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Dynamics in 2D

Net Forces in Two Dimensions

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Essential Question

How do we use vectors to model forces that do not balance?

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Learning Objective Preview

  • Be able to correctly chose the components to break a force vector into.
  • Be able to write an equation of balanced forces for forces and components perpendicular to acceleration.
  • Be able to write an equation of unbalanced forces for forces and components parallel to acceleration.
  • Be able to solve these equations for some unknown.

How do we use vectors to model forces that do not balance?

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Openstax Textbook

This lesson aligns with page 143-176 of Openstax College Physics

https://openstax.org/details/books/college-physics

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Table of Contents

  • Unbalanced Forces in Two Dimensions (6)
  • Pulling a Sled Without Friction Example (10)
  • Pulling a Sled With Friction Example (18)
  • Hanging Object In A Car Example (29)
  • Incline Without Friction Example (42)
  • Incline With Friction Example (53)

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UNBALANCED FORCES IN TWO DIMENSIONS

Either left forces don’t equal right forces, OR up forces don’t equal down forces.

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What If Forces Don’t Balance?

  • Remember: If forces balance, then this condition is called “equilibrium”, and the object cannot change velocity (speed or direction of motion).
    • This is Newton’s First Law
  • But if there are forces that are not balanced, the object has an acceleration. Likewise, if the object accelerates, the forces acting on it do not balance.
    • This is Newton’s Second Law

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What If Forces Don’t Balance?

How do we handle forces that don’t balance in two dimensions?

  • We draw a Free-Body Diagram showing the forces acting on the object.
  • In a close but separate place, we draw a vector (arrow) representing the object’s acceleration.
  • Any force that is neither perpendicular nor parallel to acceleration must be made into components.
    • One component needs to be parallel to the acceleration vector.
    • One component needs to be perpendicular to the acceleration vector.
  • We make a “balanced forces equation” for all forces and components perpendicular to acceleration. Forces perpendicular to acceleration balance each other out.
  • We make a “net force equation” for all forces and components parallel to acceleration. Forces parallel to acceleration add to ma, but forces opposed to acceleration count as negatives.

EVERYTHING YOU NEED TO KNOW ABOUT FORCES IS SUMMARIZED ON THIS SLIDE.

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Examples

We will consider these unbalanced force examples:

  • Pulling a Sled without friction
  • Pulling a Sled with friction
  • Hanging Object in a Car
  • Inclined Plane without friction
  • Inclined Plane with friction

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PULLING A SLED WITHOUT FRICTION EXAMPLE

The tension force makes an angle.

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Pulling a Sled Without Friction

A boy uses a rope to pull a sled with his sister in it over a frictionless surface. The mass of the sled-and-sister is 10 kg. The student causes the tension to be 100 N at an angle of 37o from the horizontal. What is the acceleration of the sled?

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Pulling a Sled Without Friction

A boy uses a rope to pull a sled with his sister in it over a frictionless surface. The mass of the sled-and-sister is 10 kg. The student causes the tension to be 100 N at an angle of 37o from the horizontal. What is the acceleration of the sled?

First, draw a free-body diagram showing the forces acting on the sled-and-sister.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

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Pulling a Sled Without Friction

Wait…why is normal force less than weight force? I thought that normal force always equals mg.

The normal force only needs to prevent the sled from falling through the ground. With no other forces, the normal force would have to cancel out all of the weight force.

But the normal force has some help! There is a component of the yellow tension force helping lift up on the sled, so normal force doesn’t have to apply as much force to balance weight.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

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Pulling a Sled Without Friction

A boy uses a rope to pull a sled with his sister in it over a frictionless surface. The mass of the sled-and-sister is 10 kg. The student causes the tension to be 100 N at an angle of 37o from the horizontal. What is the acceleration of the sled?

Now let’s draw an acceleration vector near the FBD. We can guess that the acceleration will probably be to the right.

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mg

FT

FN

a

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Pulling a Sled Without Friction

Now any force that is neither perpendicular to acceleration nor parallel to acceleration needs to be made into components:

  • Weight is perpendicular to acceleration – A-OK.
  • Normal force is perpendicular to acceleration – A-OK.
  • Tension is neither parallel nor perpendicular to acceleration.

This means we must make tension into components, one parallel and one perpendicular to acceleration.

