2. Dynamics
• Kinematics
It is a branch of Engineering Mechanics deals with
bodies in motion with out considering the forces
causing that motion
• Kinetics
It is a branch of Engineering Mechanics deals with
bodies in motion with considering the forces causing
that motion
3. Kinetics : Deals with bodies in motion considering the forces
causing the motion
F=ma
∑F=ma
∑F – ma = 0 ----- Dynamic equilibrium equation
∑F – ma = 0 ---- Force – Inertia method
Where
∑F is resultant of forces applied
ma – inertial force acts opposite to direction of
acceleration
4.
5. Problem: Determine the force ‘P’ that will give the body shown
in the figure an acceleration of 0.3g m/s² . Coefficient of kinetic
friction is 0.3
.
P
3125 N
4
3
θ
6. Problem: Determine the force ‘P’ that will give the body shown
in the figure an acceleration of 2g m/s² . Coefficient of kinetic
friction is 0.2
.
P
2250 N
3
4
θ= 36.87°
θ
Procedure
1. Decide the direction of
motion of block
2. Draw FBD
3. Use equilibrium equations
7. Step 1: Block moves towards right
Step 2: FBD of block
Check list for FBD
1.Weight
2. Normal Reaction
3. Frictional Force
4. Tension force
5. Forces given in the problem
6. Inertial force (ma) – acts
opposite to direction of
acceleration of the body
P Sin36.87°
P Cos36.87°
N
2250 N
F= 0.2 N
ma
8. Given a= 2 g m/s²
a = 19.62 m/s²
We know that, m=W/g
Step 3: Using equilibrium equations
∑x=0
P Cos36.87° - F –ma = 0
P Cos36.87° - 0.2N – (2250/9.81) (19.62) = 0
0.8 P – 0.2 N = 4500 ---------- Equation 1
∑y =0
P Sin36.87° + N -2250 = 0
0.6 P + N = 2250 ----------Equation 2
Solving Equations 1 and 2
P=5380.53 N
P Sin36.87°
P Cos36.87°
N
2250 N
F= 0.2 N
ma
9. Problem: Determine the acceleration of the bodies if the
coefficient of kinetic friction is 0.1 between the cable and the
fixed drum
Procedure
1. Decide the direction of motion
of blocks
2. Use belt friction equation
3. Draw FBD of each block
separately
4. Use equilibrium equations
450 N 900 N
B A
T1
T2
10. Step 1: As weight of body ‘A’
is more ‘A’ moves down
and ‘B’ moves up
450 N 900 N
Step 2: Belt friction equation
T1/T2=eµθ
Where
T1 - Tension in block having more weight
T2 – Tension in other block
µ - coefficient of friction between cable
and drum
θ= Angle of contact of cable with drum
T1 = T2 x eµθ
θ=180° = π , µ = 0.1
T1 = T2 x ex0.1x π
T1 = 1.369 T2 --- Equation 1
B A
T1
T2
11. ‘A’ moves down and
‘B’ moves up
450 N 900 N
Step 3: FBD of block ‘A’
Step 4: ∑y=0
+T1 + ma -900 = 0
+T1 + (900/9.81) a -900 = 0
T1 + 91.74 a -900 = 0 ----- Eqn2
B A
900 N
T1
ma
A
T1
T2
12. ‘A’ moves down and
‘B’ moves up
450 N 900 N
Step 3: FBD of block ‘B’
Step 4: ∑y=0
+T2 - ma -450 = 0
+T2 – (450/9.81) a -450 = 0
T2 - 45.87 a - 450 = 0 ----- Eqn3
Solving Equations 1, 2 and 3
a=1.837 m/s²
B A
450 N
T2
ma
B
T1
T2
13. .
900 N B
A
2T
T
Problem: In the system pulleys are friction less and weightless.
Find the weight of block ‘A’ that will give body ‘B’
a downward acceleration of 0.6g
Procedure
1. Decide the direction of motion of
blocks in the system
2. Establish kinematic relationship
3. Draw FBD of each block
separately
4. Use equilibrium equations
14. .
900 N
?
B
A
2T
T
Step1:Body ‘B’ moves downward (Given).Body ‘A’ moves upward
.
Step 2. Establish kinematic relation
between the bodies of the system
2T SB -T SA = 0
2T SB =T SA
Differentiating
2VB = VA
Differentiating again
2aB = aA --Kinematic relation
15. .
900 N
?
B
A
2T
T
Body ‘B’ moves downward
body ‘A’ moves upward
Step 3: FBD of block ‘B’
Step 4: ∑y=0 (Equilibrium equation)
2T + maB -900 =0
aB = 0.6g (given)
2T + (900/g) 0.6g -900 =0
2T + 540 -900 =0
T = 180N
900 N
2T
maB
B
16. .
900 N
?
B
A
2T
T
Body ‘B’ moves downward
body ‘A’ moves upward
Step 3: FBD of block ‘A’
2aB = aA -Kinematic relation
aB = 0.6g (given)
aA = 1.2g
Step 4: ∑y=0
T - maA -WA =0
T - (WA /g) 1.2g - WA =0
T - (WA) 1.2 - WA =0
T = WA (1+1.2)
Substitute T=180
180 = WA (2.2)
WA= 81.82N
T
maA
WA
A
17. .
900 N
B
A
2T
T
Compute the acceleration of block ‘B’ and the tension ‘T’ in cable
supporting block ‘A’.
