Torque Revisited — Question 3

Section 11-5 · Physics pages 1636–1637

Question 3 Some specified torque vectors cannot come from a given point force A force may be chosen freely but must act at⃗r = (1, 2, 0) m. A designer asks for torque (2, 0, 3) N m.⃗ r⃗ τ admissible torques:⃗τ ·⃗r = 0 All possible torques from one prescribed application point lie in a plane normal to its position vector. (a) Test whether the request is possible. (b) For the alternative target (2, −1, 3) N m, find the smallest-magnitude force.
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Question 3 - Solution Some specified torque vectors cannot come from a given point force⃗ r⃗ τ admissible torques:⃗τ ·⃗r = 0 (a) Derivation and calculation A necessary condition for a single applied force is⃗ τ ·⃗r = (⃗r ×⃗F) ·⃗r = 0. For the requested first target, (2, 0, 3) · (1, 2, 0) = 2̸ = 0. Therefore the first target is impossible . (b) Result and interpretation The alternative target satisfies (2, −1, 3) · (1, 2, 0) = 0. Its transverse force is⃗ F⊥=⃗τ ×⃗r r2 = (−6/5, 3/5, 1) N . Every solution is⃗F =⃗F⊥+ λ⃗r, and F 2 = F 2 ⊥+ λ2r2 ≥F 2 ⊥⇒Fmin = √ 14/5 = 1.673 N . Physical check Direct substitution gives (1, 2, 0) × (−6/5, 3/5, 1) = (2, −1, 3).

Question 3

Some specified torque vectors cannot come from a given point force

A force may be chosen freely but must act at r→=(1,2,0)m\vec r=(1,2,0)\,\mathrm m. A designer asks for torque (2,0,3)Nm(2,0,3)\,\mathrm{N\,m}.

Diagram shown in the original SVG worksheet.

  1. Test whether the request is possible.

  2. For the alternative target (2,−1,3)Nm(2,-1,3)\,\mathrm{N\,m}, find the smallest-magnitude force.

Question 3 - Solution

Some specified torque vectors cannot come from a given point force

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

A necessary condition for a single applied force is τ→⋅r→=(r→×F→)⋅r→=0.\vec\tau\cdot\vec r=(\vec r\times\vec F)\cdot\vec r=0. For the requested first target, (2,0,3)⋅(1,2,0)=2≠0.(2,0,3)\cdot(1,2,0)=2\ne0. Therefore the first target is impossible\boxed{\text{the first target is impossible}}.

(b) Result and interpretation

The alternative target satisfies (2,−1,3)⋅(1,2,0)=0(2,-1,3)\cdot(1,2,0)=0. Its transverse force is F→⟂=τ→×r→r2=(−6/5,3/5,1)N.\vec F_\perp=\frac{\vec\tau\times\vec r}{r^2} =\boxed{(-6/5,3/5,1)\ \mathrm N}. Every solution is F→=F→⟂+λr→\vec F=\vec F_\perp+\lambda\vec r, and F2=F⟂2+λ2r2≥F⟂2⇒Fmin=14/5=1.673N.F^2=F_\perp^2+\lambda^2r^2\ge F_\perp^2 \Rightarrow\boxed{F_{\min}=\sqrt{14/5}=1.673\ \mathrm N}.

Physical check

Direct substitution gives (1,2,0)×(−6/5,3/5,1)=(2,−1,3)(1,2,0)\times(-6/5,3/5,1)=(2,-1,3).