Chapter 7-1

Forces and Motion in

Two-Dimensions

Equilibrium

An object is in equilibrium when all the forces on it add up to zero

EquilibrantThe Equilibrant is the

• single additional force that, when exerted on an object, will produce equilibrium

• same magnitude as the resultant but opposite in direction

“Statics” Force Problems

With “Statics”

By definition all forces are balanced

Net Force = zero

Some Classic Problems:

Ah, the classic sign problem! Classic! Heed it well….

Practice Problems, continued

Motion along an Inclined Plane

• Please Please Please make sure you understand everything up to this point

• Now we will be applying what we know about two dimensional vectors

• You must be very clear on the use of vector components

• The coordinate system is about toTILT

Weight ComponentsThis is a very important skill:

You will use this in almost all Chap 7 problems—make sure you are clear on how to do this!

Let’s go through P1 of

The Chapter 7 Worksheet:

Analyzing forces in 2-dimensions

Let’s analyze how the weight vector can be broken down into two components: a component parallel to the plane (which tends to pull the object down the plane) and a component perpendicular to the plane (the magnitude of this component equals the Normal Force)

(#32)

A 10 kg object is sitting on an inclined plane which makes a 25 degree angle to the horizontal.a. What is the weight of the object (down/up)?

b. What angle does the weight vector make with a line which is perpendicular to the plane?c. What is the component of the weight force which is perpendicular to the plane (down/up)?

d. What is the Normal force that the plane exerts on the object (down/up)?e. What is the component of the weight force which is parallel to the plane (left/right)?

f. How much force is acting on the object, tending to pull it down the plane (left/right)?

Important to note that the thetas shown are equalbecause of similar triangles—we will use this informa-tion a lot.

Multiple forces on an angle

It’s starting to get interesting now!

These are the variables we will be using (look on the back of your yellow packet)

Type 1: Object not moving or sliding down the incline

Type 1

In this class of problems, The object is Not moving orMoving at constant velocityThe net force = zeroThe acceleration = 0Friction force = Parallel FgNormal force = Perpendicular Fg

(#37) (MP #3, #7)

If an object weighs 10 N and is placed on an inclined plane tilted at 30 degrees to the horizontal (the object is not moving):

The angle between the weight force vector and a line perpendicular to the plane is _____a. What is the component of the weight force parallel to the plane(left, right)?

b. What is the component of the weight force perpendicular to the plane (up/down)?

c. What is the Normal force (down/up)?d. What is the Friction Force (right/left)?e. μ =

f. If the object is now moving at constant velocity down the plane, what of the above forces has changed and why?

Type 2In this class of problem:There is acceleration DOWN the inclined plane.Net Force = maParallel Fg is greater than Friction ForceNormal force = Perpendicular Fg(#38) (P153 Ex Prob)

An object weighing 45 N is accelerating under its own weight at 2 m/s2 down an inclined plane set at 15 degrees to the horizontal.

a. What is the component of the weight force parallel to the plane (right/left)?

b. What is the component of the weight force perpendicular to the plane (up/down)?

c. The mass of the object is

d. What is the net Force (right/left)?e. What is the Normal Force (up/down)?f. What is the total Force down the plane?

g. What is the Friction Force (right/left)? h. μ =

i. What is the total force up the plane?

Type 3

In this class of problem, there is a force up the plane (Fh in the drawing) which just balances the parallel component of the weight force and the friction force.

(#36) (MP #4, #8)

An object weighing 5 N is being pulled up an inclined plane set at 10◦ to the horizontal. The object is not moving. The coefficient of friction is 0.1.a. What is the Normal Force (up/down)?

b. What is the Friction force (right/left)?

c. What is the parallel component of weight force (left/right)?

d. What is the force pulling or pushing up the plane on the object?

e What is the net force?f…What is the acceleration?

