Force, friction, balanced and unbalanced forces, Newton's three laws of motion and systems of objects.
Notes
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The concept of force
A force can make an object move from rest, change its speed or direction, or change its shape (kicking a ball, striking with a bat, squeezing a lemon).
Force is a vector: both magnitude and direction must be specified. The SI unit is the newton (N).
If the magnitude or direction of a force changes, its effect changes.
A spring balance measures the magnitude of a force, not only weight. Weight is the gravitational force with which the Earth pulls an object.
Balanced and unbalanced forces
Balanced forces are equal in magnitude and opposite in direction. The object stays at rest or keeps moving as before (equal pulls in tug of war).
If forces are unbalanced, a non-zero net force acts and the object accelerates.
Opposite forces of different size: net force = larger force − smaller force, in the direction of the larger.
Forces in the same direction: net force = sum of the two.
Motion depends only on the net force, even if many forces act.
Force of friction
Friction acts between the surfaces in contact, opposite to the direction of (attempted) motion.
An object starts moving only when the applied force becomes larger than the force of friction.
Without a continuing push, friction slows a moving object and stops it.
Friction depends on the nature of the surfaces. On smoother surfaces the stack of coins in Activity 6.1 travels farther, so friction is smaller there.
Activity 6.2: the spring balance reading when the block just moves at constant velocity gives the friction force.
Friction can also help motion: when you walk, the ground pushes your foot forward by friction. Grooves on shoes and treads on tyres increase friction.
Newton's first law of motion
An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon it.
If the net force is zero, acceleration is zero.
Galileo's thought experiment: on a perfectly smooth surface with no friction, a moving object would continue moving forever.
Inertia is the tendency of an object to resist a change in its state of rest or uniform motion. The first law is also called the law of inertia.
Constant velocity means no change in the magnitude or the direction of velocity.
Graphs for no net force: position-time is a horizontal line (at rest) or a straight inclined line (constant velocity); velocity-time is a horizontal line.
Newton's second law of motion
When a net force acts on an object, the object accelerates in the direction of the net force. The acceleration is proportional to the net force and inversely proportional to the mass.
a = F / m or F = ma
1 newton is the force that produces an acceleration of 1 m s-2 on a 1 kg object (1 N = 1 kg m s-2).
Activities 6.3 and 6.4 (cart and pulley): a larger force gives a larger acceleration; a larger mass gives a smaller acceleration for the same force.
Gravitational force on mass m: F = mg, with g = 9.8 m s-2 near the Earth (10 m s-2 for quick estimates). g does not depend on the mass of the object.
Applications of the second law
A fielder pulls the hands back while catching a fast ball. This increases the time over which the ball stops, reducing its acceleration and the force on the hand.
Airbags increase the time over which a passenger stops, so the force is smaller and the injury risk lower.
A coconut hitting a hard surface stops in a very short time. The ground exerts a very large force, which breaks the shell.
Landing mats, bubble wrap and hay work in the same way.
Newton's third law of motion
Whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first.
The two forces always act on different objects, so they do not cancel each other.
Examples: pushing a table makes your chair roll back; a canoeist pushes water backwards and the water pushes the paddle forward; a balloon or rocket moves opposite to the escaping gas; a person climbs a tree by pushing down on the trunk while friction pushes up.
The law applies to contact and non-contact forces (magnets, charged balloons, gravity).
The forces are equal, but the accelerations differ when masses differ (a gun recoils slowly, a bullet speeds away).
Forces on a system of objects
Two connected boxes and the string can be treated as one system.
Internal forces (such as tension in the string) need not be considered. Only external forces matter.
Acceleration of the system: a = F / (m1 + m2).
At a glance
Friction acts opposite to the direction of motion.
First law: no net force means no change in velocity.
Second law: F = ma; acceleration is along the net force.
Third law: forces come in equal and opposite pairs acting on different objects.
Numericals
Worked solutions for the examples and calculation questions in this chapter.
EXAMPLE 6.1 · NET FORCE
(a) 10 N + 6 N = 16 N towards the right
(b) 10 N − 6 N = 4 N towards the right
(c) 10 N − 6 N = 4 N towards the left
EXAMPLE 6.4 · BARBELL
Total mass = 10 + 10 + 10 = 30 kg.
F = mg = 30 × 9.8 = 294 N
She applies 294 N upward.
EXAMPLE 6.5 · BLOCK ON A FLOOR
m = 25 kg, maximum friction = 50 N.
