The Requirements of Equilibrium — Question 4

Section 12-2 · Physics pages 1798–1799

Question 4 Zero moments about two points need not enforce force balance A planar force system has zero net moment about A = (0, 0) and B = (2, 0) m. Use τB = τA −(⃗rB −⃗rA) ×⃗Fnet. A B +x Schematic; labels, not drawing lengths, specify the data. (a) What component of the net force is constrained by these observations? (b) Give a nonzero-force example satisfying both zero-moment conditions.
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Question 4 - Solution Zero moments about two points need not enforce force balance A B +x (a) Derivation and calculation Use the change-of-origin identity: τB = τA −(⃗rB −⃗rA) ×⃗Fnet. With⃗rB −⃗rA = 2ˆx and both torques zero, 0 = −2Fy =⇒ Fy = 0 . There is no constraint on Fx. (b) Result and interpretation Choose a force⃗F = 5ˆx N at any point (x, 0): τA = x(0) −0(5) = 0, τB = (x −2)(0) −0(5) = 0. Nevertheless,⃗ Fnet = 5ˆx N̸ =⃗0 . A resultant along the line joining the torque origins is invisible to both moment tests. Physical check The two torque equations supply only one independent force-component condition.

Question 4

Zero moments about two points need not enforce force balance

A planar force system has zero net moment about A=(0,0)A=(0,0) and B=(2,0)mB=(2,0)\,\mathrm m. Use τB=τA−(r→B−r→A)×F→net\tau_B=\tau_A-(\vec r_B-\vec r_A)\times\vec F_{net}.

Diagram shown in the original SVG worksheet.

  1. What component of the net force is constrained by these observations?

  2. Give a nonzero-force example satisfying both zero-moment conditions.

Question 4 - Solution

Zero moments about two points need not enforce force balance

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Use the change-of-origin identity: τB=τA−(r→B−r→A)×F→net.\tau_B=\tau_A-(\vec r_B-\vec r_A)\times\vec F_{\mathrm{net}}. With r→B−r→A=2x̂\vec r_B-\vec r_A=2\hat x and both torques zero, 0=−2Fy⇒Fy=0.0=-2F_y\quad\Longrightarrow\quad\boxed{F_y=0}. There is no constraint on FxF_x.

(b) Result and interpretation

Choose a force F→=5x̂N\vec F=5\hat x\ \mathrm N at any point (x,0)(x,0): τA=x(0)−0(5)=0,τB=(x−2)(0)−0(5)=0.\tau_A=x(0)-0(5)=0,\qquad\tau_B=(x-2)(0)-0(5)=0. Nevertheless, F→net=5x̂N≠0→.\boxed{\vec F_{\mathrm{net}}=5\hat x\ \mathrm N\ne\vec0}. A resultant along the line joining the torque origins is invisible to both moment tests.

Physical check

The two torque equations supply only one independent force-component condition.