1. Applications of Newton’s
Laws of Motion
Chapter Opener. Caption: Newton’s laws are fundamental in physics. These photos show two situations of using Newton’s laws which involve some new elements in addition to those discussed in the previous Chapter. The downhill skier illustrates friction on an incline, although at this moment she is not touching the snow, and so is retarded only by air resistance which is a velocity-dependent force (an optional topic in this Chapter). The people on the rotating amusement park ride below illustrate the dynamics of circular motion.

2.  We need VECTOR addition to add forces in the 2nd Law!
Remember that forces are VECTORS!
Newton’s 2nd Law: ∑F = ma
∑F = VECTOR SUM of
all forces on mass m
 We need VECTOR addition to add forces in the 2nd Law!
Forces add according to the rules of
VECTOR ADDITION!
Get Force Components from trig, as discussed earlier.

3. Problem Solving Procedures
Make a Sketch. For each object separately, sketch a free-body diagram, showing all the forces acting on that object. Make the magnitudes & directions as accurate as you can. Label each force.
Resolve Vectors into Components.
Apply Newton’s 2nd Law separately to each object & for each vector component.
Solve for the unknowns.
Note: This often requires algebra, such as solving 2 linear equations in 2 unknowns!

4. Example FR = [(F1)2 + (F2)2](½) = 141 N tanθ = (F2/F1) = 1  θ = 45º
Square Root!
FR = [(F1)2 + (F2)2](½) = 141 N
tanθ = (F2/F1) =  θ = 45º

5. Example If the boat moves with acceleration a, ∑F = = FR = ma
FRx = max, FRy = may

6. Example FRCx = FRCcosθ FRCy = -FRCsinθ FRBx = -FRBcosθ FRBy = -FRBsinθ
Newton’s 3rd Law 
FBR = -FRB, FCR = -FRC
FRCx = FRCcosθ
FRCy = -FRCsinθ
FRBx = -FRBcosθ
FRBy = -FRBsinθ

7. Example A box of mass m = 10 kg is pulled by an attached cord
along a horizontal smooth (frictionless!) surface of a
table. The forc exerted is FP = 40.0 N at a 30.0° angle as
shown. Calculate: a. The acceleration of the box.
b. The magnitude of the upward normal force FN exerted by the table on the box.
Figure Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2.
(b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

8. The normal force, FN is NOT equal & opposite to the weight!!
A box of mass m = 10 kg is pulled by an attached cord
along a horizontal smooth (frictionless!) surface of a
table. The force exerted is FP = 40.0 N at a 30.0° angle as
shown. Calculate: a. The acceleration of the box.
b. The magnitude of the upward normal force FN exerted
by the table on the box.
Free Body
Diagram
Figure Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2.
(b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.
The normal force, FN is NOT
equal & opposite to the weight!!

9. Components of FP FPy = FP sin() = 40sin(30.0°) = 20 N
A box of mass m = 10 kg is pulled by an attached cord along a
horizontal smooth (frictionless!) surface of a table. The force is
FP = 40.0 N at a 30.0° angle as shown.
Calculate: a. The acceleration. b. The normal force FN
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Figure Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2.
(b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

10. FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N
A box of mass m = 10 kg is pulled by an attached cord along a
horizontal smooth (frictionless!) surface of a table. The force is
FP = 40.0 N at a 30.0° angle as shown.
Calculate: a. The acceleration. b. The normal force FN
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Newton’s 2nd Law
Vertical (y): Fy = 0
FN + FPy – mg = 0; FN = mg - FPy
FN = 98 – 20 = 78 N
Figure Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2.
(b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

11. FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N
A box of mass m = 10 kg is pulled by an attached cord along a
horizontal smooth (frictionless!) surface of a table. The force is
FP = 40.0 N at a 30.0° angle as shown.
Calculate: a. The acceleration. b. The normal force FN
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Newton’s 2nd Law
Vertical (y): Fy = 0
FN + FPy – mg = 0; FN = mg - FPy
FN = 98 – 20 = 78 N
Horizontal (x): Fx = ma
FPx = ma; a = (FPx/m)
a = 3.46 m/s2
Figure Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2.
(b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

