Ch. 7 Forces and Motion in Two

Dimensions

Milbank High School

Sec. 7.1Forces 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. 7.2Projectile 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 displacementy = 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 Heighty = vyot - ½gt2

• RangeR = vxot

Projectiles launched at an Angle

• The Flight of a Ball

• Example Problem Pg. 159

Sec. 7.3Circular 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 providesthe centripetal force required for circular motion.

Net Force

• Example Problem Pg. 165