Zero Torque and Static Equilibrium Notes

Zero Torque and Static Equilibrium

When supporting a child on a plank, the upward forces exerted by the parents must equal the child's weight: F1 + F2 = mg.
This ensures zero net force, but not necessarily static equilibrium.

Definition: Static equilibrium is when an object is at rest, neither rotating nor translating.
Conditions:
- (i) Net force is zero: \Sigma Fx = 0, \Sigma Fy = 0
- (ii) Net torque is zero: \Sigma \tau = 0
These conditions are independent; one doesn't guarantee the other.

Force Condition:
- Setting upward as positive, the net force equation is: F1 + F2 - 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 F2 is positive (counter-clockwise): \tau2 = F_2L
- Torque due to gravity is negative (clockwise): \tau_{mg} = -mg(3L/4)
- Net torque equation: F_2L - mg(3L/4) = 0

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.

Using the right end as the axis of rotation:
- Net force: F1 + F2 - mg = 0
- Net torque: -F_1(L) + mg(L/4) = 0
- Solving yields: F1 = \frac{1}{4}mg and F2 = \frac{3}{4}mg

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:
- \tau1 = F1(0) = 0
- \tau2 = F2(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 = F1 + F2 - mg = 0
- F1 = mg - F2 = (90.0 \text{ kg})(9.81 \text{ m/s}^2) - 2940 \text{ N} = -2060 \text{ N}

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 = \tau1 + \tau2 = 0
-mg(d2) + Mg(d1) = 0
m = M(d1/d2) = (6.00 \text{ kg})(0.850 \text{ m}/1.11 \text{ m}) = 4.59 \text{ kg}

A "strongman" uses a lever to lift a 1 ton (910 kg) bucket of rocks.
The bucket is db = 0.85 m from the pivot, and the man applies force at dm = 3.6 m from the pivot.
Fb = wb = mg = (910 \text{ kg})(9.8 \text{ m/s}^2) = 8900 \text{ N}
\Sigma \tau = Fb db - Fm dm = 0
Fm = Fb (db/dm) = (8900 \text{ N})(0.85 \text{ m}/3.6 \text{ m}) = 2100 \text{ N}
Fp = Fb - F_m = 8900 \text{ N} - 2100 \text{ N} = 6800 \text{ N}

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 Fy = fy - mg = 0 \Rightarrow f_y = mg
- \Sigma Fx = fx - T = 0 \Rightarrow f_x = T = mg(H/V)

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}