Forces at an Angle and

Forces on Inclined Planes

Equilibrium• Recall that when the net

force on an object is zero, the object is in equilibrium.

• If an object is in equilibrium, its velocity is constant. This may be a velocity of zero or any constant velocity.

• (What would the V-T graph look like?)

Net force and Acceleration

Newton’s second Law: Fnet =ma

To solve for a net force, add up all the forces in that direction (This is where the free body diagrams are important!)

Direction of motion is positive, opposite the direction of force is negative.

Friction is always negative

Net force and Acceleration

Newton’s second Law: Fnet =ma

Thus a system with a mass of 18kg with a forward force of 6N, a frictional force of 5.8N, Fg

of176N, and Fn of 176N would have a net force of

0.2N in the direction of motion

Net force and AccelerationNewton’s second Law: Fnet =ma

Thus a system with a mass of 18kg with a forward force of 6N, a frictional force of 5.8N, Fg

of176N, and Fn of 176N would have a net force of0.2N in the direction of motion

Its acceleration would be: Fnet/m =a

0.2N/18kg=

0.01m/s2

Net force and AccelerationNewton’s second Law: Fnet =ma

Thus a system with a mass of 18kg with a forward force of 6N, a frictional force of 5.8N, Fg of176N, and Fn of 176N would have a net force of 0.2N in the direction of motion.

Its acceleration would be: 0.01m/s2

And the coefficient of friction would be found using Ff=µFn

Ff/Fn =µ = 5.8N/176N

µ =0.03

Forces at an angle

• When a force is applied parallel to motion, the full magnitude of the force goes into Fnet=ma.

• If the force is applied at an angle, only the portion that is parallel to motion may be used.

• The portion that is perpendicular must be considered in the normal force. It will either increase it or decrease it.

Example 1

• A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction force)

Example 1:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction

force)

• Step 1 Draw all the forces

Example 1:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction

force)

• Step 2: Choose direction of motion (X)

← X

Example 1:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction

force)

• Step 3: Notice that there are 2 forces acting in the “up” direction. Cannot find Fn(yet)

← X

Example 1:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction

force)

• Step 4: Since it is not moving in the Y direction, Fn + Fapp↑ must = Fg

← X

Example 1:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction

force)

• Step 4: We are ignoring friction,

so on to step 5

← X

Example 1:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its acceleration? (Ignore friction

force)

• Step 5: Fx(net)=ma

Fx(net)/m=a

23N/15kg

=1.53m/s2

← X

Example 2:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its

acceleration? (Include friction force)μk =0.14

• Everything is the same until step 4. Now we need to know Ff

Ff= μkFn

=0.24 *127.7N

=17.9N

← XFn

Example 2:A 15kg object is pulled with a force of 30N at an angle 40˚ above horizontal. What is its

acceleration? (Include friction force)μk =0.14

• Include Ff in Fnet

Fnet= Fapp+Ff

=23N-17.9N

=5.1N

← X

Fnet/m=a

5.1N/15kg=0.34m/s2

Challenge question

• What is the maximum coefficient of friction that would allow the object to move at a constant velocity with 30N of force applied at 40˚?

0.18

Practice Problem 2

• A 2000kg car is being towed out of a level field.

• The tow cable makes a 25˚angle from the car to the row truck.

• The car is being pulled out at a constant speed and the tow truck is applying 10600 N of force

• What is the coefficient of friction?

0.63

Always Use the 5 Step Process1. Draw all forces that are relevant to the situation.2. Decide on an x- direction. This should be the direction that the object will ACCELERATE. Always!!!3. Find components of all known forces that are not already in the x or y direction.

Use sin and cosin, make sure you are using SOH CAH TOA to pick the right one.4. IF the ultimate question is for acceleration, or a force in the x-

direction, THEN calculate the values of the normal force and of friction. You will be using Fy = 0 and Ff = FNcosθ to do this.

IF the ultimate question is for normal force, friction or , THEN skip to step 5. Find all the x-direction values and return to step 5. Apply Fx = max to answer the question posed in the problem.

Motion on an Inclined Plane

Set Up the System

• Sketch the force diagram

– X is in the direction of motion

– Y is perpendicular to motion

Set Up the System

• Because an object’s acceleration is usually parallel to the slope, one axis, usually the x-axis, should be in that direction.

• The y-axis is perpendicular to the x-axis and perpendicular to the surface of the slope.

• We will continue to define down as negative

Set Up the System

• Put in the force due to gravitational acceleration

With this coordinate system, there are two forces—normal and frictional forces. These forces are in the direction of the coordinate axes. However, the weight is not.

This means that when an object is placed on an inclined plane, the magnitude of the normal force between the object and the plane will usually not be equal to the object’s weight.

This is why things slide when the plane is inclined

Set up the System

Components of Weight

• Weight must be determined from the force due to gravitational acceleration, but the coordinates are not aligned with it.

Define the Components of the Weight

Remember SOH CAH TOA?

• Sin is Opposite

• Cos is Adjacent

Remember SOH CAH TOA?

• Sin is Opposite

– This means the X component is SinθFg

• Cos is Adjacent

– This means the Y component is CosθFg

Finding Net Forces

You will need to apply Newton’s laws once in the x-direction and once in the y-direction.

Because the weight does not point in either of these directions, you will need to break this vector into its x- and y-components before you can sum your forces in these two directions.

Practice Problem 1

A 35kg child is sliding down a bannister (railing), which makes a 50˚angle with the floor.

What is the component of his weight which is parallel to the bannister?

What is the component of his weight that is perpendicular to the bannister?

Which contributes to the magnitude of the frictional force?

Which contributes to the acceleration?

Practice Problem 1

Step 1 draw a free body diagram:

Practice Problem 1

Step 2 : Identify the x direction

Practice Problem 1

Step 3 : Find the x and y components

Fapp= mgsin50° = 262NFn= mgcos50° = 220N

Practice Problem 1Which affects friction (Fn) and which affects acceleration (Fapp)?

Fapp= mgsin50° = 262N

Fn= mgcos50° = 220N

Practice Problem 1aIf there is no friction, what is the acceleration?

Use Fapp in F=ma

Fapp/m=a

262/35kg=7.4m/s2

Practice Problem 1b

Suppose μk= 0.37, What is the force of friction?

Use Fn in Ff= μkFn

220* 0.37

=81N uphill

Fn

Practice Problem 1c

What is the net force?

Fnet = Fapp+ Ff

Fnet= 262N -81N

=181N downhill

Practice Problem d

What is the acceleration including friction?

Fnet = ma

Fnet /m= 181N/35kg

= 5.17m/s2

TIP FOR INCLINED PLANE PROBLEMS!

Tilt your head (or the paper) and the problem becomes what you are used to seeing!