_W\sin\alpha$. Using $\sin\alpha=3/5$, $\cos\alpha=4/5$, obtain $F_T=3F_W/7$ and $F_f=9F_W/35$.

Use the normal force before applying a friction limit

• In the ring example, normal balance gives $F_N=F_W\cos\alpha+F_T\sin\alpha=37F_W/35$. The horizontal string pushes into the plane as well as pulling uphill; the normal reaction is not simply $F_W\cos\alpha$.
• At limiting equilibrium, $\mu=F_f/F_N=9/37$. If the plane has a larger coefficient, the ring can rest with the same required friction; friction does not automatically increase to its maximum value.

Find the support interval before testing toppling

• A body resting under its weight and contact forces alone can remain upright while the vertical through its centre of mass meets its base of support 支撑底面. At the point of toppling 翻倒, the contact resultant reaches an edge; the weight's line passes through that edge. Beyond it there is an unbalanced turning moment.
• The L-shaped plate has $\bar x=2.5\ \text{m}$. On its full $6\ \text{m}$ lower edge it can balance. If turned over to stand on its short $2\ \text{m}$ top edge, its weight line lies outside the support interval, so it topples. An extra applied force changes the moment balance: weight inside the base alone is then not a sufficient test.

A horizontal push can move the contact resultant to an edge

• A uniform rectangular block of weight $F_W$ rests on a horizontal rough floor. Its base half-width is $b$, and a horizontal push $P$ acts at height $h$ above the floor. Before motion, $F_N=F_W$ and the required friction is $F_f=P$.
• Moments show that the normal reaction shifts a distance $\delta=Ph/F_W$ from the base centre towards the leading edge. Toppling begins at $\delta=b$, so $P_{\rm topple}=F_Wb/h$. Sliding begins when $P_{\rm slide}=\mu F_W$. Compare these thresholds; do not assume both limits occur together.

Decide which loss of equilibrium happens first

• For $F_W=200\ \text{N}$, $b=0.4\ \text{m}$ and $h=1\ \text{m}$, the toppling threshold is $80\ \text{N}$. With $\mu=0.3$, the sliding threshold is $60\ \text{N}$, so sliding begins first. With $\mu=0.5$, sliding would require $100\ \text{N}$, so toppling begins first.
• The thresholds coincide at $\mu=b/h=0.4$. Increasing the push height makes toppling easier, but does not change the horizontal sliding threshold in this model. These predictions concern the first motion; the subsequent motion needs a different model.

On an incline, compare two critical angles

• For a body with a symmetric base of half-width $b$ and centre of mass a perpendicular distance $c$ above its centre, the weight line reaches the downhill edge when $\tan\alpha=b/c$. With no other applied forces, impending sliding gives $\tan\alpha=\mu$. The smaller critical angle occurs first.
• For $b=0.4\ \text{m}$, $c=0.6\ \text{m}$ and $\mu=0.5$, sliding occurs at $\tan^{-1}(0.5)\approx26.6^\circ$, before toppling at $\tan^{-1}(2/3)\approx33.7^\circ$. For the cone with a hemispherical cavity, $c=61r/56$, $b=2r$, so the toppling angle satisfies $\tan\alpha=112/61$, provided the plane is sufficiently rough to prevent earlier sliding.

Keep the geometry and the force model together