3.04c - Beams, ladders and centres of mass

3.04c - Beams, ladders and centres of mass

Use moments to find support forces and weight positions in rods and rectangular laminas. You will check that contacts can supply the required forces, identify the onset of tipping and combine moments with friction in a ladder model.

Weight positions and contact forces

Does a heavier ladder need a greater minimum coefficient of friction? A uniform 5 m5\,\mathrm m ladder weighing 200 N200\,\mathrm N has its foot 3 m3\,\mathrm m from a smooth wall. Doubling its weight doubles the ground's normal force, but also doubles the friction needed. We will find whether these effects cancel.

First locate the weight. The 5 m5\,\mathrm m ladder's own weight acts 2.5 m2.5\,\mathrm m along it from the foot. Its moment arm about the foot is the horizontal distance to that point, not the 2.5 m2.5\,\mathrm m sloping distance.

You already know

Use force times perpendicular arm for a moment. A rigid body in equilibrium needs both force and moment balance. Friction at a rough contact satisfies F≤μNF\le\mu N, with equality only when limiting.

Weight is distributed throughout a real body. For its overall force and turning effect, we replace that distribution by one force.

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. A uniform rod has its centre of mass at its midpoint; a uniform rectangular lamina has its weight at its point of symmetry.

A uniform horizontal 6 m6\,\mathrm m rod therefore has weight 3 m3\,\mathrm m from either end. A rectangular lamina 4 m4\,\mathrm m wide and 2 m2\,\mathrm m high has its weight at (2,1) m(2,1)\,\mathrm m from the lower-left corner. Rotate either body and its centre moves with it: locate the centre first, then find the arm perpendicular to the vertical weight.

A non-uniform rod need not balance at its midpoint. Use a specified centre or determine its position from moment information. Knowing its outline alone does not determine its mass distribution.

Weight and contact conditions

For mass mm in a uniform gravitational field,

W=mg.W=mg.

For an unfastened pushing contact and static friction magnitude FF,

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

Here NN is the normal force, not the magnitude of the whole contact force.

Not in the formula sheet: learn it.

A smooth surface supplies only a normal force. A rough surface can also supply tangential friction. A smooth pin can supply two force components but no resisting moment. Place each force at its stated contact; a wall reaction cannot be moved to a convenient height without evidence.

The opposite force exerted on a support acts on a different body. It does not belong on the rod's force diagram.

Key idea

Choose the weight's position from the body model, and choose contact forces from the surfaces and attachments. The moment equations then use those positions.

This check distinguishes a centre position from a perpendicular moment arm.

Finding the reactions on a horizontal beam

A beam's reactions need not be equal, even when its supports are identical. The positions of the loads determine how the total support is shared.

Take moments about one support to remove its reaction. Then use vertical equilibrium to recover that reaction, and check that both are non-negative.

Worked example 1: a loaded uniform beam

Question. A uniform horizontal beam ABAB is 6 m6\,\mathrm m long and weighs 120 N120\,\mathrm N. Smooth horizontal end supports act at AA and BB. A 180 N180\,\mathrm N downward load acts 4.5 m4.5\,\mathrm m from AA. Find both upward reactions.

Plan. Take moments about AA, placing the beam's weight at its midpoint, then resolve vertically.

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

Why use 3 m3\,\mathrm m for the beam's own weight? Uniformity fixes its centre at the midpoint, independently of 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 shape of the supported body: replace the slender beam by a rectangular lamina of the same width and weight. With all forces vertical, its extra height does not change their horizontal moment arms.

Worked example 2: a rectangular lamina

Question. A uniform rectangular lamina stands in a vertical plane with horizontal width AB=6 mAB=6\,\mathrm m, vertical height 2 m2\,\mathrm m and weight 120 N120\,\mathrm N. Smooth horizontal point supports touch only its lower corners AA and BB. A downward 180 N180\,\mathrm N force acts on the top edge 4.5 m4.5\,\mathrm m horizontally from the left edge. Find both upward support reactions.

Plan. Locate the weight by symmetry and use horizontal arms for the vertical forces.

xG=3 mhorizontal centre position6RB=120(3)+180(4.5)moments about ARB=195 NsolveRA=120+180−195=105 Nvertical equilibrium\begin{aligned} x_G&=3\,\mathrm m &&\text{horizontal centre position}\\ 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 does height disappear? Shifting a vertical force along its vertical line leaves its moment about AA unchanged.

