## Force, Constraint equation

Force:- Any pull or push”, unit: kg met/sec2 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”  mS > mk      m can be less than or greater than one

If Fext = 0, f = 0  Example:- 1

Now

fs = msN = ms mg

= 0.6 x 10 x 10

fs = 60 N

fk = mkN

= 0.2 x 10 x10

fk = 20 N

Amount of frictional force = 0

Because fext = 0, = f = 0 Example:- 2

m = 10 kg As in case (i)  fs = 60 N, fk = 20 N

f = fs = 40 N

Body is at rest

Example:- 3 As in case (i) fs = 60N, fk = 20 N

f = fs = 60 = Fext →    limitation friction

Body is at rest but tends to move Example:- 4

As in case = fs = 60N, fk = 20N

Body is  in moving condition

10a = 85 - 20 f = fk = 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& 310 g - 5a = 10a

15a = 10g  T = 5a Net force on the pulley  Example:- fs = ms N

= 0.5 x 5g

fs = 25 N

fk = mk N

= 0.2 x 5g

fk = 10 N

Pulling force for  5 kg

10 g = 100 N

So both block will move

10 g –T = 10a -----(1)

T -mkN = 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

V1 = V2

a1 = -a2

x1 + x2 = l = const. V1 + V2 = 0 a1 + a2 = 0

COMPLEX CASE x1 + x3 = l1

(x1x3) + (x4x3) = l2

(x1 -x4) + (x2x4) = l3 V1 + V3 = 0

a1 + a3 = 0

a1 + a4 – 2a3 = 0

a1 + a2 – 2a4 = 0

Þ a2 = -7a1

SPECIAL CASE    V2 = V1 cos q

a2 = a1 cos q W + Y + Z = l    V1 = V2 (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 cos37o

fs = ms N

= 0.5 x 10 x 10 x 4/5

fs = 40 N

fk = mk N

= 0.3 x 10g cos37o

= 0.3 x 10x 10 x 4/5  10 g sin 37o

= 10 x 10 x 3/5

= 60 N

Now Pulling Force > therefore block will move

So,     10g sin37o fk = 10 a

10 x 10 x 3/5 – 24 = 10 a

Þ 10 a = 36 