1. PHY 101: Lecture 4C 4.1 The Concepts of Force and Mass
4.2 Newton’s First Law of Motion
4.3 Newton’s Second Law od Motion
4.4 The Vector Nature of Newton’s Second Law of Motion
4.5 Newton’ Third Law of Motion
4.6 Types of Forces: An Overview
4.7 The Gravitational Force
4.8 The Normal Force
4.9 Static and Kinetic Frictional Forces
4.10 The Tension Force
4.11 Equilibrium Applications of Newton’s Laws of Motion
4.12 Nonequilibrium Applications of Newton’s Laws of Motion

2. PHY 101: Lecture 4 Force and Newton’s Laws of Motion
4.11
Equilibrium Applications of Newton’s Laws of Motion

3. Types of Problems Equilibrium Problems Nonequilibrium Problems
Sum of forces = 0
No acceleration
Static Problems - Velocity = 0
Dynamic Problems – Velocity not 0
Nonequilibrium Problems
Sum of forces not 0
Acceleration

4. Static Equilibrium – Example 1A
You and two friends find three ropes tied together with a single knot and decide to have a three-way tug-of-war
Alice pulls to the west with 100 N of force
Bob pulls to the south with 200 N
How hard, and in which direction, should you pull to keep the knot from moving?

5. Static Equilibrium – Example 1B
T2 / T1 = tanq
= tan-1(T2/T1)
= tan-1(200/100)=63.40
T3 = T1/cosq
T3 = 100/cos(63.4)
T3 = 223 N
SFx = T1x + T2x + T3x
SFy = T1y + T2y + T3y
T1x = - T1 T2x = 0 T3x = T3cosq
T1y = 0 T2y = - T2 T3y = T3sinq
-T1 + T3cosq = 0
-T2 + T3sinq = 0
T1 = T3cosq
T2 = T3sinq

6. Static Equilibrium - Example 2A
What are tensions on each rope?
Mass Upper:
SFy = Wy + Ny + fy + Ty + Py=0
Wy=-(mU)g=-(10)(9.8)=-98
Ny = 0 fy = 0
Ty = TU-TL Py = 0
TU – TL + 0 = 0
TU – TL = 98 N
Mass Lower:
SFy = Wy + Ny + fy + Ty+Py=0
Wy=-mLg = -(10)(9.8) = -98
Ny = 0 fy = 0
Ty = TL Py = 0
TL+0 =0
TL = 98 N

7. Static Equilibrium - Example 2B
What are tensions on each rope?
TU – TL = 98 N
TL = 98 N
Tu = = 196 N

8. Static Equilibrium - Alternative Example 2
What are tensions on each rope?
Mass Upper and Mass Lower:
SFy = Wy + Ny + fy + Ty + Py=0
Wy=-(mU+mL)g=-(20)(9.8)=-196
Ny = 0 fy = 0
Ty = TU Py = 0
TU + 0 = 0
TU = 196 N
Mass Lower:
SFy = Wy + Ny + fy + Ty+Py=0
Wy=-mLg = -(10)(9.8) = -98
Ny = 0 fy = 0
Ty = TL Py = 0
TL+0 =0
TL = 98 N

9. Dynamic Equilibrium – Example 1A
A car with a weight of 15,000 N is being towed up a 200 slope at constant velocity
Friction is negligible
The tow rope is rated at 6000 N maximum tension
Will it break?

10. Dynamic Equilibrium – Example 1B
SFx = Wx + Nx + fx + Tx + Px
SFy = Wy + Ny + fy + Ty + Py
WX = -wsinq Wy = -wcosq
Nx = 0 Ny= n
Tx = T Ty = 0
T – Wsinq = 0
N – Wcosq = 0
T = (15000)sin(20) = 5130 N
Rope will not break

11. PHY 101: Lecture 4 Force and Newton’s Laws of Motion
4.12
Non-Equilibrium Applications of Newton’s Laws of Motion

12. NonEquilibrium – Example 1
What is the mass of an object that accelerates at 3.0 m/s2 under the influence of a 5.0-N net force?
Assume direction is horizontal
F = ma
m = F/a
m = 5.0 / 3.0 = 1.7 kg

13. NonEquilibrium – Example 2
Two forces act on an object of mass 4.00 kg
One force is 40.0 N in the +x-direction
The other force is 60.0 N in the + y-direction
Find magnitude and direction of the acceleration of the object
x-direction
SFx = max
40 = 4ax
40/4 = ax = 10 m/s2
y-direction
SFy = may
60 = 4ay
60/4 = ay = 15 m/s2
magnitude of a = sqrt( ) = 18 m/s2
q = tan-1(15/10) = 560

14. NonEquilibrium – Example 3
Horizontal force of 300 N is applied to a 75.0 kg box
Box slides on a level floor, opposed by a force of kinetic friction of 120 N
What is magnitude of the acceleration of box?
SFx = Wx + Nx + (fk)x + Tx + Px = max
Wx = 0
Nx = 0
(fk)x = -120
Tx = 0
Px = 300
= 75ax
ax = 180/75 = 2.4 m/s2

