What must a motion have to be two-dimensional?
Two components(typically x and y).
What is Newton's First Law of Motion?
The Law of Inertia: An object in motion stays in motion, and an object at rest stays at rest unless acted upon by an outside force.
A ball sitting on a table.
Up: Normal Force
Down: Force of gravity
Up and Down Forces cancel.
Force of a falling object due to Earth
Force of Gravity
A textbook is sitting on a table and its mass is 5 kg. What is the normal force applied by the table?
FN = -mg
= -(5 kg)(-9.8 m/s2)
= 49 N
How do you graphically add AND subtract vectors?
Addition: After drawing the first vector, begin drawing the second vector from the arrow tip of the first.
Subtraction: After drawing the first vector, begin drawing the second vector from the tail of the first vector, AND flipping the second vector.
What is Newton's Second Law of Motion?
F = ma
A book sitting on a table, under another book.
Up: Normal Force of the table
Down: Force of gravity, Force of above book on the below book
Up force = Sum of Down Forces
Force applied by a tight string or rope
Tension Force
A block with a mass of 10 kg is pulled against the ground with 20 N. The block is moving at a constant speed. What is the counteracting friction force?
FNet = 0.
FApplied + Ff = 0
Ff = -FApplied
Ff = -(20 N)
= -20 N
What is the magnitude of a vector?
r2 = rx2 + ry2
What is Newton's Third Law of Motion?
Every action has an equal and opposite reaction.
A ball in free fall that has not reached terminal velocity(still accelerating down).
Up: Air resistance
Down: Gravity
Down force overcomes as acceleration is down
Force of two objects rubbing together
Friction Force
I am pulling a wagon with a string where it's moving at a constant speed. If the force of friction acting on the wagon is 35 N, and I am pulling the string at 35o from parallel, at what force am I pulling the string?
FT,x = -Ff = -(-35 N) = +35 N
FT,x = FT*cos(theta) -> FT = FT,x/cos(theta)
FT = (35 N)/cos(35o)
= 42.7 N
If a jogger travels 100 meters 20o North of due East, and he needs to at least travel 75 meters North, will he accomplish his requirement?
No, the jogger will not accomplish his requirement.
y = r*sin(theta)
y = 100*sin(20o)
y = 34.20 m
34.20 m < 75 m
If an object with a mass of 10 kg was thrown off a building, how much force would it have when it hits the ground?
F = ma
F = (10 kg)(-9.8 m/s2)
F = -98 N or 98 N down into the ground
A car on the road moving at a constant speed.
Up: Normal Force of the road
Down: Force of gravity
Right: Force applied to accelerate the car forward
Left: Friction
Up and down forces cancel and Right and Left forces cancel.
General force given to the system
Applied Force
If a ball with a mass of 5 kg is accelerating across the floor at 2 m/s2, and the floor causes a friction force of 20 N, what is the net force on the ball? What is the applied force on the ball?
FNet = ma = (5 kg)(2 m/s2) = 10 N
FNet = FApplied + Ff -> FApplied = FNet - Ff
= 10 N - (-20 N)
= 30 N
A bat flies 30o South of due East for 50 meters, then flies 45o North of due West for 65 meters. What is the bat's displacement vector?
r1,y = 50*sin(30o) = 25 m, r1,x = 50*cos(30o) = 43.3 m
r1 = (43.3, -25)
r2,y = 65*sin(45o) = 45.96 m, r2,x = 65*cos(45o) = 45.96
r2 = (-45.96, 45.96)
rtot = r1 + r2 = (43.3, -25) + (-45.96, 45.96) = (-2.7, 21.0)
If I am standing on ice, and I apply a 100 N force on the wall, and I have a mass of 92 kg, how much will my initial acceleration be after pushing off the wall? Will I continue at this acceleration?
F = ma -> a = F/m
a = (-100 N)/(92 kg) = -1.09 m/s2
After pushing off the wall, I will not continue to accelerate because I am no longer applying a force, and therefore no force is being applied to be.
A shopping cart accelerating forward.
Up: Normal Force of the ground
Down: Force of gravity on the cart
Left: Force of Friction
Right: Force applied by the pusher
Up and Down forces cancel. Right force overcomes left force.
Force applied by two objects touching, acting perpendicular to the surface.
Normal Force
If an object with a mass of 4 kg initially starts at rest, and I apply a constant 5 N force on the object for 10 seconds, how far will the object travel?
F = ma -> a = F/m
xf = xi + vit + 1/2(a)(t2)
xf = 0 + (0)(10 s) + 1/2(5 N/4 kg)(102)
xf = 62.5 m
The object travels 62.5 meters.