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
(x1 – x3) + (x4 – x3) = l2
(x1 -x4) + (x2 –x4) = 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