(i) Push of 50 N: net force = 0, so the block stays stationary.
(ii) Net force = 55 − 50 = 5 N
a = F/m = 5 / 25 = 0.2 m s⁻²
s = ut + ½at² = 0 + ½ × 0.2 × 2² = 0.4 m
EXAMPLE 6.6 · SPORTS CAR (m = 1500 kg)
0–5 s: a = (10 − 0)/5 = 2 m s⁻²; F = 1500 × 2 = 3000 N east
5–10 s: constant velocity, a = 0, F = 0
10–15 s: a = (0 − 10)/5 = −2 m s⁻²; F = 1500 × (−2) = −3000 N (towards the west)
EXAMPLE 6.8 · RECOIL OF A GUN
Force on bullet = recoil force on gun = 2 N.
Gun: a = 2 / 5 = 0.4 m s⁻²
Bullet: a = 2 / 0.1 = 20 m s⁻²
PAUSE AND PONDER 6 · TOY CAR
Constant velocity means acceleration is zero.
Net force = ma = 0.1 × 0 = 0 N
REVISE, REFLECT, REFINE 4 · SNAKE BOAT
Forward force = 95 × 200 = 19 000 N
Backward force = 5 × 200 = 1000 N
Net force = 19 000 − 1000 = 18 000 N, forward
REVISE, REFLECT, REFINE 11 · VELOCITY-TIME GRAPH
From Fig. 6.41 the velocity rises from 10 m s⁻¹ at 0 s to 30 m s⁻¹ at 8 s.
a = (30 − 10)/8 = 2.5 m s⁻²
F = ma = 10 × 2.5 = 25 N
REVISE, REFLECT, REFINE 12 · BULLET IN A BLOCK
m = 0.05 kg, u = 100 m s⁻¹, v = 0, s = 0.5 m.
v² = u² + 2as → 0 = 10000 + 2a(0.5) → a = −10 000 m s⁻²
F = ma = 0.05 × 10 000 = 500 N (opposing the motion)
REVISE, REFLECT, REFINE 13 · PENALTY KICK
v = 108 km/h = 30 m s⁻¹; F = 800 N; m = 0.4 kg; ball starts at rest.
a = F/m = 800 / 0.4 = 2000 m s⁻²
t = v/a = 30 / 2000 = 0.015 s
REVISE, REFLECT, REFINE 14 · ROUGH PATCH
m = 2 kg, u = 10 m s⁻¹. Total opposing force = 7 + 3 = 10 N.
a = −10 / 2 = −5 m s⁻²
s = u² / (2 × 5) = 100 / 10 = 10 m
REVISE, REFLECT, REFINE 15 · TRACTOR AND HARROW
m₁ = F/a₁, m₂ = F/a₂
a = F / (m₁ + m₂) = 1 / (1/a₁ + 1/a₂) = a₁a₂ / (a₁ + a₂)
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NCERT Q&A
Answers to the Pause and Ponder questions and the Revise, Reflect, Refine exercises. Numerical working is in the Numericals tab.
Pause and Ponder
PAUSE AND PONDER 1
A weightlifter lifts a barbell. List two forces on the barbell. Are they balanced if she keeps it steady?
Answer: The gravitational force (weight) acts downward and the force applied by the weightlifter acts upward. Yes, they are balanced when the barbell is steady, since the net force is zero.
PAUSE AND PONDER 2
At the instant the arms of two arm-wrestling players tilt to the front, are the forces balanced? If not, which player applied the larger force?
Answer: No. The arms are moving (tilting), so the forces are unbalanced and there is a net force in the direction of the motion. Player S exerted the larger force.
PAUSE AND PONDER 3
An object moves with a constant velocity. Is there a net force on it?
Answer: No. Constant velocity means zero acceleration, so the net force is zero.
PAUSE AND PONDER 4
If no net force acts on an object, which situations are possible?
Answer: (i) and (ii) are possible: it stays at rest, or keeps moving with constant velocity. (iii) is not possible, because constant acceleration needs a net force.
PAUSE AND PONDER 5
Explain with an example how the net force on an object can be made zero by applying additional forces.
Answer: A box on the floor is pushed with a force equal to the friction acting on it. The two forces are equal and opposite, so the net force is zero and the box moves with constant velocity. (A book resting on a table is another case: weight is balanced by the normal force.)
