Some Examples of Static Equilibrium — Question 10

Section 12-4 · Physics pages 1850–1851

Question 10 A three-legged support tests the full horizontal gravity projection A rigid platform of weight 600 N stands on upward-only legs at A = (0, 0), B = (4, 0) and C = (0, 3) m in plan view. Its gravity projection is initially (1, 1) m. A B C G Plan view of the support triangle; G denotes the initial weight projection. (a) Find all three vertical reactions. (b) If the gravity projection moves to (3, 1) while weight stays fixed, can all three legs remain in contact?
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Question 10 - Solution A three-legged support tests the full horizontal gravity projection A B C G (a) Derivation and calculation Vertical force and two independent moment equations are RA + RB + RC = 600, 4RB = 600xG, 3RC = 600yG. For (xG, yG) = (1, 1) m, RB = 150 N , RC = 200 N , RA = 250 N . (b) Result and interpretation For (3, 1) m, the formal solution is RB = 450, RC = 200, RA = −50 (N). The required negative contact force is impossible. Equivalently, the support triangle requires xG ≥0, yG ≥0, xG 4 + yG 3 ≤1. Here 3/4 + 1/3 = 13/12 > 1, so the platform tips about edge BC. Physical check The first center lies inside the support polygon; the second lies beyond its opposite edge.

Question 10

A three-legged support tests the full horizontal gravity projection

A rigid platform of weight 600N600\,\mathrm N stands on upward-only legs at A=(0,0)A=(0,0), B=(4,0)B=(4,0) and C=(0,3)mC=(0,3)\,\mathrm m in plan view. Its gravity projection is initially (1,1)m(1,1)\,\mathrm m.

Diagram shown in the original SVG worksheet.

  1. Find all three vertical reactions.

  2. If the gravity projection moves to (3,1)(3,1) while weight stays fixed, can all three legs remain in contact?

Question 10 - Solution

A three-legged support tests the full horizontal gravity projection

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Vertical force and two independent moment equations are RA+RB+RC=600,4RB=600xG,3RC=600yG.R_A+R_B+R_C=600,\qquad4R_B=600x_G,\qquad3R_C=600y_G. For (xG,yG)=(1,1)(x_G,y_G)=(1,1) m, RB=150N,RC=200N,RA=250N.\boxed{R_B=150\ \mathrm N},\quad\boxed{R_C=200\ \mathrm N},\quad\boxed{R_A=250\ \mathrm N}.

(b) Result and interpretation

For (3,1)(3,1) m, the formal solution is RB=450,RC=200,RA=−50(N).R_B=450,\quad R_C=200,\quad R_A=-50\quad(\mathrm N). The required negative contact force is impossible. Equivalently, the support triangle requires xG≥0,yG≥0,xG4+yG3≤1.x_G\ge0,\quad y_G\ge0,\quad\frac{x_G}{4}+\frac{y_G}{3}\le1. Here 3/4+1/3=13/12>13/4+1/3=13/12>1, so the platform tips about edge BCBC.

Physical check

The first center lies inside the support polygon; the second lies beyond its opposite edge.