Transcript 7.3 Forces in Two Dimensions

7.3 Forces in Two Dimensions
Chapter 7 Objectives

Add and subtract displacement vectors to describe changes in
position.

Calculate the x and y components of a displacement, velocity, and
force vector.

Write a velocity vector in polar and x-y coordinates.

Calculate the range of a projectile given the initial velocity vector.

Use force vectors to solve two-dimensional equilibrium problems with
up to three forces.

Calculate the acceleration on an inclined plane when given the angle
of incline.
Chapter 7 Vocabulary
 Cartesian
coordinates
 polar
coordinates
 component
 projectile
 cosine
 Pythagorean
theorem
 displacement
 inclined plane
 magnitude
 parabola
 range
 resolution
 resultant
 right triangle
 scalar
 scale
 sine
 tangent
 trajectory
 velocity vector
 x-component
 y-component
Inv 7.3 Forces in Two Dimensions
Investigation Key Question:
How do forces balance in two dimensions?
7.3 Forces in Two Dimensions
 Force is also represented by x-y components.
7.3 Force Vectors
 If an object is in
equilibrium, all of the
forces acting on it are
balanced and the net
force is zero.
 If the forces act in two
dimensions, then all of
the forces in the xdirection and y-direction
balance separately.
7.3 Equilibrium and Forces
 It is much more difficult for
a gymnast to hold his arms
out at a 45-degree angle.
 To see why, consider that
each arm must still support
350 newtons vertically to
balance the force of gravity.
7.3 Forces in Two Dimensions
 Use the y-component to find the total force in
the gymnast’s left arm.
7.3 Forces in Two Dimensions
 The force in the right arm must also be 495
newtons because it also has a vertical
component of 350 N.
7.3 Forces in Two Dimensions
 When the gymnast’s arms
are at an angle, only part
of the force from each
arm is vertical.
 The total force must be
larger because the
vertical component of
force in each arm must
still equal half his weight.
7.3 The inclined plane
 An inclined plane is a straight surface,
usually with a slope.
 Consider a block sliding
down a ramp.
 There are three forces
that act on the block:
 gravity (weight).
 friction
 the reaction force acting on
the block.
7.3 Forces on an inclined plane
 When discussing forces, the word “normal”
means “perpendicular to.”
 The normal force
acting on the block is
the reaction force from
the weight of the block
pressing against the
ramp.
7.3 Forces on an inclined plane
 The normal force
on the block is
equal and
opposite to the
component of the
block’s weight
perpendicular to
the ramp (Fy).
7.3 Forces on an inclined plane
 The force parallel
to the surface (Fx)
is given by
Fx = mg sinθ.
7.3 Forces on an inclined plane
 The magnitude of
the friction force
between two sliding
surfaces is roughly
proportional to the
force holding the
surfaces together:
Ff = -mg cosθ.
7.3 Motion on an inclined plane
 Newton’s second law can be used to calculate
the acceleration once you know the
components of all the forces on an incline.
 According to the second law:
Acceleration
(m/sec2)
a=F
m
Force (kg . m/sec2)
Mass (kg)
7.3 Motion on an inclined plane
 Since the block can only accelerate along the ramp, the
force that matters is the net force in the x direction,
parallel to the ramp.
 If we ignore friction, and substitute Newtons' 2nd Law,
the net force is:
Fx = m g sin θ
a =F
m
7.3 Motion on an inclined plane
 To account for friction, the horizontal component of
acceleration is reduced by combining equations:
Fx = mg sin θ -  mg cos θ
7.3 Motion on an inclined plane
 For a smooth surface, the coefficient of friction
(μ) is usually in the range 0.1 - 0.3.
 The resulting equation for acceleration is:
Calculating acceleration
A skier with a mass of 50 kg is on a hill
making an angle of 20 degrees. The
friction force is 30 N. What is the skier’s
acceleration?
1.
2.
3.
4.
You are asked to find the acceleration.
You know the mass, friction force, and angle.
Use relationships: a = F ÷ m and Fx = mg sinθ.
Calculate the x component of the skier’s weight:
 Fx = (50 kg)(9.8 m/s2) × (sin 20o) = 167.6 N
 Calculate the force: F = 167.6 N – 30 N = 137.6 N
 Calculate the acceleration: a = 137.6 N ÷ 50 kg = 2.75 m/s2
7.3 The vector form of Newton’s 2nd law
 An object moving in three dimensions can be
accelerated in the x, y, and z directions.
 The acceleration vector can be written in a similar
way to the velocity vector: a = (ax, ay, az) m/s2.
7.3 The vector form of Newton’s 2nd law
 If you know the forces acting on an object, you
can predict its motion in three dimensions.
 The process of calculating three-dimensional
motion from forces and accelerations is called
dynamics.
 Computers that control space missions
determine when and for how long to run the
rocket engines by finding the magnitude and
direction of the required acceleration.
Calculating acceleration
A 100-kg satellite has many small rocket engines
pointed in different directions that allow it to
maneuver in three dimensions. If the engines make
the following forces, what is the acceleration of the
satellite?
F1 = (0, 0, 50) N F2 = (25, 0, –50) N F3 = (25, 0, 0) N
1.
2.
You are asked to find the acceleration of the satellite.
You know the mass, forces, and assume no friction in space.
3.
4.
Use relationships: F = net force and a = F ÷ m
Calculate the net force by adding components. F = (50, 0, 0) N
5.
Calculate acceleration: ay = az = 0 ax = 50 N ÷ 100 kg = 0.5 m/s2
 a = (0.5, 0, 0) m/s2
Robot Navigation
 A Global Positioning System
(GPS) receiver determines
position to within a few meters
anywhere on Earth’s surface.
 The receiver works by comparing
signals from three different GPS
satellites.
 About twenty-four satellites orbit
Earth and transmit radio signals
as part of this positioning or
navigation system.