MF16 - Beams, laminas and centres of mass

MF16 - Beams, laminas and centres of mass

Replace a body's distributed weight by one force through its centre of mass, then use moments to find supports or an unknown weight position. You will model rods and rectangular laminas, check contact and tipping, and test whether specified rough contacts supply enough friction.

Weight positions and contact forces

Can moving a support make a body harder to hold, even when its weight stays the same? Consider a uniform rectangular panel 2 m2\,\mathrm m wide and 3 m3\,\mathrm m high, supported at its lower-right corner and pushed rightwards by a smooth pad at its upper-left corner. Lowering that pad shortens its moment arm. We will find what happens to the required force and friction.

Its weight acts at the rectangle's centre, 1 m1\,\mathrm m horizontally left of the lower-right corner. The vertical height of that centre is not the arm of the vertical weight.

You already know

Moment equals force times perpendicular distance. Equilibrium needs zero resultant force and zero resultant moment. At a rough contact, static friction satisfies F≤μNF\le\mu N.

The force of gravity acts on all the particles of a body. For calculating the body's total weight and its moment, one effective point is enough.

Centre of mass

The centre of mass is the point through which a body's weight can be taken to act when calculating moments in a uniform gravitational field. For a uniform rod it is the midpoint; for a uniform rectangular lamina it is the intersection of the diagonals.

A uniform 6 m6\,\mathrm m rod has its weight 3 m3\,\mathrm m along the rod from either end. A uniform rectangle 4 m4\,\mathrm m wide and 2 m2\,\mathrm m high has centre (2,1) m(2,1)\,\mathrm m from its lower-left corner. Symmetry identifies these points because the mass distributions are uniform.

For a non-uniform rod, use a given centre or infer it from moments. Its midpoint need not be its centre of mass. We will locate the centre of one rod from measured reactions; combining separate bodies to calculate a composite centre is not needed.

Weight and pushing contacts

In a uniform gravitational field,

W=mg.W=mg.

For a rough unfastened contact, with friction magnitude FF and normal force NN,

N≥0,0≤F≤μN.N\ge0,\qquad 0\le F\le\mu N.

The contact force can have both normal and frictional components.

Not in the formula sheet: learn it.

A smooth contact supplies only a normal reaction. A rough contact can also supply friction along the surface. Put each force at its specified contact point: moving a wall force to an unstated height changes its moment.

Here rods and rectangular laminas have coplanar forces perpendicular to the rod or to a lamina edge. The horizontal and vertical edges of the panel allow the horizontal and vertical forces used later.

Key idea

A body's centre determines where its weight acts. Its surfaces and attachments determine what other forces can act, and where.

Separate a valid weight model from an unjustified symmetry assumption.

Finding the reactions on a horizontal beam

Support forces share a beam's weight and any added load according to their positions. Take moments about one support to remove its reaction from that equation.

Use the body's midpoint for its own weight only when uniformity justifies it. Keep an added load separate.

Worked example 1: a load near one end

Question. A uniform horizontal beam ABAB, 6 m6\,\mathrm m long and weighing 120 N120\,\mathrm N, rests on smooth horizontal end supports. A downward 180 N180\,\mathrm N load acts 4.5 m4.5\,\mathrm m from AA. Find both upward reactions.

Plan. Take moments about AA, then use vertical force balance.

6RB=120(3)+180(4.5)moments about ARB=195 NsolveRA=120+180−195=105 Nvertical equilibrium\begin{aligned} 6R_B&=120(3)+180(4.5) &&\text{moments about }A\\ R_B&=195\,\mathrm N &&\text{solve}\\ R_A&=120+180-195=105\,\mathrm N &&\text{vertical equilibrium} \end{aligned}

Why place the beam's weight 3 m3\,\mathrm m from AA? Its own uniform mass distribution remains unchanged by the added load.

Answer. RA=105 NR_A=105\,\mathrm N, RB=195 NR_B=195\,\mathrm N, both upwards.

Check. About BB, 6(105)=120(3)+180(1.5)=630 N m6(105)=120(3)+180(1.5)=630\,\mathrm{N\,m}.

Change only the added load's position, moving it to the other side of the midpoint.

Worked example 2: moving the load

Question. The same uniform 6 m6\,\mathrm m, 120 N120\,\mathrm N beam has smooth end supports and a downward 180 N180\,\mathrm N load now 1.5 m1.5\,\mathrm m from AA. Find both upward reactions.

