In the previous lesson we applied Newton's Laws to objects moving horizontally and objects moving vertically.  (To refresh your memory on the vertical applications, just recall the practice exercises on the elevator and the sanding block on a wall.)

In this lesson we will deal with objects on ramps or inclined planes.

Practice exercise 1

In this exercise we will draw a free body diagram that describes a block sliding down an inclined plane.  The object is to identify all the forces and the components of the forces perpendicular to and parallel to the plane.

Although our still picture may not suggest it at first, you must realize that the block is actually sliding down the plane.  If this were not the case, the friction force would not be acting up the plane! 

Can you see what makes the block slide down?  If not, go to the next picture.

Since gravity pulls straight downward to the centre of the earth, a component of the weight lies along the plane (see orange vector).  We will call this F-parallel or where

This force is causing the block to slide.

Of course, if the angle were smaller, then the orange vector would be smaller and the block might just sit there.  Why would it sit there?  Because of friction.  And friction requires a normal force.  Can you see what causes the normal force?  Go to the next picture for the answer.

You should be saying to yourself that this picture doesn't really show the normal force.  However, we do see a perpendicular force which is the component of the weight that is perpendicular to the plane.  We can write where is the also the force that the block exerts perpendicular to the plane.  Can you see how this causes the normal force?  Go the next picture.

The normal force is the reaction force (equal and opposite in direction) to .  It follows that

Fn = mg cos θ.

The final picture in this exercise has the trigonometric terms written in it.

Practice exercise 2

This exercise has two animations that you can interact with.  By clicking on the step button at the bottom of the first picture you will work through (in a slightly different way) the information that was presented in Practice exercise 1.

Up to this point point in time we have not said why it is that the angle between the plane and the ground is the same as the angle between the perpendicular and parallel components of weight vector.  The next interactive animation show this is so. 

Practice Exercise 3

A bundle of shingles sits on a plank while a worker raises one end so that the shingles slide down the plank to a co-worker at the other end.  At the very, very instant that the shingles are about to slide, the plank makes an angle of 30o with the ground.  If the mass of the bundle is 35.0 kg, what is the coefficient of static friction?

Solution

Step 1--read and understand the story of the problem

Do you remember how, in primary and elementary school, you use to put your eraser on your ruler and then raise one end of your ruler until the eraser started to slide?  This exercise is like that. 

Step 2--list the givens and the unknown

m= 35.0 kg

g = -9.8 N/kg  (you will probably be more comfortable if we dispense with the negative for "g".  You will see what we mean as you do the exercise.)

θ  =  30o

μs= ?

Step 3--plan a strategy

If down the plane is negative and up the plane is positive, then, strictly speaking we should be writing .  And this indeed would work mathematically if g is substituted as -9.8.  However, let's try, at least for a little while, to work with magnitudes only, and deal with direction as needed.

Step 4--carry out the plan

First find the frictional force:

You should be able to see that if we had used and substituted -9.8 instead of 9.8, the answer would have been the same.  It's a matter of preference how you want to attack this, but you must be consistent.

Back to the problem.  Since 172 N is the frictional force just as the bundle starts to move, we can compute the coefficient of static friction:

Step 5--communicate

The coefficient of static friction is 0.58.

Let's do another...

Practice exercise 4

Continuing with the story in  exercise 3, what will be the acceleration if the plank is raised to an angle of 45o with the ground and μk = 0.43?

Step 1--Read and understand the problem

If you do a rough sketch you will see that increasing the angle makes the component larger, and Fn smaller.  If Fn becomes smaller, then so does the friction force.  Moreover, the bundle now begins to move so that the coefficient of friction is μk which is less than μs.  The end result is that shingles accelerate down the plank.

Step 2--list the givens and the unknown

m= 35.0 kg

g = 9.8 N/kg 

θ  =  45o

μk= 0.43

a  = ?

Step 3--plan a strategy

Step 4--carry out the plan

Σ (parallel forces)  =  .

Next, substitute for   and Ff, but be mindful that is down the plane, so, put a "-" in front of mg sin θ because its direction is down the plane.  Also,  Fnet is zero because a = 0  (we've beat the life out of that).

