NEWTONʼS FIRST LAW - 1634maths · 2013. 6. 23. · Newtonʼs First Law of Motion states that:...
13
NEWTONʼS FIRST LAW Newtonʼs First Law of Motion states that: Every object remains in a state of rest or move with constant velocity in a straight line unless forces acts on it to change that state Only when these external forces have a non-zero resultant with the body change from its previous state of rest or of constant velocity. (i) A body at rest If forces act on a body and it does not move, the forces must balance. Hence, if a number of forces act on a body and it remains at rest, the resultant force in any direction must be zero. Example 1 A body is at rest when subjected to the forces shown in the diagram. Find X and Y. (ii) A body in motion A body can only change its velocity, i.e. increases its speed, slow down or change direction, if a resultant force acts upon it. Thus, if a body is moving with constant velocity, there can be no resultant force acting on it. Example 2 A body moves horizontally at a constant subject to the forces subject to the forces shown. Find P and S. NEWTONʼS LAW OF MOTION 1 50 N Y 10 N 10 N 50 N X 5 ms −1 5 ms −1 S S 2000 N P 300 N
Transcript of NEWTONʼS FIRST LAW - 1634maths · 2013. 6. 23. · Newtonʼs First Law of Motion states that:...
NEWTONʼS FIRST LAW
Newtonʼs First Law of Motion states that:
Every object remains in a state of rest or move with constant velocity in a straight line unless forces acts on it to change that state
Only when these external forces have a non-zero resultant with the body change from its previous state of rest or of constant velocity.
(i) A body at rest
If forces act on a body and it does not move, the forces must balance. Hence, if a number of forces act on a body and it remains at rest, the resultant force in any direction must be zero.
Example 1A body is at rest when subjected to the forces shown in the diagram. Find X and Y.
(ii) A body in motion
A body can only change its velocity, i.e. increases its speed, slow down or change direction, if a resultant force acts upon it. Thus, if a body is moving with constant velocity, there can be no resultant force acting on it.
Example 2A body moves horizontally at a constant subject to the forces subject to the forces shown. Find P and S.
" NEWTONʼS LAW OF MOTION
1
50 N
Y 10 N
10 N50 N
X
5 ms−1
5 ms−1
S S
2000 N
P300 N
Example 3A car of mass 800 kg is traveling at a constant velocity of along a level road. The resistance to motion is 1500 N. Taking , determine:
(a) the force produced by the engine(b) the normal reaction acting on the car
Example 4A trunk is pulled across a horizontal floor by a force of 50N inclined at above the horizontal. The trunk has a mass of 40 kg and travels with a constant velocity. Determine the exact values of:
(a) the resistance to motion(b) the normal reaction
" NEWTONʼS LAW OF MOTION
2
30 ms−1
g = 10 ms−2
30◦
NEWTONʼS SECOND LAW
Newtonʼs Second Law of Motion states that:
When a force of F Newtons acts on an object of mass m kg, it produces an acceleration, , given by F = ma
When a resultant force acts on a body, it causes acceleration. The acceleration is proportional to the force. The same force will not produce the same acceleration in all bodies.
The acceleration produced by a force depends upon the mass of the body on which it acts. The unit of mass is the kilogram (kg). A force of 1N produces an acceleration of in a body of mass 1kg. In general terms, a force F newtons on a body of mass m kg produces an acceleration of , where:
This is a vector equation, so the acceleration produced is in the direction of the applied force, or of the resultant force if there is more than one force acting. Newtonʼs Second Law can be summarized by the equation F = ma, which is often referred to as the equation of motion.
Example 1A body of mass 8kg is acted upon by a force of 10N. Find the acceleration.
Example 2A body of mass 2kg, subject to forces as shown in the diagram, accelerates uniformly in the direction indicated. Find the acceleration and the value of P.
" NEWTONʼS LAW OF MOTION
3
a ms−2
1 ms−2
a ms−2
F = ma
P
10 N
20 N
4 N
a ms−2
Example 3In each of the following situations of the forces acting on the body causes it to accelerate as indicated. Find the magnitude of the unknown forces X and Y.
