Torque — Question 8

Section 10-7 · Physics pages 1506–1507

Question 8 A torque requirement leaves the radial force undetermined A force acts at⃗r = (0.30, 0.40) m and must produce τz = 5.0 N m. Its in-plane direction is free.⃗ r⃗ Fmin The least-force solution is perpendicular to the radius. (a) Find the force of minimum magnitude meeting the requirement. (b) Describe the family of other forces with exactly the same torque.
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Question 8 - Solution A torque requirement leaves the radial force undetermined⃗ r⃗ Fmin (a) Derivation and calculation Resolve the force along the radial and positive tangential unit vectors: r = √ 0.302 + 0.402 = 0.50 m, ˆer = (0.6, 0.8), ˆeθ = (−0.8, 0.6). τz = rFθ ⇒Fθ = 5 0.50 = 10 N,⃗ Fmin = 10ˆeθ = (−8, 6) N . (b) Result and interpretation The radial component is unrestricted by torque:⃗ F = (−8, 6) + c(0.6, 0.8) N , c ∈R. |⃗F|2 = 102 + c2 ≥100,⃗ r × (cˆer) =⃗0. The minimum occurs at c = 0 because the components are orthogonal. Physical check The proposed minimum gives 0.30(6) −0.40(−8) = 5 N m exactly.

Question 8

A torque requirement leaves the radial force undetermined

A force acts at r→=(0.30,0.40)m\vec r=(0.30,0.40)\,\mathrm m and must produce τz=5.0Nm\tau_z=5.0\,\mathrm{N\,m}. Its in-plane direction is free.

Diagram shown in the original SVG worksheet.

  1. Find the force of minimum magnitude meeting the requirement.

  2. Describe the family of other forces with exactly the same torque.

Question 8 - Solution

A torque requirement leaves the radial force undetermined

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Resolve the force along the radial and positive tangential unit vectors: r=0.302+0.402=0.50m,êr=(0.6,0.8),êθ=(−0.8,0.6).r=\sqrt{0.30^2+0.40^2}=0.50\ \mathrm m,\quad \hat e_r=(0.6,0.8),\quad\hat e_\theta=(-0.8,0.6). τz=rFθ⇒Fθ=50.50=10N,F→min=10êθ=(−8,6)N.\tau_z=rF_\theta\Rightarrow F_\theta=\frac{5}{0.50}=10\ \mathrm N, \qquad\boxed{\vec F_{\min}=10\hat e_\theta=(-8,6)\ \mathrm N}.

(b) Result and interpretation

The radial component is unrestricted by torque: F→=(−8,6)+c(0.6,0.8)N,c∈ℝ.\boxed{\vec F=(-8,6)+c(0.6,0.8)\ \mathrm N},\quad c\in\mathbb R. |F→|2=102+c2≥100,r→×(cêr)=0→.|\vec F|^2=10^2+c^2\ge100,\qquad \vec r\times(c\hat e_r)=\vec0. The minimum occurs at c=0c=0 because the components are orthogonal.

Physical check

The proposed minimum gives 0.30(6)−0.40(−8)=50.30(6)-0.40(-8)=5 N m exactly.