PAUSE AND PONDER 6
A toy car of mass 100 g moves with a constant velocity of 0.5 m s⁻¹. What is the net force on it?
Answer: Zero, because its velocity is constant.
PAUSE AND PONDER 7
Two children of different masses sit on identical swings. For which child is a larger force needed to give the same initial acceleration?
Answer: For the heavier child. By F = ma, for the same acceleration a larger mass needs a larger force.
PAUSE AND PONDER 8
How are glass items packed with bubble wrap or hay protected from damage?
Answer: If the package falls or is jolted, the wrap or hay increases the time over which it stops. This reduces the acceleration and the force on the glass, so it is less likely to break.
PAUSE AND PONDER 9
Why does a fireperson sometimes struggle to hold a hose with water issuing from it?
Answer: The hose pushes water forward at high speed. By the third law, the water pushes the hose backward with an equal force, so the hose is hard to hold steady.
PAUSE AND PONDER 10
A spacecraft moves where gravity is negligible. How can it change its velocity?
Answer: By firing its engine. The engine expels gas in one direction, and the gas pushes the spacecraft in the opposite direction, changing its velocity.
Revise, Reflect, Refine
QUESTION 1
A table is moved across the floor at constant velocity by a horizontal force F. How much is the frictional force?
Answer: Equal to F, in the opposite direction. The net force is zero because the velocity is constant.
QUESTION 2
On a smooth frictionless surface, complete the statements.
Answer: (i) With no net force, the velocity remains the same. (ii) A net force along the direction of motion makes the speed increase. (iii) A net force opposite to the direction of motion makes the speed decrease.
QUESTION 3
Block P has opposing forces of 4 N and 5 N. Block Q moves with constant velocity. Which statement is correct?
Answer:(i). P has a net force of 1 N and Q has no net force.
QUESTION 4
95 oarsmen row the boat forward and 5 row the wrong way, each applying 200 N. Find the net force.
Answer: 18 000 N in the forward direction (19 000 N − 1000 N).
QUESTION 5
When a net force acts on an object, it accelerates:
Answer:(iv). In the direction of the force, with acceleration proportional to the force.
QUESTION 6
From the position-time graphs in Fig. 6.37, a net force acts on:
Answer:(iii) Object C. Its position-time graph is a curve, so its velocity is changing. A is a straight line (constant velocity), B is horizontal (at rest) and D is a straight line sloping down (constant velocity).
QUESTION 7
A sailor jumps from a small boat to the shore. Does the boat move?
Answer: Yes, the boat moves backward, away from the shore. The sailor pushes the boat backward, and by the third law the boat pushes the sailor forward with an equal force.
QUESTION 8
Why is a landing mat or sand bed placed for a high jumper?
Answer: The mat increases the time taken by the athlete to stop. This reduces the acceleration and therefore the force on the athlete, so the risk of injury is smaller.
QUESTION 9
A loaded hand cart collides with an identical empty cart. During the collision:
Answer:(iv). Both carts exert forces of equal magnitude on each other (third law).
QUESTION 10
Plot the force-mass graph from the acceleration-mass graph in Fig. 6.40.
Answer: From the graph, mass × acceleration is constant (for example 1 kg × 10 m s⁻² and 5 kg × 2 m s⁻²), so F = 10 N for every mass. The force-mass graph is a horizontal straight line at 10 N.
QUESTION 11
Find the force on the 10 kg object using the velocity-time graph in Fig. 6.41.
Answer: 25 N (acceleration = 2.5 m s⁻²).
QUESTION 12
A 50 g bullet at 100 m s⁻¹ penetrates 50 cm into a block. Estimate the stopping force.
Answer: 500 N.
QUESTION 13
A 0.4 kg football kicked at 108 km h⁻¹ with a force of 800 N. Find the contact time.
Answer: 0.015 s.
QUESTION 14
A 2 kg object at 10 m s⁻¹ meets friction of 7 N plus an extra 3 N opposing force. How far does it travel?
Answer: 10 m.
QUESTION 15
A tractor pulls a harrow (m₁) with acceleration a₁ and a trolley (m₂) with a₂ using the same force F. Find the acceleration for the trolley with the harrow on it.
Answer: a = a₁a₂ / (a₁ + a₂).
QUESTION 16
Why does the compass needle move but the bar magnet does not?
Answer: The forces on the two are equal in magnitude. The compass needle has a very small mass, so it has a large acceleration and moves. The bar magnet has a much larger mass, so its acceleration is too small to notice.