## Force, Constraint equation

Force:- Any pull or push”, unit: kg met/sec^{2} or Newton (N)

Types of forces:-

1- Field Forces:- No need of contact

Always attraction type

Direction:- Always along (-y) axis or straight vertically down ward.

(**2). Contact forces: **There must be some contact

(i) Normal force (N):- This force act when two surfaces of different object are in contact.

Direction:- perpendicular to contact surfaces.

(ii). Frictional force(*f*):- This force also act when two surfaces of different objects in contacts.

Direction:- Parallel to contact surfaces

Also F = m N here m is coefficient of friction.

Now

N → Perpendicular to contact surface

So F ^ N

**Understanding of frictional force**

The force which resist the motion”

m_{S} > m_{k } m can be less than or greater than one

I*f* F_{ext} = 0, *f* = 0

**Example:- 1 **

Now

*f _{s}* =

*m*N

_{s} = m_{s} mg

= 0.6 x 10 x 10

*f _{s} *= 60 N

*f _{k}* =

*m*N

_{k}= 0.2 x 10 x10

*f _{k}* = 20 N

Amount of frictional force = 0

Because *f _{ext}* = 0, =

*f*= 0

**Example:- 2 **

m = 10 kg As in case (i) *f _{s} *= 60 N,

*f*= 20 N

_{k}*f* = *f _{s}* = 40 N

Body is at rest

**Example:- 3 **

As in case (i) *f _{s} *= 60N,

*f*= 20 N

_{k}*f* = *f _{s}* = 60 =

*F*→ limitation friction

_{ext}Body is at rest but tends to move

**Example:- 4 **

As in case = *f _{s}* = 60N,

*f*= 20N

_{k}Body is in moving condition

10a = 85 - 20

*f *= *f _{k}* = 20 N

(3) Attachment forces indirect force

Direction: Direction of tension (T) is always away from the body.

**Example**:-

10 g –T = 10 a ---- (1)

N = 5 g ------(2)

T = 5a --------(3)

**Solving 1& 3**10 g - 5a = 10a

15a = 10g

T = 5a

Net force on the pulley

**Example**:-

*f _{s}* =

*m*

_{s}_{ }N

= 0.5 x 5g

*f _{s }*= 25 N

*f _{k} *= m

*N*

_{k }= 0.2 x 5g

*f*_{k} = 10 N

Pulling force for 5 kg

10 g = 100 N

So both block will move

10 g –T = 10a -----(1)

T -m_{k}N = 5a ------(2)

T – 10 = 5a

N = 5g -----(3)

**Solving (1) & (2)**

10g –T = 10a

1-10 = 5a

90 = 15a Þ2

Similarly

T -10 = 5 x 6 T= 40N

## Constraint Equation

**CONSTRAINT EQUATION **

Here in place of constraint we will apply a trick

or T. a = coust.

Note: T a 1/*v*

10g –T = 70 a

7T -10g = 10a

70g -7T = 490 a

60g -500a

Constraint Equations

**Simple Case **

Conclusion

V_{1} = V_{2}

a_{1} = -a_{2}

*x*_{1} + *x*_{2} = *l* = const.

V_{1} + V_{2} = 0

a_{1 }+ a_{2} = 0

**COMPLEX CASE**

*x*_{1 }+ *x*_{3} = l_{1}

(*x*_{1} – *x*_{3}) + (*x*_{4} – *x*_{3}) = l_{2}

(*x*_{1 }-*x*_{4}) + (_{x}_{2} –*x*_{4}) = l_{3}

V_{1} + V_{3} = 0

a_{1} + a_{3 }= 0

a_{1} + a_{4} – 2a_{3} = 0

a_{1} + a_{2} – 2a_{4} = 0

Þ a_{2 }= -7a_{1}

**SPECIAL CASE**** **

V_{2} = V_{1} cos q

a_{2} = a_{1} cos q

W + Y + Z = *l*

V_{1 }= V_{2} (1 + sin q)

**PSUDO FORCE **

T sinq = ma

T cosq = mg

Example Elevator or lift

T + 20 - 4g = 4a

6g – T - 30 = 6a

2g -10 = 10a

10 = 10a

T + 20 - 4g = 4 x1

T = 24N

Thrust on the pulley

F = 2T = 2 x 24 = 48 N

**ROUGH INCLINED PLANE**

N = 10g cos37^{o}

*f _{s}* =

*m*

_{s}_{ }N

= 0.5 x 10 x 10 x 4/5

*f*_{s} = 40 N

*f _{k}* =

*m*

_{k}_{ }N

= 0.3 x 10g cos37^{o}

= 0.3 x 10x 10 x 4/5

10 g sin 37^{o}

= 10 x 10 x 3/5

= 60 N

Now Pulling Force > therefore block will move

So, 10g sin37^{o }–*f _{k} *= 10

*a*

10 x 10 x 3/5 – 24 = 10 *a*

Þ 10 *a *= 36