unit 3 physics - · pdf filethe school for excellence 2017 succeeding in the vce – unit...

Unit 3 Physics

succeeding in the vce, 2017
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The School For Excellence 2017 Succeeding in the VCE Unit 3 Physics Page 1
INCLINED PLANES An inclined plane is a surface inclined at an angle above the horizontal. This is a topic that integrates many of the topics covered so far in this booklet. A normal force N is the force that a surface exerts on an object that is in contact with it. It acts at right angles to a surface and changes as the force exerted on the surface changes. The normal force N acting on an object on an inclined plane is equal to the component of the weight force perpendicular to the incline. As the incline becomes steeper, the normal force becomes smaller. For an object that is stationary on a rough inclined plane, the frictional force acts up the incline and is equal in magnitude to the component of the weight force down the slope. Consider the following example:
EXAMPLE 9 A sphere of mass 12 kg is allowed to roll down a plane inclined at 30 to the horizontal. There is a constant frictional force of 20 N acting on the sphere. What is its acceleration? Solution Draw a diagram, showing all the information contained in the question. The diagram should also show all the forces that act on the sphere. Forces acting on the ball are: Weight force (downwards).
Normal reaction force (perpendicular to the plane).
Frictional force (opposite direction to motion i.e. up the plane parallel to surface).
30 mg = 120 N
F Normal
F Friction = 20 N
m = 12 kg
The forces seem to be pointing in all sorts of inconvenient directions. (In what direction does the sum of all these forces, i.e. net force, act? Why, down the plane, of course. We know that the sphere will accelerate down the plane, so the net force must be directed down the plane.) The next step is to remove any vector that happens to point in an inconvenient direction and replace it with its components (that will point in convenient directions). So, the weight vector must go, and be replaced by components parallel and perpendicular to the plane.

30 mg
F Normal
F Friction
= 20 N
30 mgcos30
mgsin30 = 60 N
The component of the weight that is perpendicular to the plane is equal in magnitude to the normal reaction force, so these two forces cancel. We are left, therefore, with the component of the weight parallel to the plane and the frictional force.
30
F Friction = 20 N
mgsin30 = 60 N
The sum of these two forces is 40 N down the plane.
30
F net = 40 N
m = 12 kg
We can now work out acceleration:
maFnet a)12(40
2/3.3 sma So the sphere accelerates at 3.3 m/s2 down the plane.

QUESTION 27 A model car of 1.0 kg mass runs down a smooth 30 decline in which friction can be ignored. Calculate: (a) The normal force that acts on the car. (b) The net force that acts on the car. (c) The cars acceleration as it travels down the slope. Solution QUESTION 28 A smooth incline rises 3 metres vertically for every 5 metres of its length. A body weighing 80 N is placed on the incline, which is assumed to be smooth. Find: (a) The resolved parts of its weight parallel and perpendicular to the incline. (b) The least force necessary to hold the body in position. Solution

PROJECTILE MOTION A projectile is any object that has been thrown through the air. A force must necessarily set the object in motion initially but, while it is moving through the air, no force other than gravity acts on it (we shall ignore air resistance for now). Thus, a brick can be a projectile but a rocket or an airplane cannot. The path, or trajectory, of the projectile is curved (it is, in fact, parabolic).
trajectory
v
mg
projectile
At any given point in the motion, the velocity vector is always a tangent to the path. Note also that the vector mg = weight force
= the only force acting on the object = net force
Here is something to remember about net forces: If the net force is in the same direction as the velocity of the object, the object speeds up:
v
F net
object speeds up

If the net force is in the opposite direction to the velocity of the object, the object slows down:
v
F net
object slows down
But, if the net force acts at right angles to the velocity vector, then the speed of the object in the direction of that vector does not change.
v
F net
speed in the direction of v remains constant
Now consider the following diagram:
v
mg = F net
v H
v V
m

Notice that the velocity vector, v, has two components: A horizontal component, vH.
A vertical component vV. The net force, mg, has: An effect on vV.
No effect on vH. Thus, there is no acceleration horizontally (which is to say that vH remains constant throughout the motion) but there is indeed a vertical acceleration. The next diagram shows the path taken by a projectile that has been thrown horizontally. The position of the projectile is shown at equal time intervals. Notice that it travels at constant velocity horizontally (for it covers equal distances in equal time intervals) but it is accelerating vertically (it covers greater distances in successive equal time intervals). As one would expect, it is moving at constant speed horizontally, but it is speeding up vertically.

The diagram that follows shows how the velocity vector changes as the projectile moves along its trajectory. Also shown are the horizontal and vertical components of the velocity vector. Of course, the horizontal component stays constant but the vertical component changes.

We usually handle projectile motion problems by breaking up the motion into horizontal and vertical components. For the horizontal component, we use:
tdv
For the vertical component, we use the constant acceleration formulae:
atuv
221 atutd
tvud
2
aduv 222

unit 3 physics - · pdf filethe school for excellence 2017 succeeding in the vce – unit...

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