Let us observe the particle of mass m in fixed (X, Y) coordinate as in Fig. 8.6 (a). We can write the equation of motion according to Newton’s second law of motion as

Now (x, y) moving coordinate is introduced at particle of mass m, and the particle under the forces

and – m ..a (inertia force) is in rest or equilibrium. So summation of all the forces acting on the particle of mass m is equal to zero.

The equilibrium mentioned above is known as dynamic equilibrium. The moving coordinate (x, y) and

is the principle of D’Alemberts which is the apparent transformation of dynamic problems to statics problem. Also summation of moments about any point P will be zero.

**SOLVED EXAMPLES**

**Example 8.1.** What acceleration a of the collar along the horizontal guide will result in a steadystate 15° deflection of the pendulum from the vertical ? The slender rod of length l and the particle each have mass nz. Friction at pivot P is negligible.

**Solution.**

** Ʃ **f _{x} = 0,

F_{1} – 2ma = 0

F_{1} = 2ma

** Ʃ **f _{y} = 0,

F_{2} -2mg = 0

F_{2} = 2mg

Applying D’ Alembert’s principle.

Taking moment about point P

**Ʃ **M_{P} = 0,

mg x 1/2 sin 15°– ma x 1/2 cos 15° – ma l cos + mg x l sin 15° = 0

g tan 15° = a

a = 0.2679 g.Ans.

**Example 8.2. **A bicyclist applies the brakes as he descends the 1 oo incline. What deceleration a would cause the dangero us condition of tipping about the front wheel A ? The combined center of mass of the rider and bicycle is at G.

**Solution. **Applying D’ Alembert’s Principle

For condition of tipping about A, R8

Taking moment about A,

**Ʃ **M_{A} = 0,

ma x 900 / 1000 – mg cos 10^{0} x 625 / 1000 + mg sin 10^{0} x 900 / 1000 = 0

or 0.9 ma – 0.6155 mg + 0.15628 + 0.15628 mg = 0

or a = (0.6155 – 0.15628) g / 0.9 = 0.5102g.Ans.

**Example 8.3. **The 1650 kg car has its mass center af G. Calculate the normal fo rces NA and N8 between the road and the fro nt and rear pairs of wheels under conditions of maximum acceleration. The mass of the wheels is small compared with the total mass of the car. The coefficient of static friction between the road and the rear driving wheels is 0.8.

**Solution.** Applying D’ Alemberts Principle,

N_{A} + N_{B} = 1650 g

– N_{A} x 1200 / 1000 + N_{B} x 1200 /1000 – 0.8 x 400 /1000 = 0

or N_{A} = 0.734 N_{B}

Solving (1) and (2), we get

N_{B} = 1650 x 9.81 / 1.734 = 9335 N.Ans

N_{A} = 6852N.Ans