CONSTRAINT MOTION:

A)     Due to inextensible taut strings.

B)     Due to normal reactions.

A) Due to inextensible taut strings:

As the length of the string is constant, so velocity, components of its ends along the string are equal.

B) Due to normal reactions:

The bodies in contact do not have any relative motion in the normal direction. Relative motion can occur only in perpendicular to normal direction.

Illustration 1:

A) Find m if B moves as shown in figure (6).

                  Fig (6)

Solution:

The ‘V’ is broken down into two components – along the perpendicular to string and parallel to string. Now from the above principle mVCosq. (Ans).

B)

            Fig (7)

A and B are attached using a rod. Now motion occurs in the direction perpendicular to normal direction as shown.

Tip: Remember the following rules:

1)      Velocity components along a string are equal.

2)      No motion (relative) occurs in normal direction.

3)      Acceleration component in normal direction are equal.

Illustration 2:

Example 5.20 page 190 D C Pandey

Find the Constraint relation between a₁, a₂, and a₃.

                                                Fig (8)

Solution:

Points 1, 2, 3, 4 are movable let their displacements from a fixed line be x₁, x₂, x₃, and x₄.

We have x₁+x₄ = l₁ (length of first string)                                            ---------------- (1)

And (x₂ – x₄) + (x₃ – x₄) = l₂ (length of second string)

Or x₂ + x₃ – 2x₄ = l₂                                                                             ----------------- (2)

Or double differentiating with respect to time we get

a₁ + a₄ = 0                                                                                         ------------------- (3)

a₂ + a₃ – 2a₄ = 0                                                                                ------------------ (4)

But since a₄ = -a₁          from equation (3)

We have a₂ + a₃ + 2a₁ = 0

This is the required constant relation between a₁, a₂, and a₃.

Dumb Question:

1)      Why  and ?

Ans: If the rope is given inextensible then l₂ and l₁ does not change with time. So their derivatives are zero.
Illustration 3:
In the adjacent figure (9) mass of A, B and C are 2kg, 5kg, and 4kg respectively. Find
A) Acceleration of the system.

B) Tension in the string. Neglect friction (g = 10 m/s²)

                               Fig (9)

Solution:

A) Take A+B as whole system then,

7gSin60 – T = 7a

T – 4gSin30 = 4a

B) For the tension in the string between A and B

FBD of A-

                   Fig (10)

For the tension in the String between B and C

FBD of C-

                  Fig (11)

Tip: Always remember that the two end string over a pulley; always move up and opposite velocities with respect to pulley only.

6) SPRINGS:

Hook’s law for ideal springs F_sp = -Kx

Where, K ® Spring constant

            x® elongation

Dumb Question:

1)      What is the force applied on wall in a situation like:

       Ans:

       Let the force applied be equal to F¹ then FBD of spring:

       So F – F¹ = ma

       Now since we are considering light springs, m®0

       So, F¹ = F

       Hence the force applied by a light spring on connecting bodies at both ends is                                                always equal and opposite, and either of this force is known as spring force.

For the same material, .    Where l is natural/relaxed length.

Spring force does not change instantly except when it is cut [So in all other cases it is a non impulsive force]   

Illustration:

Find the acceleration (initial) of 1 and 2 if the upper spring is cut?

            Fig (12)

When in equilibrium:

FBD of Body (2)

  Fig (13)

F₂ = F₁ + m₂g

FBD of Body (1)

                                                             Fig (14)

F₁ = m₁g

So, F₁ = 2g (N); F₂ = 4g (N).

Now if upper one is cut then FBD looks like this:

By the property of spring force F₂ suddenly becomes zero while F₁ remains unchanged.

So initially