Plan. Change the load's arm in the moment equation, then resolve vertically.

6RB=120(3)+180(1.5)moments about ARB=105 NsolveRA=120+180−105=195 Nvertical equilibrium\begin{aligned} 6R_B&=120(3)+180(1.5) &&\text{moments about }A\\ R_B&=105\,\mathrm N &&\text{solve}\\ R_A&=120+180-105=195\,\mathrm N &&\text{vertical equilibrium} \end{aligned}

Why do the reactions exchange? The load's position is the mirror image of the first, while the beam and supports are unchanged.

Answer. RA=195 NR_A=195\,\mathrm N, RB=105 NR_B=105\,\mathrm N, both upwards.

Check. RA+RB=300 NR_A+R_B=300\,\mathrm N; about BB, 6(195)=120(3)+180(4.5)=1170 N m6(195)=120(3)+180(4.5)=1170\,\mathrm{N\,m}.

Both reactions are non-negative, so neither demands an impossible pulling force from an unfastened support.

Watch out

The incorrect shortcut RA=RB=(120+180)/2R_A=R_B=(120+180)/2 ignores the off-centre load. Identical supports need not carry equal forces.

Use a fresh beam with its load closer to the left support.

Checking contact and the onset of tipping

A reaction equation can predict that a support would have to pull. For an unfastened contact, that means the assumed static configuration fails.

The boundary occurs when one reaction is zero. The other support becomes the prospective tipping point.

Worked example 3: a limiting overhang load

Question. A uniform 5 m5\,\mathrm m horizontal rod weighs 100 N100\,\mathrm N. Unfastened smooth supports act at its left end AA and at CC, 3 m3\,\mathrm m from AA. A downward load P≥0P\ge0 acts at the right end. Find the greatest permissible PP and identify the contact lost.

Plan. Express the reactions in terms of PP, then impose non-negative contact.

3RC=100(2.5)+5Pmoments about ARA=100+P−250+5P3=50−2P3vertical equilibriumRA≥0 ⟹ P≤25pushing contact\begin{aligned} 3R_C&=100(2.5)+5P &&\text{moments about }A\\ R_A&=100+P-\frac{250+5P}{3} =\frac{50-2P}{3} &&\text{vertical equilibrium}\\ R_A\ge0&\ \Longrightarrow\ P\le25 &&\text{pushing contact} \end{aligned}

Why check AA? The reaction at CC increases with PP; the one at AA decreases.

Answer. Greatest load 25 N25\,\mathrm N; contact at AA is about to be lost.

Check. About CC at the boundary, 100(0.5)=25(2)100(0.5)=25(2), and RC=125 NR_C=125\,\mathrm N.

Change only the second support's position to 4 m4\,\mathrm m from AA, leaving a shorter overhang.

Worked example 4: the support moves outwards

Question. The same uniform 5 m5\,\mathrm m, 100 N100\,\mathrm N rod now has supports at AA and at CC, 4 m4\,\mathrm m from AA. A downward load P≥0P\ge0 acts at the right end. Find the greatest permissible PP and identify the contact lost.

Plan. Use the changed support arm, then check the decreasing reaction.

4RC=100(2.5)+5Pmoments about ARA=100+P−250+5P4=150−P4vertical equilibriumRA≥0 ⟹ P≤150contact condition\begin{aligned} 4R_C&=100(2.5)+5P &&\text{moments about }A\\ R_A&=100+P-\frac{250+5P}{4} =\frac{150-P}{4} &&\text{vertical equilibrium}\\ R_A\ge0&\ \Longrightarrow\ P\le150 &&\text{contact condition} \end{aligned}

Why can this support arrangement take more load? The weight has a larger stabilising arm about CC, while the load has a smaller tipping arm.

Answer. Greatest load 150 N150\,\mathrm N; contact at AA is about to be lost.

Check. About CC, 100(1.5)=150(1)100(1.5)=150(1), and RC=250 NR_C=250\,\mathrm N at the boundary.

A larger load would make the algebraic RAR_A negative. It predicts the loss of the assumed contact, rather than a downward force that the unfastened support can exert.

Problem-solving

For a boundary load, set the disappearing reaction to zero and take moments about the remaining support. Then check that the other reaction is physically possible.

This is loss of contact, not limiting friction. No friction equality is involved.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account