12. Example: Pulling against friction
A box, mass m = 10 kg, is pulled along a horizontal surface
by a force FP = 40.0 N applied at a 30.0° angle above
horizontal. The coefficient of kinetic friction is k = 0.3.
Calculate a. the acceleration, b. the normal force FN.
Figure 5-5.
Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

13. A box, mass m = 10 kg, is pulled along a horizontal surface
by a force FP = 40.0 N applied at a 30.0° angle above
horizontal. The coefficient of kinetic friction is k = 0.3.
Calculate a. the acceleration, b. the normal force FN.
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Figure 5-5.
Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

14. FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N
A box, mass m = 10 kg, is pulled along a horizontal surface
by a force FP = 40.0 N applied at a 30.0° angle above
horizontal. The coefficient of kinetic friction is k = 0.3.
Calculate a. the acceleration, b. the normal force FN.
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Newton’s 2nd Law
Vertical (y): Fy = 0
FN + FPy – mg = 0; FN = mg - FPy
FN = 98 – 20 = 78 N
Figure 5-5.
Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

15. FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N
A box, mass m = 10 kg, is pulled along a horizontal surface
by a force FP = 40.0 N applied at a 30.0° angle above
horizontal. The coefficient of kinetic friction is k = 0.3.
Calculate a. the acceleration, b. the normal force FN.
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Newton’s 2nd Law
Vertical (y): Fy = 0
FN + FPy – mg = 0; FN = mg - FPy
FN = 98 – 20 = 78 N
Figure 5-5.
Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.
Friction:
Ffr = kFN = (0.3)(78) = 23.4 N

16. FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N
A box, mass m = 10 kg, is pulled along a horizontal surface
by a force FP = 40.0 N applied at a 30.0° angle above
horizontal. The coefficient of kinetic friction is k = 0.3.
Calculate a. the acceleration, b. the normal force FN.
Components of FP
FPx = FP cos() = 40cos(30.0°) = 34.6 N
FPy = FP sin() = 40sin(30.0°) = 20 N
Newton’s 2nd Law
Vertical (y): Fy = 0
FN + FPy – mg = 0; FN = mg - FPy
FN = 98 – 20 = 78 N
Figure 5-5.
Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.
Friction:
Ffr = kFN = (0.3)(78) = 23.4 N
Horizontal (x): Fx = ma
FPx - Ffr = ma; a = (FPx - Ffr)/m) = 1.14 m/s2

17. Example: Two boxes and a pulley
2 boxes are connected by a cord running
over a pulley. The coefficient of kinetic
friction between box A & the table is
k = 0.2. Ignore mass of cord & pulley &
friction in the pulley, which means a
force applied to one end of the cord has
the same magnitude at the other end.
Find the acceleration, a, of the system,
which has the same magnitude for both
boxes if the cord doesn’t stretch. As box
B moves down, box A moves to right.
Find the tension FT in the cord.
a  a 
∑F = ma For EACH mass
separately! x & y components
plus friction Ffr = μkFN

18. Example: Two boxes & a pulley
Coefficient of kinetic friction is k = 0.2.
Find the acceleration, a, of the system
& the tension FT in the cord.
∑F = ma For EACH mass
separately! x & y components
plus friction Ffr = μkFN
mA: ∑Fx = mAa = FT – Ffr
∑Fy = 0 = FN – mAg
FN = mg, Ffr = μkFN = μkmAg
So: mAa = FT – μkmAg (1)
mB: ∑Fy = mBa = mBg – FT
So: mBa = mBg – FT (2)
Solve (1) & (2) to find a & FT
a  a 

19. Example: Two boxes and a pulley
Solutions are:
a = (mBg – Ffr)/(mA + mB)
a = 1.4 m/s2
FT = Ffr + mAa
FT = 17 N
a  a 
∑F = ma For EACH mass
separately! x & y components
plus friction Ffr = μkFN

20. Conceptual Example: To push or to pull a sled?
Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the same angle θ in each case.
∑F = ma
Pulling
Pushing
Figure 5-6.
Answer: Pulling decreases the normal force, while pushing increases it. Better pull.
y forces: ∑Fy = 0
FN – mg –Fy = 0
FN = mg + Fy
Ffr (max) = μsFN
x forces: ∑Fx = ma
y forces: ∑Fy = 0
FN - mg + Fy = 0
FN = mg - Fy
Ffr (max) = μsFN
x forces: ∑Fx = ma