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mg

FT

FN

a

θ

FTcosθ

FTsinθ

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Pulling a Sled Without Friction

Now we make two equations:

  • Balanced forces equation: Forces perpendicular to acceleration balance. Normal and FTsinθ are both up, and mg is down.
    • So we get the equation FN + FTsinθ = mg
  • Net force equation: Forces parallel to acceleration make ma. Forces against acceleration count as negative. The only force parallel to acceleration is FTcosθ.
    • So we get the equation FTcosθ = ma.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

a

θ

FTcosθ

FTsinθ

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Pulling a Sled Without Friction

A boy uses a rope to pull a sled with his sister in it over a frictionless surface. The mass of the sled-and-sister is 10 kg. The student causes the tension to be 100 N at an angle of 37o from the horizontal. What is the acceleration of the sled?

We can use the net force equation to solve for acceleration:

FTcosθ = ma

(100)(0.8) = (10)a, so a = 8 m/s2

How do we use vectors to model forces that do not balance?

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mg

FT

FN

a

θ

FTcosθ

FTsinθ

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PULLING A SLED WITH FRICTION EXAMPLE

Friction depends on normal force, but normal force is NOT equal to mg.

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

First, draw a free-body diagram showing the forces acting on the sled. The diagram is almost the same; there is just one new force.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

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Pulling a Sled (Friction)

Because the normal force is less than mg, the friction force is not μmg. The kinetic friction force is still μFN, but normal force is not the same as mg this time.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

Now let’s draw an acceleration vector near the FBD. Again, it is probably to the right.

mg

FT

FN

Ff

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a

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Pulling a Sled (Friction)

Now any force that is neither perpendicular to acceleration nor parallel to acceleration needs to be made into components:

  • Weight is perpendicular to acceleration – A-OK.
  • Normal force is perpendicular to acceleration – A-OK.
  • Friction force is parallel to acceleration – A-OK.
  • Tension is neither parallel nor perpendicular to acceleration.

This means we must make tension into components, one parallel and one perpendicular to acceleration.

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θ

FTcosθ

FTsinθ

mg

FT

FN

Ff

a

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Pulling a Sled (Friction)

Now we make two equations:

  • Balanced forces equation: Forces perpendicular to acceleration balance. Normal and FTsinθ are both up, and mg is down.
    • So we get the equation FN + FTsinθ = mg
  • Net force equation: Forces parallel to acceleration make ma. Forces against acceleration count as negative. The force that “helps” acceleration is FTcosθ, and the force “against” acceleration is friction.
    • So we get the equation FTcosθFf = ma.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

a

θ

FTcosθ

FTsinθ

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

Before we can solve for acceleration, we need friction (μFN). To get friction, we need the normal force. So first, we will solve the balanced-forces equation to get the normal force.

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

a

θ

FTcosθ

FTsinθ

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

FN + FTsinθ = mg

FN + (100)(0.6) = (10)(10)

FN + 60 = 100

FN = 40 N

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

a

θ

FTcosθ

FTsinθ

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

Now find the strength of the friction force:

Ff = μFN

Ff = (0.5)(40)

Ff = 20 N

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

a

θ

FTcosθ

FTsinθ

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Pulling a Sled (Friction)

Again the boy pulls with 100 N angled at 37o on a sled (total mass 10 kg), but now there is kinetic friction with coefficient 0.5 between the sled and the ground. Find the acceleration.

Now find the acceleration using the net force equation:

FTcosθFf = ma

(100)(0.8)(20) = (10)a

(80)(20) = (10)a

a = 6 m/s2

How do we use vectors to model forces that do not balance?

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mg

FT

FN

Ff

a

θ

FTcosθ

FTsinθ

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HANGING OBJECT IN A CAR EXAMPLE

It swings back when the car accelerates, but not when the car has a constant velocity.

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Hanging Object in a Car

Some people have objects hanging from the inside mirrors in their cars. The picture below is a trinket that someone has hanging from their car mirror—it is supposed to be Thor’s Hammer.

An object like this can be used to determine the acceleration of the car, because hanging objects will make angles when the car accelerates.

The animation on the next slide shows what this looks like.

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Hanging Object in a Car

The top of the animation shows the object inside the car. The bottom of the animation shows the car accelerating along a road.

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Hanging Object in a Car

When the car accelerates forward, the object “swings backward”, and when the car has backward acceleration (slowing down), the object “swings forward”.

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Hanging Object in a Car

The backward acceleration at the end has greater magnitude than the initial forward acceleration, so the object’s angle is greater for the greater acceleration.