Type 4

In this class of problem, there is a force up the plane which is greater than the parallel component of the weight force plus the friction force. There is a net force up the plane and an acceleration up the plane

An object weighing 5 N is being pulled up an inclined plane set at 10 degrees to the horizontal with a force of 10N. The coefficient of friction is 0.1.a. What is the Normal Force (up/down)?b. What is the Friction force (right/left)?c. What is the parallel component of weight force (right/left)?d. What is the force pulling up the plane on the object?e What is the net force (right/left)?

f..What is the mass?

g..What is the acceleration (right/left)?

Type 5Another class of 2-dimensional problem:

The object is on the horizontal but it is being pulled by a rope on an angle.

Normal Force is less than FgBecause the y-component of the pulling Force, in effect, lightens the object

If the object is not moving or is moving at constant velocity, then Friction Force = x-component of pulling Force (#35) (MP #5, #6)

A 4 kg object is being pulled at constant velocity across a horizontal surface with a force of 10 N by a rope which is at a 37 degree angle to the horizontal.a. What is the object’s weight (up/down)?

b. What is the vertical (y-) component of the pulling force (up/down)?

c. What is the normal force (up/down)?

d. What is the Friction Force (right/left)?

e. What is the coefficient of friction?

f…What is the Net Force?g. What is the acceleration?

Type 6A variation on the previous scenario:

If the object is accelerating, thenFriction Force is less than the x-component of the pulling force and the object accelerates in the direction of motion, in the direction of the Net Force.

A 4 kg object is being pulled across a horizontal surface with a force of 10 N by a rope which is at a 37 degree angle to the horizontal. The coefficient of friction is 0.1a. What is the object’s weight (up/down)?b. What is the vertical (y-) component of the pulling force (up/down)?c. What is the normal force (up/down)?d. What is the Friction Force (right/left)?

e. What is the Net Force (right/left)?

f. What is the acceleration (right/left)?

Let’s consider some Classic Problems:

The skier—a classic problem—anything (a box, a skier) accelerating down an

incline with friction (Type 3)

Here’s the vector representation

Here’s the Picture:The skier is accelerating downThe slope

Strategy

FREAK OUT!!!!

NO, NO, NO----

Resolve all forces •Parallel to the plane (all X forces)•Perpendicular to the plane (all Y forces)

List of variables:m = 62 kgθ = 37◦μ = 0.15v0 = 0.0 m/st = 5.0 s

Since you need the Normal Force, FN before you Can calc the Friction Force, Ff

Resolve the weight vector, Fg first[The Normal Force, FN is the Y-component of FgAnother way of saying that is that FN is the component of the weight which is perpendicular to the plane]

Now calc the Friction Force, Ff, using FN that you just calculated

List of variables:m = 62 kgθ = 37◦μ = 0.15v0 = 0.0 m/st = 5.0 s

We are no longer interested in the Y-components of anything.All of our interest is in resolving forces on the X-axis.

We just calculated the only force that points up the X-axis:The Friction Force.

Two things point down the X-axis:•The X-component of the weight, Fg in the X-direction•The ma which is the Net Force accelerating the skiier downhill

Find the Net Force (=ma) by taking the difference between Ff and Fgx:

Find the acceleration by dividing the Net Force by the mass of the skier

List of variables:m = 62 kgθ = 37◦μ = 0.15v0 = 0.0 m/st = 5.0 s

Calculate the velocity using v = v0 +`at

List of variables:m = 62 kgθ = 37◦μ = 0.15v0 = 0.0 m/st = 5.0 s

Other Classic Problems:

Tension (Force) applied horizontally to an object whose weight points straight down:

Other Classic Problems:

A rope attached on an angle to an object that is being pulled horizontally (Type 1):

Ans: 51 ◦,140N,0.19

Other Classic Problems:

(Type 4) Object pulled up a plane:MP,p84: A crate weighing 1800 N is pulled along a floor by a force of

990 N. The floor is inclined 15◦ to the horizontal and th crate moves with constant velocity. What is the coefficient of sliding friction between the crate and the floor? Ans. 0.31

End Chapter 7-1