Transcript Ch. 7 Forces and Motion in Two Dimensions

Ch. 7
Forces and Motion in Two
Dimensions
Milbank High School
Sec. 7.1
Forces in Two Dimensions
• Objectives
– Determine the force that produces equilibrium
when three forces act on an object
– Analyze the motion of an object on an inclined
plane with and without friction
What is meant by two dimensions?
• Consider a golf ball being hit out of a sand
trap
– It has a horizontal force AND a vertical force
– We can solve for many different things using a
combination of forces and vectors
• Height of the ball
• Time in the air
• Velocity when it hits the ground
Equilibrant
• A force exerted on an object to produce
equilibrium
• Same magnitude as the resultant force but
opposite in direction
Solving problems in two
dimensions
• Draw it out!
• Rearrange vectors to form a triangle if
possible
• Solve for the resultant vector
– Opposite in direction
– Example Problem Pg. 151
Sec. 7.2
Projectile Motion
• Objectives
– Recognize that the vertical and horizontal
motions of a projectile are independent
– Relate the height, time in the air, and the
initial velocity of a projectile using its vertical
motion, then determine the range.
– Explain how the shape of the trajectory of a
moving object depends upon the frame of
reference from which it is observed.
Projectiles have independent
motions!
• Projectiles have two velocities, one in the
“x” direction, and one in the “y” direction
• x is always constant
• y will be changing due to the acceleration
due to gravity
Displacement
• y displacement
y = yo - 1/2gt2
• x displacement
x = vxot
v = 25m/s 
Velocity of projectiles launched
horizontally
• vx = initial velocity
• vy = (-g)t
• v = resultant velocity vector
• Example Pg. 157
Effects of air resistance
• We ignore the effects of air resistance for
these problems
• Sometimes it would make a large
difference, other times it wouldn’t
• Many projectiles modified so that they
reduce air resistance
Projectiles launched at an Angle
•
•
•
•
Usually given angle of launch and velocity
What do we have to find?
Maximum height
Range
– Horizontal distance
• Flight time
– hang time
Projectiles Launched at an Angle
• Two initial velocity components
• vxo
• vyo
How do we find these?
vx = vo(cosθ)
vy = vo(sinθ)
Projectiles Launched at an Angle
• tup = vyo/g
• ttotal = 2(tup)
• Peak Height
y = vyot - ½gt2
• Range
R = vxot
Projectiles launched at an Angle
• The Flight of a Ball
• Example Problem Pg. 159
Sec. 7.3
Circular Motion
• Objectives
– Explain the acceleration of an object moving
in a circle at constant speed
– Describe how centripetal acceleration
depends upon the object’s speed and the
radius of the circle
– Recognize the direction of the force that
causes centripetal acceleration
– Explain how the rate of circular motion is
changed by exerting torque on it.
Uniform Circular Motion
• Movement of an object at constant speed
around a circle with a fixed radius
• Merry-go-round
Circumference = 2*pi*Radius
Vectors
Acceleration
• Which direction?
• Always towards
the center
Centripetal Force
• “Center seeking”
• Net force towards the center that causes
the object to try to seek the center
• What force is pulling it in?
• As a bucket of water is
tied to a string and spun
in a circle, the force of tension
acting upon the bucket provides
the centripetal force required for
circular motion.
Net Force
• Example Problem Pg. 165