(a)" " " " " (b)" " " " " (c)
(d)" " " " " (e)" " " " " (f)
" NEWTONʼS LAW OF MOTION
4
mass 10kg mass 10kg mass 10kg
100 N
X
100 N
X
50 N
Y
X
10 N2 ms−2 2 ms−2
3 ms−2
mass 5kg mass 5kg mass 10kgY
X
50 N
20 N
2 ms−22 ms−2
X
Y10 N
30 N
50 N
10 ms−2
X 57 N
157 N
GRAVITY AND WEIGHT
A body falling under gravity experiences an acceleration . From Newtonʼs Laws it is clear that this acceleration must be caused by a force acting on the body. This force is called the weight of the body.
Example 1Find:" (a) the weight in Newtons of a box of mass 5kg" (b) the mass of a stone of weight 294N
Example 2A box of mass 5kg is lowered vertically by a rope. Find the force in the rope when the box is lowered with an acceleration of
Example 3A pack of bricks of mass 100kg is hoisted up the side of a house. Find the force in the lifting rope when the bricks are lifted with an acceleration of
" NEWTONʼS LAW OF MOTION
5
g = 10ms−2
w = mg w - weight forcem - mass of the objectg - acceleration due to gravity
1
4ms−2
4 ms−2
Example 4In each of the following situations, the forces acting on the body cause it to accelerate as indicated. In (a), (b) and (c), find the magnitude of the unknown forces X and Y.
(a)" " " " " (b)" " " " " (c)
In (d), (e) and (f), find the mass m.
(d)" " " " " (e)" " " " " (f)
In (g), (h) and (i), find the magnitude of the acceleration a.
(g)" " " " " (h)" " " " " (i)
" NEWTONʼS LAW OF MOTION
6
mass 10kg mass 10kg mass 10kg
mass m mass m mass m
mass 10kg mass 9kg mass 4kg
10g N
X
10g N
X
5g N
X
Y10 N
mg N
20 N
mg N
180 N 7 N 20 N
mg N
Y
10g N
50 N
9g N
90 N 16 N 4 N
4g N
4 ms−2 4 ms−2
5 ms−2
6 ms−22 ms−2 1 ms−2
a ms−2a ms−2 a ms−2
F = ma and INCLINED FORCES
Examples:
1.(a)" " " " " " " " (b)
2.(a)" " " " " " " " (b)
3.(a)" " " " " " " " (b)
" NEWTONʼS LAW OF MOTION
7
5 kg10 kg
15 kg
10 kg
m kg m kg
30◦
30◦
35◦ 45◦
20◦
10◦
a ms−2 a ms−2
7 ms−22 ms−2
6.5 ms−2 2√2 ms−2
50√2 N
100 N5 N
100 N
P
30 N
P
18 N 300 N
4. A particle of mass 5kg is pulled up a smooth slope inclined at to the horizontal. The applied force has a value of 50N and acts parallel to the slope. Calculate the acceleration of the particle.
5. An object of mass 20kg slides down a slope inclined at to the horizontal. The resistance to motion is 60N and acts parallel to the slope. Taking , determine the value of the acceleration of the object in surd form.
6. A particle of mass 100g is projected up a smooth inclined plane with an initial speed of . The slope is inclined at to the horizontal. Find:
" (a) the deceleration of the particle" (b) the distance traveled up the slope before the particle comes to rest
7. A body has a mass of 30kg and is accelerated up a slope inclined at to the horizontal. The acceleration of the body is and the frictional resistance is 300N. Calculate the force parallel to the slope causing this acceleration.
" NEWTONʼS LAW OF MOTION
8
30◦
45◦
g = 10ms−2
19.6 ms−1
3 ms−230◦
20◦
NEWTONʼS THIRD LAW
Newtonʼs Third Law of Motion states that:
To every action there is an equal and opposite reaction
Example 1A man of mass 90kg is standing in a lift of mass 300kg which is acceleration upwards at . Find the tension in the lift cable and the reaction between the man and the floor of the lift.
