**5) CONSTRAINT MOTION: **

<!–[if !supportLists]–> A) Due to inextensible taut strings.

<!–[if !supportLists]–> 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).

/v:imagedata /v:shape

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)

/v:imagedata /v:shape

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:

<!–[if !supportLists]–> 1) Velocity components along a string are equal.

<!–[if !supportLists]–> 2) No motion (relative) occurs in normal direction.

<!–[if !supportLists]–> 3) Acceleration component in normal direction are equal.

Illustration 2:

Example 5.20 page 190 D C Pandey

Find the Constraint relation between a_{1}, a_{2}, and a_{3}.

/v:imagedata /v:shape

Fig (8)

Solution:

Points 1, 2, 3, 4 are movable let their displacements from a fixed line be x_{1}, x_{2}, x_{3}, and x_{4}.

We have x_{1}+x_{4} = l_{1} (length of first string) —————- (1)

And (x_{2} – x_{4}) + (x_{3} – x_{4}) = l_{2} (length of second string)

Or x_{2} + x_{3} – 2x_{4} = l_{2} —————– (2)

Or double differentiating with respect to time we get

a_{1} + a_{4} = 0 ——————- (3)

a_{2} + a_{3} – 2a_{4} = 0 —————— (4)

But since a_{4} = -a_{1} from equation (3)

We have a_{2} + a_{3} + 2a_{1} = 0

This is the required constant relation between a_{1}, a_{2}, and a_{3}.

Dumb Question:

<!–[if !supportLists]–> 1) Why /v:imagedata and /v:imagedata ? /v:shape/v:shape

Ans: If the rope is given inextensible then l_{2} and l_{1} 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/s^{2})

/v:imagedata /v:shape

Fig (9)

Solution:

A) Take A+B as whole system then,

/v:imagedata /v:shape

7gSin60 – T = 7a

T – 4gSin30 = 4a

/v:imagedata /v:shape

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

FBD of A-

/v:imagedata /v:shape

Fig (10)

/v:imagedata /v:shape

For the tension in the String between B and C

FBD of C-

/v:imagedata /v:shape

Fig (11)

/v:imagedata /v:shape

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:

<!–[if !supportLists]–> 1) What is the force applied on wall in a situation like:

/v:imagedata /v:shape

Ans:

Let the force applied be equal to F^{1} then FBD of spring:

/v:imagedata /v:shape

So F – F^{1} = ma

Now since we are considering light springs, m®0

So, F^{1} = 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, /v:imagedata . Where l is natural/relaxed length. /v:shape

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?

/v:imagedata /v:shape

Fig (12)

When in equilibrium:

FBD of Body (2)

/v:imagedata /v:shape

Fig (13)

F_{2} = F_{1} + m_{2}g

FBD of Body (1)

/v:imagedata /v:shape

Fig (14)

F_{1} = m_{1}g

So, F_{1} = 2g (N); F_{2} = 4g (N).

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

/v:imagedata /v:shape

By the property of spring force F_{2} suddenly becomes zero while F_{1} remains unchanged.

So initially /v:imagedata /v:shape

/v:imagedata /v:shape