Newton’s Laws of Motion - WordPress.com...Newton’s, what is the reading on the scale when the elevator is not moving? •W = mg = 55 x 10 = 550 N •If the elevator begins to accelerate

Newton’s Laws of Motion Book page 44 - 47

©cgrahamphysics.com 2016

Inertia… • Moving objects have inertia a property of all objects to resist a change in motion

• Mass: a measure of a body’s inertia

• Two types of mass: - inertial mass 𝑚𝑖 - gravitational mass 𝑚𝑔

• Since acceleration of free fall is independent of the mass of an object 𝑚𝑖 = 𝑚𝑔

• Weight: The gravitational force that Earth exerts on a body

• Weight: varies with location, but inertial and gravitational mass stays constant


Newton’s 1st Law If a body is at rest it wants to remain at rest and if the body is moving in a straight line with uniform motion it will continue to move with uniform motion unless acted upon by an external force. External force = unbalanced

• This does not take friction into account • Friction increases with increasing speed


Example of the first law

• What are the forces acting on a ball thrown towards a partner?

• Aristotle view: Galileo view

thrust

W W


What does the 1st Law tell us?

• An object moving at constant velocity is in EQLB. All forces that act on it cancel out.

• Anything that is changing direction must have a resultant force acting on it – accelerating, decelerating or changing direction

• Also called “Law of Inertia” b/c it has the tendency of matter described to keep moving in the way it is already moving


Inertial frame of reference A system on which no forces act

Laws of Physics are valid

There is no experiment to distinguish between an

observer at rest and one that is moving at constant velocity

I am still

Stand still


Newton’s 2nd Law

• If an unbalanced force acts on an object, the object accelerates the greater the mass the smaller the acceleration for a constant force

𝑎 ∝1

𝑚

If the force doubles, acceleration doubles 𝑎 ∝ 𝐹𝑛𝑒𝑡

• Combining both observations: 𝐹𝑛𝑒𝑡 = 𝑚𝑎


𝐹𝑛𝑒𝑡 = 𝑚𝑎 • The acceleration is directly proportional to the force acting

and is in the same direction as the applied force.

• If we express the acceleration in terms of rate of change of velocity, Newton’s 2nd law can be rewritten:

• 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝑚∆𝑣

∆𝑡

• 𝐹𝑛𝑒𝑡 =∆𝑝

∆𝑡 another form of Newton’s 2nd

Law

Important: The net force is the sum of all forces acting on the object


Example

• In an extreme test on its braking system under ideal road conditions, a car traveling initial at 26.9𝑚𝑠−1[S] comes to a stop in 2.61s. The mass of the car with the driver is 1.18 x 103𝑘𝑔. Calculate a) the car’s acceleration and b) the net force required to cause that acceleration

Solution

• a) 𝑎 =𝑣−𝑢

𝑡=

0−26.9

2.61= −10.3𝑚𝑠−2

• b) 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 1.18 × 10−3 × 10.3 = 1.22 × 104𝑁[𝑁]

Newton’s Laws hold true in an inertial frame a car going around a curve is an example of a non – inertial

frame


Example • A man is riding in an elevator. The combined mass of the man and

the elevator is 7.00× 102kg. Calculate the magnitude and direction of the elevator’s acceleration, if the tension in the supporting cable is 7.50 × 103𝑁.

Assume up, if not sign will be negative

T

W W

T

ma + mg = T

𝑎 =𝑇−𝑚𝑔

𝑚=

7.50 ×103−7.00×102×10

7.00×102 = 0.71𝑚𝑠−2up!


Continued… • Find the tension in the cable if the acceleration is down

• T = mg – ma = m(g – a) = 7.00× 102(10 - 0.71) = 6.5 × 103N

T

W

T

W

W = T + ma


Net force in free fall

• The net force of an freely falling object is W = mg

• But 2nd law says 𝐹𝑛𝑒𝑡 = 𝑚𝑎

• Hence mg = ma g = a

• This also shows the equivalence between inertial and gravitational mass

• Assume gravitational force of Earth on object is 𝐹𝑔 = 𝑘𝑚𝑔

where k is a constant

• The acceleration is given by Newton’s 2nd law: 𝐹𝑔 = 𝑘𝑚𝑔 = 𝑚𝑖𝑔

where 𝑚𝑖 = inertial mass

• But experiments show that g is a constant and has the same value for all objects

k = g and 𝑚𝑔𝑔 = 𝑚𝑖𝑔

𝑚𝑔 = 𝑚𝑖


Definition of force

1N is the force which produces an acceleration of 1𝒎𝒔−𝟐 in a mass of 1kg. ©cgrahamphysics.com 2016

Example • Assume your weight is

55kg and you are standing on a scale in an elevator. If the scale is calibrated in Newton’s, what is the reading on the scale when the elevator is not moving?

• W = mg = 55 x 10 = 550 N

• If the elevator begins to accelerate upward at 0.74𝑚𝑠−2, what will be the reading on the scale?

• N = mg + ma = m(g + a) N = 55 (10 + 0.74) = 5.9 x102N

N

W

ma ma

W

N


Example

• The diagram shows a block of wood of mass 1.0 kg attached via a pulley to a hanging weight of mass 0.5kg. Assuming that there is no friction between the block and the bench and taking g to be 10 𝑚𝑠−2, calculate the acceleration of the system and the tension in the string

1.0kg

0.5kg

T ma T = 1a

T

W

T

W ma ma

T + 0.5a = 0.5g

a + 0.5a = 0.5g 1.5 a = 5

𝑎 =5

1.5= 3.3𝑚𝑠−2

T = 3.3N ©cgrahamphysics.com 2016

Example

• A person of mass 70 kg is strapped into the front seat of a car, which is travelling at a speed of 30𝑚𝑠−1. The car brakes and comes to rest after travelling a distance of 180m. Estimate the average force exerted on the person during the braking process.

• Solution

• 𝑣2 = 𝑢2 + 2𝑎𝑠

𝑎 =𝑣2 − 𝑢2

2𝑠=

0 − 302

2 × 180= −2.5𝑚𝑠−2

• F = ma = 70 x 2.5 = 175N

Minus sign slowing down


Newton’s Laws of Motion - WordPress.com...Newton’s, what is the reading on the scale when the elevator is not moving? •W = mg = 55 x 10 = 550 N •If the elevator begins to accelerate

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