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Hanging Object in a Car

The angle can be used to measure the acceleration. The diagram below shows the object on the string when the car is in the middle of its initial acceleration. Note that the angle appears to be about 26.5o.

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Hanging Object in a Car

On the left is a diagram showing the object on its string, along with the angle θ.

On the right, we have a free-body diagram which also shows the angle θ attached to the tension force.

Note that an acceleration vector is also drawn near the FBD. The car (and object’s) acceleration was to the right.

How do we use vectors to model forces that do not balance?

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θ

θ

FT

mg

a

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Hanging Object in a Car

We now make components out of the tension force because the tension force is neither parallel nor perpendicular to the acceleration.

Weight is perpendicular to acceleration so it requires no components.

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θ

θ

FT

mg

a

FTsinθ

FTcosθ

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Hanging Object in a Car

The forces perpendicular to acceleration balance each other:

FTcosθ = mg (because FTsinθ is up and mg is down)

The forces parallel to acceleration add to make ma.

FTsinθ = ma (there is only one force, and it helps acceleration)

How do we use vectors to model forces that do not balance?

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θ

θ

FT

mg

a

FTsinθ

FTcosθ

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Hanging Object in a Car

We know the angle θ and gravity, but we don’t know the value FT or the object’s mass m. These are preventing us from solving for acceleration.

However, you can always divide one equation by the other equation in order to eliminate unknown quantities that multiply.

How do we use vectors to model forces that do not balance?

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θ

θ

FT

mg

a

FTsinθ

FTcosθ

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Hanging Object in a Car

Here’s what it looks like here:

FTsinθ = ma

FTcosθ = mg

Then, the FT on top and bottom on the left cancels, and the m on top and bottom on the right cancels.

How do we use vectors to model forces that do not balance?

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θ

θ

FT

mg

a

FTsinθ

FTcosθ

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Hanging Object in a Car

Also, sine over cosine becomes tangent, so we get:

tanθ = a/g

We can multiply both sides by g, and we get a = g tanθ

For our angle of 26.5o, a = g tanθ = (10)tan(26.5o) = 5 m/s2.

How do we use vectors to model forces that do not balance?

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θ

θ

FT

mg

a

FTsinθ

FTcosθ

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Hanging Object in a Car

When the car slows down, the object seems to settle on 45o.

Can you figure out the magnitude of the car’s acceleration in this case? Hint: it will be a familiar acceleration.

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INCLINE WITHOUT FRICTION EXAMPLE

You’ll be surprised which force we break into components.

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Incline Without Friction

Suppose there is a straight ramp with an angle θ to the horizontal. A cart with frictionless bearings is placed on the ramp and released from rest. What acceleration will the cart have?

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θ

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Incline Without Friction

  • First, we draw a free-body diagram.
  • There are only two forces
    • Weight
    • Normal Force
  • Let’s also draw an acceleration vector.
    • We can be sure acceleration points down the incline.

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θ

FN

mg

a

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Incline Without Friction

  • We next need to make components, which is VERY COUNTERINTUITIVE this time.
  • We are used to making components that are horizontal and vertical for vectors that are diagonal, but here…
  • Normal force is perpendicular to acceleration – A-OK
  • Weight force is neither parallel nor perpendicular to acceleration – Components!

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θ

FN

mg

a

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Incline Without Friction

  • The components of weight must be parallel to the acceleration and perpendicular to the acceleration.
    • That means one of the components is parallel to the plane, and the other is parallel to the plane.
  • Remember that components make a right triangle with the vector itself
    • The components are the legs
    • The vector is the hypotenuse.

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θ

FN

mg

a

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Incline Without Friction

  • Using your eyes, you can see that the angle θ that is at the bottom right of the incline is also the same angle that is at the top of the components triangle.
  • Now we can make component equations
    • The component opposite the angle is “vector sine angle”
    • The component adjacent to the angle is “vector cosine angle”.

How do we use vectors to model forces that do not balance?

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

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Incline Without Friction

Forces perpendicular to acceleration balance each other.

  • In this case, it is normal force and the perpendicular component of weight that balance.
  • Therefore FN = mg cosθ
  • Note that normal force is not the same as mg in every case. Normal force only equals mg if:
    • The surface is flat,
    • There is no vertical motion, AND
    • There are no other up or down forces.

How do we use vectors to model forces that do not balance?

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

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Incline Without Friction

Forces parallel to acceleration add to ma.