Step 5--communicate

The angle that allows constant speed down the plane is 23o.   Isn't it interesting that for our givens the plank had to be lowered by 7o to maintain constant speed once the motion had begun?  Isn't it also neat that for zero acceleration μk = tan θ?  You will re-visit this idea when you do the laboratory activity in Lesson 06.

Let's do another...

Practice exercise 7

Without pushing at any time, a skier skis down a frictionless 20.0 m slope, hits a frictionless horizontal portion and coasts to a stop a certain distance up another slope where the coefficient of kinetic friction is 0.20.  The first slope is inclined at an angle of 35o and the second at 30o.  How far up the second incline will she go?

Solution

Step 1--read and understand the story of the problem

The picture really tells the story.  We have to find d2, the distance up the second slope.

Step 2--list the givens and the unknown

d1  =  20.0 m

μk1 = 0

θ1   =  35o

μk2 = 0.20

θ1   =  30o

d2  =  ?

While d2 is the main unknown we also have to deal with h1 and h2.

Step 3--plan a strategy

The key information, at least at the beginning, is that there is no friction on the first slope and on the horizontal portion.  We can therefore apply the law of conservation of mechanical energy and say that the kinetic energy at the bottom of the second incline is exactly equal to the gravitational potential energy at the top of the first.  On the second incline some of the kinetic energy is "robbed" by friction.  This "lost" energy can be calculated because the friction force does work against the motion of the  skier.  This energy is  determined by Wf = Ff  x  d2.  So the kinetic energy at the bottom of the second slope changes into Wf and gravitational potential energy at the top of the second incline.  Writing the appropriate equations for the law of conservation of energy will allow the determination of d2.

Step 4--carry out the plan

Let's start by writing expressions for h1 and h2:

h1  =  d1  sin 35  = 20 m  x  sin 35  =  11.5 m

h2  =  d2  sin 30  = 0.5 d2   (this is as far as we can go with h2 because d2 is unknown)

Next write the gravitational potential energy equation for the top and bottom of the first incline:

(Ek) bottom  =  (Eg) top  (because there is no friction)

½ m v2           =   mgh1   which allows the cancellation "m", and that's a good thing because we were not told the mass of the skier.  Continuing:

v2  =  2gh1  =  2  x  9.8 m/s2  x  11.5 m    =   225 m2/s2  .

Let's not bother to find the square root of both sides, because as you will see in a moment, it will just be a waste of time.

Because the horizontal portion is frictionless, the kinetic energy at the bottom of the second incline is the same as the kinetic energy at the bottom of the first.  That is

(Ek)bottom 2  =  (Ek)bottom 1  =  ½ m v2   =  ½ m (225 m2/s2)

Now write the  gravitational potential energy equation for the top and bottom of the second incline.  Remember, the kinetic energy at the bottom is transformed into energy lost due to friction (Wf) plus a gain in gravitational potential energy (mgh2).

(Ek)bottom 2      =  Wf                +  mgh2

(Ek)bottom 2      =  Ff  x  d2        +  mgh2

½ m (225 m2/s2)   =  μk2 Fn  x  d2   +  mg (0.5d2)

Fn = mg cos θ

Which means that   ½ m (225 m2/s2)   =  μk2 Fn  x  d2   +  mg (0.5d2)  can be written as

½ m (225 m2/s2)   =  μk2 mg cos θ  x  d2   +  mg (0.5d2)

Again we conveniently cancel the mass "m" and solve for d2:

½  (225 m2/s2)   =  μk2 x 9.8 m/s2 x cos 35 x d2   +  9.8 m/s2x (0.5d2)

113 m2/s2     =  0.20 x 9.8 m/s2 x 0.819 d2  + 4.9 m/s2 d2 

113 m2/s2     =   1.6 m/s2  d2   +  4.9  m/s2  d2

113 m2/s2    =   d2  (1.6 m/s2  +  4.9 m/s2)

113 m2/s2    =  d2  x  6.5 m/s2 

d2                 =   113 m2/s2   ÷  6.5 m/s2 

d2                 =  17 m

Step 5--communicate

The skier comes to a stop 17 m up the second slope.