1. Chapter 3 Force and Motion
4th Edition

2. What You Will Learn Newton’s 3 Laws of Motion Kinematics
In SI and Engineering English units
gc usage to clarify units
Free body diagrams
Kinematics
Speed vs. time diagrams
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3. Newton’s Laws of Motion
Newton’s First Law.
The speed v of a body remains constant unless the body is acted upon by an external force F.
Newton’s Second Law.
The acceleration a of a body is parallel to and directly proportional to the net force F applied to it and inversely proportional to its mass m, or a  F/m.
Newton’s Third Law.
The mutual forces of action and reaction between two bodies are equal, opposite, and collinear.
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4. Newton’s 1st Law First law implies that speed v = constant
An object at rest will stay at rest unless an unbalanced force acts upon it.
An object in motion will not change its speed unless an unbalanced force acts upon it.
Throw up a very heavy ball in an open sport’s car and a stationary observer sees the ball execute a parabola
In the car, an observer sees it go straight up and back down
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5. Newton’s 2nd Law
An object will accelerate if the unbalanced force on the object is not zero
The direction of the acceleration is the same as the direction of the unbalanced force. The magnitude of the acceleration is directly proportional to the unbalanced force, and inversely proportional to the mass of the object and written F  ma.
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6. Newton’s 3rd Law
For every action, there is an equal and opposite reaction.
Free Body Diagrams, FBD, are used for systems in static equilibrium such as bridges and buildings and thus occupy much of the activities of civil engineers
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7. Back to Newton’s 2nd Law
Recall the definition of force as proportional to mass × acceleration or F  ma
What is the missing proportionality constant?
Suppose we knew a force F1 corresponding to a mass m1 being accelerated by a1. Then F1  m1a1
Therefore
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8. More Newton’s 2nd Law
If we say the length scale is the meter, the mass scale is the kilogram and the time scale is the second, and F is defined to be in newtons (N).
To complete the definition let’s say F1 is 1 N when m1 is 1 kg and a1 is 1 m/s2.
Then m1a1/F1 = 1 [dimensionless] and F = ma, which is the way you have previously seen Newton’s 2nd Law.
Now we know the units of the force and its magnitude. This is called the SI or MKS (meter-kilogram-second) system.
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9. Still More Newton’s 2nd Law
If we say the length scale is in feet (ft), the mass is in pounds mass (lbm) and the time scale in seconds (s), and F is defined to be in pounds force (lbf).
To complete the definition let’s say F1 is 1 ft when m1 is 1 lbm and a1 is m/s2.
Then gc = m1a1/F1 = 1 × /1 [lbm][ft/s2]/[lbf] and F = ma/gc. This is called the Engineering English unit system
Now we know the units of the force and its magnitude. Note gc = [lbm][ft]/[lbf s2] and not to be confused with g = acceleration due to gravity, ft/s2
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10. Ok, More Newton’s 2nd Law
What force in N is needed to accelerate a frictionless motorcycle of 250. kg by 7.50 m/s2?
Need: F on a mass of 250. kg accelerated by 7.50 m/s2
Know: Newton's 2nd Law
How: In SI units F = ma
Solve: F = ma = 250. × 7.50
= × 103 [kg][m/s2] = 1.88 × 103 N
Note: 1 N = is about the weight of a small apple on earth!
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11. Even More of Newton’s 2nd Law
What’s the force on a frictionless wheel-barrow of 50.0 lbm being accelerated at 2.10 ft/s2? What’s the weight of the barrow?
Need: a) F on a mass of 50. lbm accelerated by 2.10 ft/s2; b) wt of wheel-barrow
Know: Newton's 2nd Law, acceln due to gravity g = 32.2 ft/s2; gc = 32.2 [lbm][ft]/[lbf s2]
How: Engineering English units F = ma/gc
Solve:
a) F = ma/gc = 50. × 2.10/32.2
= 3.26 [lbm ft/s2][lbf s2/lbm ft] = 3.3 lbf;
b) Wt = mg/gc = 50. × 32.2/32.2 = 50. lbf
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12. Newton’s 3rd Law
If you push on something it pushes back with the same force (but not the same acceleration since the pushed masses may be different)
If you fire a 15.0 gram bullet from a 5.0 kg rifle. If the initial acceleration of the bullet from the cm barrel is m/s, what is the recall force you experience from firing the gun at your shoulder?
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13. More Newton’s 3rd Law Need: Recoil force from the rifle.
Know: Rifle weighs 5.0 kg, the bullet kg. The barrel (down which the bullet accelerates) is 0.60 m and the bullet leaves the muzzle at m/s.
How: Assume the acceleration in the barrel is uniform and use Newton’s 3rd law to calculate the recoil force.
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14. Newton’s 3rd Law - Example
Accept (for the moment) that the acceleration of the bullet = (½v2)/barrel length, or a = ½ × /0.60 [m/s]2/[m] = 8.3 × 105 m/s2.
Therefore the Force on the bullet is
F = ma = × 8.3 × 105 [kg][m/s2] = 1.3 × 104 N
The reaction on the rifle is in the opposite direction to that on the bullet = × 104 N. (If you also want the initial acceleration of the rifle it is × 104/5.0 = -2.6 × 103 [kg m/s2][1/kg] = -2.6 × 103 m/s2)
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15. Free Body Diagram (FBD) Example
FBDs are used to keep track of forces
A stationary crate weighing 100. N sits on a 30.0 degree ramp
Resolve the forces on the crate and the slope parallel to the hill and normal to it.
Need: FBD for crate on a 30 degree slope
Know: Forces on the crate and the slope are equal and opposite
How: Resolve forces parallel and normal to the slope
100. N Fr
30º
87.N
50.N
Ramp
Crate
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16. Free Body Diagram Example
Solve: The weight of the block resolves to 100. N cos30º = 86.6 N normal to the slope and 100.sin30º = 50.0 N parallel to the slope.
The friction force Fr on the block must be equal and opposite to the force on the block. The frictional force is thus 50.0 N.
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17. Kinematics
The study of motion without consideration of the associated forces.
In this elementary treatment we will only use 1-dimensional methods.
In > 1-D, speed is not equivalent to velocity because velocity is a vector with components in each direction.
In 1-D speed is the magnitude of velocity and is a scalar.
We will graphical solutions only. Basically no equations are used and the method emphasizes understanding not formula plugging.
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18. Cartesian Geometry
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19. Kinematics and Cartesian Geometry
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20. Kinematics, v-t and Constant a
Max speed
v, Speed
Start accn
Slope is
acceleration
Area under curve is
distance travelled.
Time
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21. Kinematics Example
You accelerate from rest at a constant 5.0 m/s2 for 13 s. What’s 1) Your final speed and 2) how far will you have travelled?
Slope = v/t = 5.0 m/s2
v = v = 5.0 × 13 [m/s2][s]
= 65 m/s
Distance travelled = ½ v t
= ½ × 65 × 13 [ m/s][s]
= 4.2 × 102 m
Speed, m/s
5.0 m/s2 v t
0 s
Time, s
13 s
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22. Summary Newton’s 3 Laws: Kinematics:
1) Bodies that remain undisturbed continue until external forces act on them
2) Force proportional to mass × acceleration and how to manipulate units for gc formulation
3) Action and reaction are equal and opposite
Kinematics:
Central position of speed vs. t diagrams.
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