In our everyday life, we observe that some effort is required to put a stationary object into motion or to stop a moving object. We ordinarily experience this as a muscular effort and say that we must push or hit or pull on an object to change its state of motion. The concept of force is based on this push, hit or pull.
A force can be used to change the magnitude of velocity of an object or to change its direction of motion. It can also change the shape and size of objects (e.g., stretching a rubber band).
Friction is a force that always opposes the motion of objects. It arises due to contact between two surfaces. To accelerate an object, an unbalanced force must be applied that is greater than the opposing force of friction.
Q: Two forces of 10 N and 15 N act on a body in opposite directions. What is the net force acting on the body and what is its direction?
Solution:
Net Force = Larger Force - Smaller Force
Net Force = $15 \text{ N} - 10 \text{ N} = 5 \text{ N}$.
The direction of the net force is in the direction of the 15 N force. Since the net force is non-zero, this is an unbalanced force.
Galileo deduced that objects move with a constant speed when no force acts on them. Based on Galileo's ideas, Sir Isaac Newton presented three fundamental laws that govern the motion of objects. The first law is stated as:
"An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force."
In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why the first law of motion is also known as the law of inertia.
Do all bodies have the same inertia? We know that it is easier to push an empty box than a box full of books. A heavier or more massive object offers larger inertia. Quantitatively, the inertia of an object is measured by its mass. Thus, mass is a measure of the inertia of a body.
Q: Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Solution:
When the bus suddenly brakes, the lower part of your body comes to rest with the bus, but the upper part continues moving forward due to inertia of motion. Conversely, when the bus accelerates from rest, your feet move forward with the bus, but your upper body tends to remain at rest due to inertia of rest, causing you to fall backwards.
The first law of motion indicates that an unbalanced force is required to change velocity (cause acceleration). But how is this acceleration related to the force? Newton's second law gives us a quantitative measure of force.
Before stating the law, we need to introduce a property called momentum ($p$). Newton introduced momentum as the product of mass ($m$) and velocity ($v$).
Momentum has both direction and magnitude. Its direction is the same as that of velocity. The SI unit of momentum is kilogram-metre per second ($kg \cdot m/s$).
The Second Law of Motion:
"The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force."
Let an object of mass $m$ be moving with initial velocity $u$. A constant force $F$ accelerates it to final velocity $v$ in time $t$.
Initial momentum, $p_1 = mu$
Final momentum, $p_2 = mv$
Change in momentum $\propto p_2 - p_1 \propto m(v - u)$
Rate of change of momentum $\propto \frac{m(v - u)}{t}$
Since $F$ is proportional to the rate of change of momentum:
By choosing the unit of force appropriately, the constant of proportionality $k$ becomes 1. Thus:
The unit of force is $kg \cdot m/s^2$, which is also known as Newton ($N$). One Newton is the force that produces an acceleration of $1 \text{ m/s}^2$ on an object of mass 1 kg.
Q: A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force.
Solution:
Mass $m = 5 \text{ kg}$
Initial velocity $u = 3 \text{ m/s}$
Final velocity $v = 7 \text{ m/s}$
Time $t = 2 \text{ s}$
Force $F = m \frac{v - u}{t} = 5 \left(\frac{7 - 3}{2}\right) = 5 \left(\frac{4}{2}\right) = 5 \times 2 = 10 \text{ N}$.
The first two laws of motion tell us how an applied force changes the motion and provide us with a method of determining the force. The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first.
"To every action, there is an equal and opposite reaction."
These two forces are always equal in magnitude but opposite in direction. They act on different objects and never on the same object. For example, when you walk, you push the ground backwards (action). The ground exerts an equal and opposite force on your foot in the forward direction (reaction).
Even though the action and reaction forces are always equal in magnitude, these forces may not produce accelerations of equal magnitudes. This is because each force acts on a different object that may have a different mass (recall $a = F/m$). For instance, a fired bullet accelerates massively, while the heavy gun recoils with much less acceleration.
Q: If action is always equal to the reaction, explain how a horse can pull a cart.
Solution:
The horse pushes the ground backwards with its feet (Action). According to Newton's Third Law, the ground pushes the horse forward with an equal and opposite force (Reaction). It is this reaction force from the ground that enables the horse (and the attached cart) to move forward, provided this force is greater than the opposing frictional force of the cart's wheels.
Suppose two objects, two balls A and B of masses $m_A$ and $m_B$, are traveling in the same direction along a straight line at different velocities $u_A$ and $u_B$. If they collide and there is no external unbalanced force, the total momentum of the two objects remains unchanged (or conserved) before and after the collision.
Mastering these typical numericals will guarantee you can solve any exam question from this chapter.
Q: A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.
Solution:
Initial velocity $u = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$
Final velocity $v = 0 \text{ m/s}$
Time $t = 4 \text{ s}$
Mass $m = 1000 \text{ kg}$
Force $F = m(v - u) / t$
The negative sign indicates that the force exerted by the brakes is opposite to the direction of motion.
Q: Which would require a greater force: accelerating a 2 kg mass at 5 m/s² or a 4 kg mass at 2 m/s²?
Solution:
For the first mass: $F_1 = m_1 a_1 = 2 \text{ kg} \times 5 \text{ m/s}^2 = 10 \text{ N}$
For the second mass: $F_2 = m_2 a_2 = 4 \text{ kg} \times 2 \text{ m/s}^2 = 8 \text{ N}$
Therefore, accelerating the 2 kg mass at 5 m/s² requires a greater force.
Q: A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg).
Solution:
Initial velocity $u = 0$
Distance $s = 400 \text{ m}$
Time $t = 20 \text{ s}$
Using $s = ut + \frac{1}{2} a t^2$:
Mass $m = 7 \text{ tonnes} = 7000 \text{ kg}$
Force $F = ma = 7000 \times 2 = 14000 \text{ N}$
Q: From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.
Solution:
Mass of rifle $m_1 = 4 \text{ kg}$
Mass of bullet $m_2 = 50 \text{ g} = 0.05 \text{ kg}$
Before firing, both are at rest. Initial total momentum = 0.
After firing: Velocity of bullet $v_2 = 35 \text{ m/s}$. Let recoil velocity of rifle be $v_1$.
Final total momentum = $m_1 v_1 + m_2 v_2 = 4 \times v_1 + 0.05 \times 35 = 4v_1 + 1.75$
According to conservation of momentum: Initial Momentum = Final Momentum
The negative sign indicates the recoil velocity is opposite to the bullet's direction.
Q: Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.
Solution:
$m_1 = 100 \text{ g} = 0.1 \text{ kg}, m_2 = 200 \text{ g} = 0.2 \text{ kg}$
$u_1 = 2 \text{ m/s}, u_2 = 1 \text{ m/s}$
$v_1 = 1.67 \text{ m/s}$, let $v_2$ be the velocity of the second object.
Total momentum before collision = $m_1 u_1 + m_2 u_2 = 0.1(2) + 0.2(1) = 0.2 + 0.2 = 0.4 \text{ kg m/s}$
Total momentum after collision = $m_1 v_1 + m_2 v_2 = 0.1(1.67) + 0.2(v_2) = 0.167 + 0.2v_2$
