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Newton's Laws Of Motion
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Constraint Motion

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 Fsp = -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

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