15. NonEquilibrium – Example 4
A boy pulls a box of mass 30 kg with a force of 25 N at an angle of 300 above the horizontal
(a) Ignoring friction, what is the acceleration of the box?
(b) What is the normal force exerted on the box by the ground?
SFx = Wx + Nx + (fk)x + Tx + Px = max
SFy = Wy + Ny + (fk)y + Ty + Py = may
Wx = 0 Wy = mg = 30(-9.8) = -294
Nx = 0 Ny = N
Tx = Ty = 0
(fk)x (fk)y = 0
Px = 25 cos30 = 21.7 Py = 25 sin30 = 12.5
(a) x-direction
= 30ax
ax = 21.7/30 = 0.72 m/s2
(b) y-direction
N = N = 30ay = 0
N = newtons

16. NonEquilibrium – Example 5A
A 1500 kg car is pulled by a tow truck
The tension in the tow rope is 2500 N
A 200 N friction force opposes the motion
What is the car’s acceleration

17. NonEquilibrium – Example 5B
SFx = Wx + Nx + (fk)x + Tx + Px = max
SFy = Wy + Ny + (fk)y + Ty + Py = may
Wx = 0 Wy = mg = 1500(-9.8) =
Nx = 0 Ny = N
Tx = 2500 Ty = 0
(fk)x = -200 (fk)y = 0
Px = 0 Py = 0
= 1500ax
ax = 2300/1500 = 1.53 m/s2

18. NonEquilibrium – Example 6
The Atwood machine consists of two masses suspended from a fixed pulley
m1 = 0.55 kg
m2 = 0.80 kg
(a) What is the acceleration of the system?
(b) What is magnitude of tension in string?

19. NonEquilibrium – Example 6a
Mass 1:
SFy = Wy + Ny + (fk)y + Ty + Py = may
Wy = -m1g Ny = 0
Ty = T (fk)y = 0
Py = 0
-0.55g T + 0 = 0.55ay
Mass 2:
Wy = -m2g Ny = 0
-0.80g T + 0 = -0.80ay
Subtract equations: 0.25g = 1.35ay
ay = 1.8 m/s2

20. NonEquilibrium – Example 6b
-0.55g + T = 0.55ay
T = 0.55ay g
T = 0.55(1.8) (9.8) = 6.4 N

21. NonEquilibrium Alternative Example 6A
The Atwood machine consists of two masses suspended from a fixed pulley
m1 = 0.55 kg
m2 = 0.80 kg
(a) What is the acceleration of the system?
(b) What is magnitude of tension in string?

22. NonEquilibrium Alternative Example 6B
Pull = m2g
Pull =
m1g a
Lift masses m1 and m2 so that they are horizontal
Gravity on m1 becomes a pull to the left
Gravity on m2 becomes a pull to the right
Treat two masses and rope between as a single system
Tension in the rope is an Internal force and is ignored in Newton’s Second Law

23. NonEquilibrium Alternative Example 6C
Mass 1 + Mass2:
SFx = Wx + Nx + (fk)x + Tx + Px = max
Wx = 0 Nx = 0
Tx = 0 (fk)x = 0
Px = -m1g + m2g
-0.55g+0.80g = ( )ax
ax = .25(9.8) / 1.35 = 1.8 m/s2
Mass 1:
Wx = -m1g Nx = 0
Tx = T (fk)x = 0
Px = 0
-0.55g +0 + T = 0.55(1.8)
T = (9.8) = 6.4 N

24. NonEquilibrium – Example 7A
Three blocks are pulled along a frictionless surface by a horizontal force of F = 18.0 N
(a) What is the acceleration of the system?
(b) What are tension forces in the strings?

25. NonEquilibrium – Example 7B
Mass 1:
SFx = Wx+Nx+(fk)x+Tx+Px =m1ax
Wx = 0 Nx = 0
Tx = T1 (fk)x = 0
Px = 0
T1 + 0 = m1ax
T1 = m1ax
T1 = 1.0ax (Equation 1)
Mass 2:
SFx=Wx+Nx+(fk)x+Tx+Px=m2ax
Wx = 0 Nx = 0
Tx = -T1+T2 (fk)x = 0
Px = 0
T1 + T2 + 0 = m2ax
-T1 + T2 = m2ax
-T1 + T2 = 2.0ax (Equation 2)

26. NonEquilibrium – Example 7C
Mass 3:
SFx = Wx Nx+(fk)x+Tx+Px =m3ax
Wx = 0 Nx = 0
Tx = -T2 (fk)x = 0
Px = P
T2 + P = m3ax
-T2 + P = m3ax
-T = 3.0ax (Equation 3)
Add Equations 1 and 2
T2 = 3.0ax (Equation 4)
Add Equations 3 and 4
18.0 = 6.0ax
ax = 18.0/6.0 = 3.0 m/s2
Substitute into Equation 1
T1 = 1.0(3.0) = 3.0 N
Substitute into Equation 3
T2 = 18.0 – 3.0(3.0) = 9.0 N

