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