Newton’s First Law of Motion

An object at rest will remain at rest and a body in motion will continue its uniform motion unless acted upon by a resultant or net force.

It is important to note that in physics the​​ uniform motion​​ means​​ the body is in same speed and direction.

By this, it can be deduced that a body may experience resultant force equal to zero in two conditions.​​ 

  • One when body is at rest (v=0)

  • Second, when the body is moving at a constant velocity (same speed and direction).​​ 

Moreover, the definition of mass can be defined with the help of 1st​​ law of motion.​​ 

Mass (inertia) is the property of an object that resists any change in its current state of rest or uniform motion.

Let’s now consider some examples below to learn how 1st​​ law of newton works in our daily life.​​ 

Example 1: Sliding Book

Scenario: Imagine a book placed on a table. The table is suddenly pulled out from under the book.

Explanation: When the table is pulled out, there's no horizontal force acting on the book. According to Newton's First Law, the book will remain at rest in its position in the air. It will continue to float momentarily until gravity pulls it down.

Example 2: Car Braking Incident

Scenario: The man inside the car was not wearing a seatbelt. The car was moving at a speed of 100 kmph.

Explanation: When the car suddenly applied the brakes, the car experienced a decelerating force. However, the man inside the car was still moving at 100 kmph due to inertia (property of an object that resists any change in its current state). Since there was no force acting specifically on the man, he continued to move forward at 100 kmph. Without a seatbelt to stop his forward motion, he was propelled out of the windscreen when the car stopped abruptly.

Example 3: Bus Braking Incident:

Scenario: A person standing inside a bus loses balance and falls backward when the bus suddenly starts moving.

Explanation: When the bus is at rest, the person and everything inside the bus are at rest. When the bus suddenly starts moving forward, the lower portion of the person's body, which is in​​ contact with the bus​​ (meaning it has a contact force acting on its lower part of body in the form of friction force by the floor of the bus on him), starts moving along with the bus. However, the upper portion of the person's body tends to stay at rest due to inertia. Since there is no force applied to keep the upper body in motion with the bus, the person loses balance and falls backward.

In all these examples, the concept of inertia, which is a fundamental aspect of Newton's First Law of Motion, is clearly demonstrated. Objects tend to maintain their state of​​ motion (whether at rest or in motion at a constant velocity) unless acted upon by a resultant/net force.

 

Is it necessary for a body to be at rest, the polygon of forces acting on it must be closed?

When a body is at rest, the concept of the "polygon of forces" refers to the graphical method used to represent multiple concurrent forces acting on an object. In equilibrium (which includes the state of rest), the vector sum of all the forces acting on the object is​​ equal to zero. This condition ensures that the object is not accelerating and remains at rest.

The "polygon of forces" method involves drawing vectors to scale, representing all the forces acting on the object. These vectors are placed head-to-tail in a sequence. If the object is in equilibrium, the polygon formed by these vectors must be closed.​​ In other words, the starting point of the first vector should meet the ending point of the last vector, forming a closed geometric shape, typically a polygon.​​ 

The closure of the polygon signifies that the forces are in balance and that the object is in equilibrium. If the polygon does not close, it indicates that the forces are not balanced, and the object would be in a state of motion or acceleration, not at rest. Therefore, a closed polygon of forces is a visual representation of the balance of forces in an object at rest or in equilibrium.

 

Q1: ​​ 9702 Mar 23/P12/Q12, M/J 14/P11/Q11, O/N 09/P12/Q11, O/N 09/P11/Q12

The diagrams show two ways of hanging the same picture.

 ​​ ​​ ​​ ​​ ​​​​ In both cases, a string is attached to the same​​ points on the picture and looped symmetrically

 ​​ ​​ ​​ ​​ ​​​​ over a nail in a wall. The forces shown are those that act on the nail.

 ​​ ​​ ​​ ​​ ​​​​ In diagram 1, the string loop is shorter than in diagram 2.

 ​​ ​​ ​​ ​​ ​​​​ Which information about the magnitude of the forces is correct?

 ​​ ​​ ​​ ​​ ​​ ​​​​ A​​ R1 = R2T1 = T2

 ​​ ​​ ​​ ​​ ​​ ​​​​ B​​ R1 = R2T1> T2

 ​​ ​​ ​​ ​​ ​​​​ ​​ C​​ R1> R2T1< T2

 ​​ ​​ ​​ ​​ ​​​​ ​​ D​​ R1< R2T1 = T2

QUESTION BREAKDOWN

Context:​​ Two picture frames are being hung on a wall using strings looped symmetrically.