27. NonEquilibrium Alternative Example 7A
Mass 1:
SFx=Wx+Nx+(fk)x+Tx+Px=m1ax
Wx = 0 Nx = 0
Tx = T1 (fk)x = 0
Px = 0
T1 + 0 = m1ax
T1 = (1.0)(3.0) = 3.0 N
Mass 1, Mass 2, and Mass 3:
SFx = Wx+Nx+(fk)x+Tx+Px =max
Wx = 0 Nx = 0
Tx = 0 (fk)x = 0
Px = P
P = (m1+m2+m3)ax
18.0 = 6.0ax
ax=18.0 / 6.0=3.0 (Equation 1)

28. NonEquilibrium Alternative Example 7B
Mass 3:
SFx = Wx + Nx + (fk)x + Tx + Px =m3ax
Wx = 0 Nx = 0
Tx = -T2 (fk)x = 0
Px = P
– T2 + P = m3ax
T2 = P – m3ax
T2 = 18 – (3.0)(3.0) = 9.0 N

29. NonEquilibrium – Example 8A
Pulleys are frictionless
There is no friction between mass m3 and the table
What is the acceleration of the system?
m1=0.25 kg, m2 = 0.50 kg, m3 = 0.25 kg

30. NonEquilibrium – Example 8B
Mass 1:
SF =Wy+Ny+(fk)y+Ty+Py=m1a
Wy = -m1g Ny = 0
Ty = T1 (fk)y = 0
Py = 0
-m1g T1 + 0 = m1a
-m1g+T1=m1a (Equation 1)
Mass 2:
SF =Wy+Ny+(fk)y+Ty+Py=m2a
Wy = -m2g Ny = 0
Ty = T2 (fk)y = 0
Py = 0
-m2g T2 + 0 = m2(-a)
-m2g+ T2=-m2a (Equation 2)

31. NonEquilibrium – Example 8C
Mass 3:
SFx=Wx+Nx+(fk)x+Tx+Px=ma
Wx = 0 Nx = 0
Tx = -T1 + T2 (fk)x = 0
Px = 0
– T1 + T2 + 0 = m3a
– T1+T2=m3a (Equation 3)
Subtract equation 2 from equation 1
(m2-m1)g-T2+T1=(m1+m2)a (Eq. 4)
Add equations 3 and 4
(m2-m1)g=(m1+m2+m3)a
a = (m2-m1)g/(m1+m2+m3)
a = ( )(9.8)/( )
a = 2.45 m/s2

32. NonEquilibrium Alternative Example 8A
Mass 1:
SF =Wy+Ny+(fk)y+Ty+Py=m2a
Wy = 0 Ny = 0
Ty = T1 (fk)y = 0
Py = -m1g
T1 – m1g = m1a
-m1g+ T1=m1a
T1=m1a+m1g = (0.25)( )
T1 = 3.1 N
Mass 1, Mass2, and Mass3
SFx=Wx+Nx+(fk)x+Tx+Px=ma
Wx = 0 Nx = 0
Tx = 0 (fk)x = 0
Px = -m1g+m2g
m1g+m2g = ma
-m1g+m2g=ma
a = g(m2-m1)/(m1+m2+m3)
a=(9.8)( )/1=2.45 m/s2

33. NonEquilibrium Alternative Example 8B
Mass 3:
SF =Wy+Ny+(fk)y+Ty+Py=m2a
Wy = 0 Ny = 0
Ty = T2 (fk)y = 0
Py = -m3g
T2 – m3g = -m3a
-m3g+ T2=-m3a
T2=-m3a+m3g = (0.25)( )
T2 = 1.8 N

34. NonEquilibrium Alternative Example 9A
Weight of the block on table is 422 N
Weigh of the hanging block is 185 N
Ignoring all friction and assume the pulley is massless
(a) Find the acceleration of the two blocks
(b) Find the tension in the cord
Mass of table block is m1 = 422/9.8 =43.06 kg
Mass of hanging block is m2 = 185/9.8 = kg

35. NonEquilibrium Alternative Example 9B
Mass 1:
SFx=Wx+Nx+(fk)x+Tx+Px=m1a
Wx = 0 Nx = 0
Tx = T (fk)x = 0
Px = 0
0+0+0+T+0 = m1a
T = m1a
T = 43.06(2.99) = N
Mass 1 and Mass2:
SFx=Wx+Nx+(fk)x+Tx+Px=ma
Wx = 0 Nx = 0
Tx = 0 (fk)x = 0
Px = m2g
m2g = (m1 + m2)a
m2g = (m1 + m2)a
a = -g(m2/(m1+m2)
a=(9.8)(18.88)/(61.94)=2.99 m/s2