Zero Torque and Static Equilibrium Notes

Zero Torque and Static Equilibrium

Introduction

• When supporting a child on a plank, the upward forces exerted by the parents must equal the child's weight: $F<em>1 + F</em>2 = mg$.
• This ensures zero net force, but not necessarily static equilibrium.

The Need for Zero Torque

• If one parent lets go, the plank rotates, indicating a non-zero net torque.
• Static equilibrium requires both zero net force and zero net torque.

Conditions for Static Equilibrium

• Definition: Static equilibrium is when an object is at rest, neither rotating nor translating.
• Conditions:
  • (i) Net force is zero: $\Sigma F<em>x = 0$, $\Sigma F</em>y = 0$
  • (ii) Net torque is zero: $\Sigma \tau = 0$
• These conditions are independent; one doesn't guarantee the other.

Applying the Conditions to the Plank

• Force Condition:
  • Setting upward as positive, the net force equation is: $F<em>1 + F</em>2 - mg = 0$
• Torque Condition:
  • Choose an axis of rotation (e.g., the left end of the plank).
  • Torque due to $F_1$ is zero because it acts through the axis.
  • Torque due to $F<em>2$ is positive (counter-clockwise): $\tau</em>2 = F_2L$
  • Torque due to gravity is negative (clockwise): $\tau_{mg} = -mg(3L/4)$
  • Net torque equation: $F_2L - mg(3L/4) = 0$

Solving for Forces

• From the torque equation: $F_2 = \frac{3}{4}mg$
• Substitute into the force equation to find $F_1 = \frac{1}{4}mg$
• The force nearest the child is greater.

Arbitrariness of Axis Choice

• In static equilibrium, net torque is zero regardless of axis location.
• Choose the most convenient axis, often at an unknown force to eliminate it from the torque equation.

Example: Child on a Plank (Alternative Solution)

• Using the right end as the axis of rotation:
  • Net force: $F<em>1 + F</em>2 - mg = 0$
  • Net torque: $-F_1(L) + mg(L/4) = 0$
  • Solving yields: $F<em>1 = \frac{1}{4}mg$ and $F</em>2 = \frac{3}{4}mg$

Example: Diving Board

• A diving board of length $L = 5.00$ m is supported by two pillars, one at the left end and the other $d = 1.50$ m away.
• A diver of mass $m = 90.0$ kg stands at the far end.
• Torques:
  • $\tau<em>1 = F</em>1(0) = 0$
  • $\tau<em>2 = F</em>2(d)$
  • $\tau_3 = -mg(L)$
• $\Sigma \tau = 0 + F_2(d) - mg(L) = 0$
• $F_2 = mg(L/d) = (90.0 \text{ kg})(9.81 \text{ m/s}^2)(5.00 \text{ m}/1.50 \text{ m}) = 2940 \text{N}$
• Forces:
  • $\Sigma F = F<em>1 + F</em>2 - mg = 0$
  • $F<em>1 = mg - F</em>2 = (90.0 \text{ kg})(9.81 \text{ m/s}^2) - 2940 \text{ N} = -2060 \text{ N}$

Example: Cat on a Plank

• Plank mass $M = 6.00$ kg, supported by two sawhorses.
• Center of mass is $d_1 = 0.850$ m left of sawhorse B.
• Cat is $d_2 = 1.11$ m right of sawhorse B when the plank tips.
• $\Sigma \tau = \tau<em>1 + \tau</em>2 = 0$
• $-mg(d<em>2) + Mg(d</em>1) = 0$
• $m = M(d<em>1/d</em>2) = (6.00 \text{ kg})(0.850 \text{ m}/1.11 \text{ m}) = 4.59 \text{ kg}$

Example: Lifting a Load

• A "strongman" uses a lever to lift a 1 ton (910 kg) bucket of rocks.
• The bucket is $d<em>b = 0.85$ m from the pivot, and the man applies force at $d</em>m = 3.6$ m from the pivot.
• $F<em>b = w</em>b = mg = (910 \text{ kg})(9.8 \text{ m/s}^2) = 8900 \text{ N}$
• $\Sigma \tau = F<em>b d</em>b - F<em>m d</em>m = 0$
• $F<em>m = F</em>b (d<em>b/d</em>m) = (8900 \text{ N})(0.85 \text{ m}/3.6 \text{ m}) = 2100 \text{ N}$
• $F<em>p = F</em>b - F_m = 8900 \text{ N} - 2100 \text{ N} = 6800 \text{ N}$

Forces with Vertical and Horizontal Components

• Consider a wall-mounted lamp (sconce) with a curved rod bolted to the wall.
• A lamp of mass $m$ is suspended a horizontal distance $H$ from the wall.
• A horizontal wire is a vertical distance $V$ above the bottom of the rod.
• Torque condition (axis at the bottom of the rod):
  • $\Sigma \tau = T(V) - mg(H) = 0$
  • $T = mg(H/V)$
• Force conditions:
  • $\Sigma F<em>y = f</em>y - mg = 0 \Rightarrow f_y = mg$
  • $\Sigma F<em>x = f</em>x - T = 0 \Rightarrow f_x = T = mg(H/V)$

Example: Sconce Forces

• Lamp mass $m = 2.00$ kg, $V = 12.0$ cm, $H = 15.0$ cm.
• $T = mg(H/V) = (2.00 \text{ kg})(9.81 \text{ m/s}^2)(15.0 \text{ cm}/12.0 \text{ cm}) = 24.5 \text{ N}$
• $f_x = T = 24.5 \text{ N}$
• $f_y = mg = (2.00 \text{ kg})(9.81 \text{ m/s}^2) = 19.6 \text{ N}$

Example: Person on a Ladder

• An 85-kg person stands on a ladder.
• The floor exerts normal force $f<em>1$ and friction $f</em>2$.
• The wall exerts normal force $f_3$.
• $\tau<em>{net} = f</em>3(a) - mg(b) = 0 \Rightarrow f_3 = mg(b/a) = 150 \text{ N}$
• $F<em>{net x} = 0 \Rightarrow f</em>2 - f<em>3 = 0 \Rightarrow f</em>2 = f_3 = 150 \text{ N}$
• $F<em>{net y} = 0 \Rightarrow f</em>1 - mg = 0 \Rightarrow f_1 = mg = 830 \text{ N}$