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: FUnits: N (newton)


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 (10,10); Fc (-20,20)_Resultant Force: Fr (-10,0)No, the system is not in equilibrium for the resultant vector (component method) does not have (0.0) coordinatesTry with the polygon method!


Equilibrant force:

Is the force that balance 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 forceFa = 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 aFa – Ff = m aa = (Fa – Ff ) / m a = (60N – 10N) / 15kga = 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 aFa – Ff = m aa = (Fax cos 30°– Ff ) / m a = (60N)(cos 30°) – 10N / 15kga = 2.8 m/s2


Impulse: It can be consider the causeof 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 ΔvΔp = 0.168 kg (-20 m/s – 28 m/s)Δp = 0.168 kg (-48 m/s)Δp = – 8.1 kg m/sΔp = – 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 V2

2 – V12 = 2ad

Fa – Ff = m a a = (V22 – V1

2 ) / 2dFa = m a + Ff 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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