3
4
θ= 36.87°
θ
Procedure
1. Decide the direction of motion
of blocks in the system
2. Establish kinematic relationship
3. Draw FBD of each block
separately
4. Use equilibrium equations
18. Step 1: Decide the direction of
motion of blocks
Given µ=0.2
900 N
B
A
2T
T
Assume block ‘A’ at rest
∑y=0
T=900N
For block ‘B”
∑y=0
NB = 1080
2T = 2x900 = 1800 (acting upwards)
810 + 216 = 1026 (acting downwards)
1800 > 1026 So, Block ‘B’ moves up
and block ‘A’ moves down.
T
A
3
4
θ= 36.87°
θ
900 N
B
2T
1350 N
θ
19. Block ‘B’ moves up and block
‘A’ moves down.
900 N
B
A
2T
T
Step3: FBD of block ‘B’
Step 4: ∑x=0
2T-810-216-maB = 0
2T-810-216-(1350/9.81) aB = 0
2T-810-216- 137.61aB = 0
2T- 137.61aB - 1026 = 0 ---Eqn 3
Solve Eqns 1, 2 and 3
3
4
θ= 36.87°
θ
B
2T
1350 N
θ
maB
20. Block ‘B’ moves up and block
‘A’ moves down.
900 N
B
A
2T
T
Step 2. Establish kinematic
relation between the bodies of
the system
2T SB -T SA = 0
2T SB =T SA
Differentiating
2VB = VA
Differentiating again
2aB = aA ---- Equation 1
maA
3
4
θ= 36.87°
θ
21. Block ‘B’ moves up and block
‘A’ moves down.
900 N
B
A
2T
T
Step 3:FBD of block ‘A’
Step4: ∑y=0
T + maA -900 =0
T + (900/9.81) aA -900 =0
T + 91.74 aA -900 =0 – Equation 2
T
maA
A
3
4
θ= 36.87°
θ
900 N
maA
22. .
750 N B
A
2T
T
Problem: In the system pulleys are friction less and weightless.
Find the weight of block ‘A’ that will give body ‘B’
a downward acceleration of 0.5g
30. 0°
18m
Two blocks A and B are released from rest on a 30° incline plane when
they are 18m apart. The Coefficient of friction under block A is 0.2 and
Coefficient of friction under block B is 0.4. Compute elapsed time until
blocks touch.
35. Determine the acceleration of the bodies shown in the fig. if the
coefficient of friction is 0.2 at all contact surfaces. Body ‘A’ weighs
900N and ‘B’ weighs 1350N.
B
A
36. Determine the acceleration of the bodies shown in the fig. if the
coefficient of friction is 0.2 at all contact surfaces. Body ‘A’ weighs
900N and ‘B’ weighs 1350N.
B
A
37. Determine the acceleration of the bodies shown in the fig. if the
coefficient of friction is 0.2 at all contact surfaces. Body ‘A’ weighs
900N and ‘B’ weighs 1350N.
B
A
38.
39.
40.
41.
42. Problem: Determine the force ‘P’ that will give the body shown
in the figure an acceleration of 0.2g m/s² . Coefficient of kinetic
friction is 0.20
.
P
2500N
3
4
θ
43. .
850 N B
A
2T
T
Problem: In the system pulleys are friction less and weightless.
Find the weight of block ‘A’ that will give body ‘B’
a downward acceleration of 0.45g
Procedure
1. Decide the direction of motion of
blocks in the system
2. Establish kinematic relationship
3. Draw FBD of each block
separately
4. Use equilibrium equations
44. .
875 N
B
A
2T
T
Compute the acceleration of block ‘B’ and the tension ‘T’ in cable
supporting block ‘A’.
3
4
θ= 36.87°
θ
Procedure
1. Decide the direction of motion
of blocks in the system
2. Establish kinematic relationship
3. Draw FBD of each block
separately
4. Use equilibrium equations
Given µ=0.25
45.
46. .
425 N 875 N
B A
T1
T2
Determine the acceleration of the bodies if the coefficient of
kinetic friction is 0.1 between the cable and the fixed drum
47. .
425 N 875 N
B A
T1
T2
Determine the acceleration of the bodies if the coefficient of
kinetic friction is 0.1 between the cable and the fixed drum
Procedure
1. Decide the direction of motion
of blocks
2. Use belt friction equation
3. Draw FBD of each block
separately
4. Use equilibrium equations
48. Step 1: As weight of body ‘A’
is more ‘A’ moves down
and ‘B’ moves up
450 N 900 N
Step 2: Belt friction equation
T1/T2=eµθ
Where
T1 - Tension in block having more weight
T2 – Tension in other block
µ - coefficient of friction between cable
and drum
θ= Angle of contact of cable with drum
T1 = T2 x eµθ
θ=180° = π , µ = 0.1
T1 = T2 x ex0.1x π
T1 = 1.369 T2 --- Equation 1
B A
T1
T2
49. 40°
22m
Two blocks A and B are released from rest on a 40° incline plane when
they are 22m apart. The Coefficient of friction under block A is 0.20 and
Coefficient of friction under block B is 0.30. Compute elapsed time until
blocks touch.
50. Determine the acceleration of the bodies shown in the fig. if the
coefficient of friction is 0.2 at all contact surfaces. Body ‘A’ weighs
900N and ‘B’ weighs 1350N.
B
A