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

Check. Their sum is 300 N300\,\mathrm N, and moments about BB again give 630 N m630\,\mathrm{N\,m} in each sense.

If a lamina is inclined, first locate its centre using the two side directions. For adjacent side lengths LL upwards to the left at angle θ\theta above horizontal and bb upwards to the right perpendicular to it, the centre's horizontal coordinate relative to the lower corner is

xG=−L2cos⁡θ+b2sin⁡θ.x_G=-\frac L2\cos\theta+\frac b2\sin\theta.

This follows by moving halfway along each side from that corner. Its vertical weight has arm ∣xG∣|x_G|; the diagonal to the centre is not the arm. The corner positions similarly give the height of any specified horizontal wall reaction. This projection is derived from geometry and is not printed in the formula sheet.

Watch out

Equal supports do not imply equal reactions. Nor should an added load move the body's own weight point: include the body's weight and the added load separately.

Move the load across a fresh rectangular lamina and see the sharing reverse.

Checking contact and the onset of tipping

A negative upward reaction at an unfastened support is a warning about the model. That support cannot pull the beam down, so the assumed contact configuration cannot persist.

At the onset of tipping, one reaction reaches zero. The remaining contact is the prospective pivot; these static equations describe the boundary, not the motion afterwards.

Worked example 3: the largest supported load

Question. A uniform horizontal rod is 6 m6\,\mathrm m long and weighs 120 N120\,\mathrm N. Unfastened smooth point supports act at its left end AA and at CC, 4 m4\,\mathrm m from AA. A downward load P≥0P\ge0 acts at the right end. Find the greatest PP for static contact, and identify the contact lost.

Plan. Express both reactions in terms of PP, then require non-negative reactions.

4RC=120(3)+6Pmoments about ARC=90+32PsolveRA=120+P−RC=30−12Pvertical equilibriumRA≥0 ⟹ P≤60contact can only push\begin{aligned} 4R_C&=120(3)+6P &&\text{moments about }A\\ R_C&=90+\frac32P &&\text{solve}\\ R_A&=120+P-R_C=30-\frac12P &&\text{vertical equilibrium}\\ R_A\ge0&\ \Longrightarrow\ P\le60 &&\text{contact can only push} \end{aligned}

Why test RAR_A? RCR_C stays positive for all P≥0P\ge0, whereas RAR_A decreases towards zero.

Answer. Greatest load 60 N60\,\mathrm N; contact at AA is about to be lost, with tipping about CC.

Check. At the boundary, moments about CC give 120(1)=60(2)120(1)=60(2), and RC=180 NR_C=180\,\mathrm N.

Change only the second support's position, from 4 m4\,\mathrm m to 5 m5\,\mathrm m from AA. Its shorter overhang should allow a larger load.

Worked example 4: moving the support

Question. The same uniform 6 m6\,\mathrm m, 120 N120\,\mathrm N rod has supports at its left end AA and at CC, now 5 m5\,\mathrm m from AA. A downward load P≥0P\ge0 acts at the right end. Find the greatest PP for static contact, and identify the contact lost.

Plan. Repeat the reaction calculation with the changed support arm.

5RC=120(3)+6Pmoments about ARC=72+65PsolveRA=120+P−RC=48−15Pvertical equilibriumRA≥0 ⟹ P≤240contact condition\begin{aligned} 5R_C&=120(3)+6P &&\text{moments about }A\\ R_C&=72+\frac65P &&\text{solve}\\ R_A&=120+P-R_C=48-\frac15P &&\text{vertical equilibrium}\\ R_A\ge0&\ \Longrightarrow\ P\le240 &&\text{contact condition} \end{aligned}

Why is the limiting load larger? About CC, the weight now has arm 2 m2\,\mathrm m and the load only 1 m1\,\mathrm m.

Answer. Greatest load 240 N240\,\mathrm N; contact at AA is about to be lost.

Check. Boundary moments about CC: 120(2)=240(1)120(2)=240(1). The remaining reaction is 360 N360\,\mathrm N.

For the first support arrangement, P=80 NP=80\,\mathrm N would give RA=−10 NR_A=-10\,\mathrm N. This predicts failure of that configuration, not a real downward pull by the support.

Watch out

Tipping uses a reaction reaching zero. Limiting friction uses F=μNF=\mu N. They are different boundaries; do not insert a friction equality into a smooth-support problem.

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