Example 2An engine of mass 10 tonnes is pulling a track of mass 3 tonnes. The resistance of forces acting on the engine and the truck are 4000N and 1500N respectively. The driving force of the engine is 14000N. Find the acceleration of the system and the tension in the coupling between the engine and the truck.
" NEWTONʼS LAW OF MOTION
9
T T
4000 N1500 N
14 000 N
0.6 ms−2
CONNECTED PARTICLES
In the following examples the strings are all considered to be light and inextensible.
Note also that, when a surface is said to be smooth, it is to be assumed that the surface offers no resistance to the motion of a body across it.
Example 1Consider a body of mass 3kg at rest on a smooth horizontal table. This body is connected by a light string, which passes over a smooth pulley at the edge of the table, to another body of mass 2kg hanging freely. As the pulley is smooth, the tension in the string on both sides of the pulley will be the same. Find the acceleration of the two bodies and the tension in the string.
The string is said to be light, so its weight can be ignored. As the string is inextensible, when the system is released from rest the two bodies will have equal acceleration along the line of the string.
" NEWTONʼS LAW OF MOTION
10
3 kg
2 kg
T
T
Example 2Particle of mass 4kg and 2kg are connected by a light string passingover a smooth fixed pulley. The particles hang freely and are releasedfrom rest. Find acceleration of the two particles and the tension in the string.
Example 3
A body A rests on a smooth horizontal table. Two bodies of mass 2kg and 10kg, hanging freely, are attached to A by string which pass over smooth pulley at the edges of the table. The two string are taut. When the system is released from rest, it accelerates . Find the mass of A.
" NEWTONʼS LAW OF MOTION
11
2 kg
2 ms−2
A
2 kg10 kg
T1
T1
T2
T2
4 kg
2g N
4g N
T
T TT
a
a
Example 4Two particles of mass 2kg and 3kg are connected by a light inextensible string passing over a fixed smooth pulley. Initially the system is at rest with the strings taut and vertical with both particles at a height of 2m above the ground. When the system is released, find the time which elapses before the 3kg mass hits the ground and the maximum height reached by the 2kg mass.
To find the maximum height reached by the 2kg mass, we have to find the speed at which both particles are traveling when the 3kg hits the ground.
Once the 3kg mass hits the ground, the string is no longer taut, the 2kg mass now moves freely upwards (against gravity) until it reaches its maximum height before going down again and make the string taut again.
" NEWTONʼS LAW OF MOTION
12
initial position
aaT T
TT
2g N 3g N2m
? m
4 m 4 m
position when 3 kg mass hits the ground
position when 2kg mass going down making the string taut again
SITUATION INVOLVING FRICTION
Examples
1. A block of mass 5kg moves on a rough horizontal plane with coefficient of friction 0.2 under the action of a horizontal force of 30N. If the block starts from rest, find the distance it travels in the first 3 seconds of motion.
2. A particle of mass 6kg, moving at on a smooth horizontal surface, goes onto a rough horizontal surface with a coefficient of friction 0.25. Find the distance it moves across the rough surface before coming to rest.
3. The diagram shows a small block of mass 2kg able to move on a rough plane of a length of 8m inclined at" to the horizontal. The block is attached by means of a light string passing over a smooth pulley at " the top of the plane to a particle of mass 5kg hanging freely. The coefficient of friction between the block " and the plane is 0.2. The system is released from rest with the block at the bottom of the plane. Find the " time which elapses before it reaches the top.
4. A particle, moving at on a smooth horizontal surface, goes onto a rough horizontal surface and is brought to rest in a distance of 20m. Find the coefficient of friction involved.
5. A block of mass 6kg moves on a rough horizontal surface ( ) under the action of a horizontal force. It accelerates from rest to a speed of in a distance of 12 m, continues for a time at this speed and then decelerates to rest in a distance of 2 m. Find the magnitude and direction of the horizontal force required during each stage of the journey.
" NEWTONʼS LAW OF MOTION
13
8 ms−1
6 ms−1
4 ms−1
20◦
µ = 0.25
20◦
8 m
2g N
R
a
a
5g N
T
T
F