  • In this case, the only force parallel to acceleration is the parallel component of weight, which could also be called the “downhill force” mg sinθ.
  • Therefore mg sinθ = ma
  • The mass cancels in all terms, so we get a = g sinθ.

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

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Incline Without Friction

Important take-aways:

  • Force components are not always horizontal and vertical. But force components are always parallel and perpendicular to acceleration.
  • The acceleration of an object rolling down an incline only depends on gravity and the incline angle: a = g sinθ
    • The steeper the incline, the greater the θ. This makes the sin θ greater, so a steeper incline results in greater acceleration.
  • Acceleration is constant if the incline is flat (has the same angle or slope) at all points.
    • So on flat ramps, we can use our constant acceleration equations: v = at + v0, x = ½ at2 + v0t + x0, and v2 = 2aΔx + v2.
    • Curved ramps, on the other hand, have an acceleration that changes and we can’t use these equations.

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Incline Without Friction

Here’s an animation showing the fact that steeper ramps give objects greater acceleration. Do you remember this animation from your lesson on Acceleration?

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Incline Without Friction

Here’s an animation showing the fact that a ramp, as it becomes less steep, gives an object a decreasing acceleration.

Acceleration is the slope of the velocity graph. See how the slope of the graph decreases as the ramp becomes less steep.

Also note how it is possible for acceleration to become less even as velocity is still becoming greater.

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INCLINE WITH FRICTION EXAMPLE

One extra force doesn’t make the problem that much more complicated.

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Incline WITH Friction

Suppose now our incline has friction, and the coefficient of kinetic friction between the incline and the block is μ.

The free-body diagram is exactly the same (including the components of the weight force), except that there is a new friction force that needs to be drawn.

Note that friction is not always horizontal, but friction is always parallel to the surface that exerts the friction force.

How do we use vectors to model forces that do not balance?

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

Ff

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Incline WITH Friction

Note that there is no reason to make components of friction, because friction is parallel with acceleration.

Note that the balanced forces equation is still the same, because we have the same perpendicular forces: Normal, and the perpendicular component of weight.

So the balanced forces equation is still FN = mg cosθ.

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

Ff

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Incline WITH Friction

But this time we have two forces parallel to acceleration: the parallel component of weight (again) and friction.

Friction opposes the object’s acceleration, so friction goes into the Fnet = ma equation as a negative:

mg sinθFf = ma

How do we use vectors to model forces that do not balance?

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

Ff

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Incline WITH Friction

mg sinθFf = ma

For kinetic friction, the friction is still FF = μFN.

But we already found that normal force is equal to the perpendicular component of weight: FN = mg cosθ.

So plugging this in, we get friction is FF = μmg cosθ.

So now our parallel force equation (at the top) becomes mg sinθμmg cosθ = ma.

Mass is in every term, so cancel: a = g sinθ – μg cosθ

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θ

FN

mg

a

θ

mg sinθ

mg cosθ

Ff

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Incline WITH Friction

a = g sinθ – μg cosθ

Let’s understand what is happening in this equation. As the ramp gets steeper, the angle θ increases.

  • As θ increases, sinθ also increases (look at the graph to the right), so that increases the “downhill force” that helps the block go.
  • Also, as θ increases, cosθ decreases. This decreases the friction holding the block back.

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θ

0

30

60

90

0.5

1.0

sinθ

cosθ

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Incline WITH Friction

Now here’s the curved ramp again, this time the block has friction on the ramp. (here, we made μ = 1.)

The weight, normal, and friction forces are shown as they change during the block’s motion.

Notice that the components of weight are also shown as outline arrows.

We also see a velocity meter and a velocity vs. time graph.

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Incline WITH Friction

Please make sure you understand what those two orange outline arrows represent before going on to the next slide.

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Perpendicular component

of weight cancels out

normal force

Parallel component

of weight provides

the downhill force

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Incline WITH Friction

Before the white arrow, the downhill force is greater than the friction force. Therefore, the net force is downhill and the block speeds up.

After the white arrow, the friction force is greater than the downhill force, so the net force is backward and the block slows down.

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Incline WITH Friction

The friction force becomes greater as the ramp becomes less steep, because less steep ramps exert greater normal force.

The downhill force becomes less as the ramp becomes less steep, because the weight force and the ramp itself become “less parallel”.

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Dynamics in 2D

THE END

How do we use vectors to model forces that do not balance?

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Professional Development for Teachers | Tutoring for Students

www.phine-physics.com