__Static equilibrium__

Two telephone poles are separated by 40 m and connected by a massless wire. A bird of mass 0.5 kg lands on the wire midway between the poles, causing the wire to sag 2.0 m below horizontal. What is the tension in the wire?

Solution:

- Concepts:

Static equilibrium - Reasoning:

All parts of the system are at rest. F_{total}= τ_{total}= 0 - Details of the calculation:

2F_{y}= mg, F_{y}= 9.8*0.25 N = 2.45 N.

sinθ = 2/(4 + 400)^{½}.^{}F = F_{y}/sinθ = 24.6 N is the tension in the wire.

**Problem:**

One end of a spring of negligible mass is attached to the ceiling. When a 250 g mass is placed on the free end (without stretching the spring) and then released, the mass descends 20.0 cm before it changes direction and begins to ascend. What is the spring constant k (in N/m)?

Solution:

- Concepts:

Hooke's law, equilibrium - Reasoning

The force exerted by a spring obeys Hooke's law. - Details of the
calculation:

When the mass is hanging on the spring in equilibrium, i.e. when all oscillations have damped out, then the magnitude of the force F = kx from the spring exactly cancels the magnitude of the gravitational force F = mg.

We have mg = kx, or k = mg/x.

m = 0.25 kg is given, we need to know the equilibrium stretch x.

The mass initially oscillates between x = 0 and x = 20 cm. If we wait long enough the oscillation will damp out and it will come to rest at x = 0.1 m.

k = (0.25kg)(9.8 m/s^{2})/(0.1 m) = 24.5 N/m.

**Problem:**

Two
identical springs with spring constant k = 1 N/m and equilibrium length l
= 0.25 m are connected by a middle string of length L = (3/8) m and support a
weight w = 0.5 N.

Two strings of length 1 m are loosely connected as shown.

Find the position of the weight below the support and show that it moves up when
the middle string is cut.

Solution:

- Concepts:

Hooke's law: F = -kx, k = spring constant, equilibrium - Reasoning

Consider two isolated springs with constants k.

Connect them in series. The effective spring constant is k' = k/2. (The same force produces twice the displacement.)

Connect them in parallel. The effective spring constant is k'' = 2k. (The same force produces half the displacement.) - Details of the
calculation:

Let the length of the unstretched springs be l (m).

The distance from the support to the weight w in case (a) is

d(m) = 2l + 3/8 m + w/k' = 2l + 3/8 m + 1 m = (½ + 3/8 + 1) m = (15/8) m.

The distance from the support to the weight w in case (b) is

d'(m) = l + 1 m + w/k'' = (¼ + 1 + ¼) m = (12/8) m

d - d' = (3/8) m, d > d'.

Consider a particle moving along the x axis under the influence of the potential

U(x) = ½k(x/a)^{2} - ⅓k(x/a)^{3}

where k and a are constants.

(a) Plot the potential.

(b) Find the equilibrium point(s).

(c) Determine whether the equilibrium point(s) are stable or unstable.

Solution:

- Concepts:

Static equilibrium - Reasoning:

We are asked to find the equilibrium point(s), where F = 0. An equilibrium point is stable if for small displacements from the equilibrium point the force points towards the equilibrium point. - Details of the calculation:

(a) Let k = 1.

(b) Let x' = x/a.

F = -dU/dx' = -kx' + kx'^{2}= 0. x' = x‘^{2}, x/a = 0, 1 are the equilibrium points.

(c) dF/dx' = -k + 2kx'.

dF/dx'|_{x'= 0}= -k --> stable equilibrium (restoring force).

dF/dx'|_{x'= 1}= k --> unstable equilibrium.

Two small positively charged spheres are suspended from a common point at the
ceiling by the insulating light strings of equal length. The first sphere has
mass m_{1} and charge q_{1} while the second one has mass m_{2}
and charge q_{2}. If the first string makes an angle θ_{1} with
the vertical, find the angle θ_{2} that the second string makes with the
vertical.

- Concepts:

Equilibrium - Reasoning:

In equilibrium the center of mass of the two-pendulum system must lie directly beneath the suspension point.

- Details of the calculation:

For the center of mass of the two-pendulum system to lie directly beneath the suspension point. We need m_{1}l sinθ_{1}= m_{2}l sinθ_{2}.

θ_{2}= sin^{-1}((m_{1}/m_{2})sinθ_{1})

Or draw a vector diagram. Balance the forces.

For example: m_{1}g sinθ_{1}= F sinΦ = m_{2}g sinθ_{2}.

__Dynamic equilibrium__

A hollow sphere of radius R = 0.5 m rotates about a vertical axis
through its center with an angular velocity of ω = 5/s.^{ }Inside the
sphere a small block is moving together with the sphere at the height of R/2.
(Let g = 10 m/s^{2}.)**
**(a) What is the minimal coefficient of friction to fulfill this
condition?

Solution:

- Concepts:

Dynamic equilibrium - Reasoning:

If the block moves along a horizontal circle of radius R sinα, then the net force acting on the block is pointed to the center of this circle.The vector sum of the normal force exerted by the wall N, the frictional force S, and the weight mg is equal to the centripetal force mω

^{2}R sinα. - Details of the calculation:

(a) For the horizontal and vertical components we of F_{net}we have

mω^{2}R sinα = N sinα - S cosα, mg = N cosα + S sinα.

Solving for S and N we obtain

S = mg sinα [1 - ω^{2}R cosα/g], N = mg [cosα**+**ω^{2}R sin^{2}α/g]The block does not slip down if μ

_{α }= ≥ |S/N| = sinα |1 - ω^{2}R cosα/g]|/[cosα**+**ω^{2}R sin^{2}α/g].

If ω^{2}R cosα/g < 1 then S points upward, the frictional force prevents the block from sliding down. This is the case for ω = 5/s. We have α = 60^{o}and μ_{α min}= 0.2259.

(b) If ω^{2 }R cosα/g >1 then S points downward, the frictional force prevents the block from sliding up. This is the case for ω = 8/s, and μ_{αmin}= 0.1792.^{ }(c) In case (a) the direction of the frictional force prevents the block from sliding down. μ_{α min}increases when α increases by a small amount and decreases when α decreases by a small amount. If the block is displaced upwards, it slips downward, if it is displaced downward, it stays. If ω increases be a small amount, the block remains in equilibrium, if ω decreases by a small amount, it slips downwards.

In case (b) the direction of the frictional force prevents the block from sliding up. μ_{α min}increases when α decreases by a small amount and decreases when α increases by a small amount. If the block is displaced upwards, it stays, if it is displaced downward, it slips upward. If ω decreases by a small amount, the block remains in equilibrium, if ω increases by a small amount, it slips upward.