Physics

Physics

PHS 5043 Forces & EnergyNewton’s Laws

Force:

Any agent capable of changing the shape of an object or changing its state of rest or motion

Symbol: F

Units: N (newton)

N = kg m/s2


Force:

Vector quantity

(magnitude and direction)

Forces can be added as any other vector (polygon, parallelogram and component method)


Force:

At any given moment in time, all objects are subjected to the simultaneous action of many forces


Equilibrium:

Objects are said to be in equilibrium when they are “at rest” or when they have “uniform rectilinear motion”


Equilibrium:

When an object is in equilibrium all external forces add up to zero.

If we add all forces vectors (regardless of the method), we should obtain a zero value resultant


Practice:

Given the following forces values, determine if the object is in equilibrium.

Fa = 30N, 270°

Fb = 10N, 45°

Fc = 20N, 135°

_Coordinates: Fa (0, -30); Fb (7.1,7.1); Fc (-20,20)

_Resultant Force: Fr (-12.9,-2.9)

No, the system is not in equilibrium for the resultant vector (component method) does not have (0.0) coordinates

Try with the polygon method!


Equilibrant force:

Is the force that balances the resultant force It brings the system (object) to a state of

equilibrium


Practice:

Given the following forces values, determine the magnitude and direction of the equilibrant force

Fa = 3N, 90°

Fb = 4N, 0°

_Coordinates: Fa (0, 3); Fb (4, 0), therefore Fr (4, 3)

_Magnitude Fr = 5N (Pythagoras), direction = 37°

_Feq = Fr but in opposite direction

Therefore, the equilibrant force would be Feq = 5N, 217°


Inertia:

Property that causes an object to remain in its state of rest or uniform rectilinear motion, that is, to remain in equilibrium.


Newton’s First Law:

“If no external or unbalanced force acts on an object, it maintains its state of rest or its constant velocity in a straight line”


Newton’s Second Law:

“The change of momentum of an object is proportional to the net applied force”


Newton’s Second Law:

F = m a

F = Δp / Δt F: Net force (N)

F = mΔv / Δt m: Total mass (kg)

F = m a a: acceleration (m/s2)


Practice:

A 15 kg sled is pulled horizontally with a constant force of 60N. What acceleration would this force impart on the sled if the frictional force (sled-snow) is 10N?

F = m a

Fa – Ff = m a

a = (Fa – Ff ) / m

a = (60N – 10N) / 15kg

a = 3.3 m/s2


Practice:

A 15 kg sled is pulled horizontally with a constant force of 60N at 30° with respect to ground. What acceleration would this force impart on the sled if the frictional force (sled-snow) is 10N?

F = m a

Fa – Ff = m a

a = (Fax cos 30°– Ff ) / m

a = (60N)(cos 30°) – 10N / 15kg

a = 2.8 m/s2


Impulse: It can be consider the cause

of the motion of an object Its effect is the change of

Momentum of a moving object It is defined as the product of

Force and time, or more conveniently

as the variation of momentum


Practice:

A 168 g hockey puck hits the boards of the ice rink perpendicularly at a velocity of 28 m/s and rebounds along same trajectory at 20 m/s. What impulses did the boards impart to the puck?

Δp = m Δv

Δp = 0.168 kg (-20 m/s – 28 m/s)

Δp = 0.168 kg (-48 m/s)

Δp = – 8.1 kg m/s

Δp = – 8.1 N s


Newton's second law & momentum:

You (50 kg) ride a bike (10 kg) at 10 m/s and speed up to 20 m/s, after 100 m. If frictional force (air & ground) is 30 N, what force did you apply during the acceleration?

F = m a

Fa – Ff = m a

Fa = m a + Ff V22 – V1

2 = 2ad

a = (V22 – V1

2 ) / 2d

a = 1.5 m/s2

Fa = (60 kg) (1.5 m/s2) + 30 N

Fa = 120 N


Newton’s Third Law:

“If object A exerts a force on object B (action), then object B exerts a force equal in magnitude, but opposite in direction, on object A (reaction)”


Newton’s Third Law:

Rockets and airplanes are examples of the application of Newton’s third law

Both forces act simultaneously (always)

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