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

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