Focus:​​ Identify the correct information about the magnitude of forces involved (specifically, normal​​ reaction force and tension in the strings).

SOLUTION

  • How to know this question is about the application of 1st​​ Law of Newton?

The question says that the frame is hanging, a condition which aligns with the principles of Newton's First Law.

 

  • Understanding​​ the Forces

Normal Reaction Force (R): This force is exerted by the wall on the nail and supports the weight of the picture frame. For both frames to remain​​ stable, the normal reaction force (R) must balance the weight of the frame (W). Therefore,​​ 

R=W

Tension​​ in Strings (T1 & T2): Tension in the strings depends on the angles at which the strings are hung. Smaller angles require a higher tension to compensate for the smaller vertical component, while larger angles require less tension due to a significant vertical component. This can be learned by the following derivation.​​ 

Consider a frame hanging, we are considering a general scenario here.​​ 

 

Since the frame is hanging, meaning it is in equilibrium,​​ not​​ moving vertically or horizontally, the sum of forces acting on​​ a body along the y – axis is zero.​​ 

Fy=0

+R-Tcosθ-Tcosθ=0

R=Tcosθ+Tcosθ

R=2Tcosθ

T=R2cosθ

By this expression, it may be concluded that​​ 

  • When​​ θ​​ is small, the value of​​ cosθ​​ is large, so the tension in the string would be greater.​​ 

  • When​​ θ​​ is large, the value of​​ cosθ​​ is small, so the tension in the string would be smaller.

 

  • ​​ Comparison of Picture Frames:

Picture 1: The string loop is shorter, resulting in smaller angles. The higher tension (T1) is required to balance the smaller vertical component.

Picture 2: The string loop is longer, leading to larger angles. The lower tension (T2) is needed because of the significant vertical component.

  • Correct Option:

Eliminating Options: By understanding that normal reaction force must equal the weight of the frame​​ (R = W)​​ and recognizing the influence of angles on tension, we can eliminate options C and D.

Correct Answer: The correct​​ answer is option B, where tension​​ T1​​ is greater than tension​​ T2​​ due to the differences in angles and string lengths.

 

Q2: 9702 M/J 23/ P11/ Q13

The diagram shows a ball of weight W hanging in equilibrium from a​​ string.

The string is at an angle​​ θ​​ to the vertical. The tension in the string is T. The ball is held away from the wall by a horizontal force P from a metal rod.

​​ Which relationship between the magnitudes of T, P and W is correct?

 ​​ ​​ ​​ ​​ ​​​​ A ​​ ​​ ​​​​ P = T cos​​ and W = T sin

 ​​ ​​ ​​ ​​ ​​​​ B​​  ​​ ​​​​ T = P + W

 ​​ ​​ ​​ ​​ ​​​​ C ​​  ​​​​ T2​​ = P2​​ + W2

 ​​ ​​ ​​ ​​​​ ​​ D​​ W = P tan​​ and W = T cos

QUESTION BREAKDOWN

Context:​​ Ball is hanging in equilibrium by the string attached and metal rod.​​ 

Focus:​​ Identify the correct equations defining the relationship​​ between forces T, P, & W.​​ 

SOLUTION

  • How to know this question is about the application of 1st​​ Law of Newton?

The question says that the ball is equilibrium, a condition which aligns with the principles of Newton's First Law.

 

  • Understanding the Forces

Tension in the spring causes forces two act on the ball in two directions. One is upwards, and the second is leftwards. ​​ 

Force P is the force applied by the metal rod on the ball horizontally (along the positive x- axis).​​ 

Whereas, weight always acts vertically downward, acting towards the centre of the Earth.​​ 

 

Since ball is in equilibrium, the sum of forces acting on the ball horizontally and vertically are equal to zero.​​ 

It is important to note that the angle​​ θ, which is marked in the diagram with help of angle properties that alternative angles are equal, is with the y -axis, therefore the horizontal component of tension in the string is ​​ Tcos θ​​ and the vertical component of the string is​​ Tsin θ.​​ 

Let us first consider the forces acting vertically.

Fy=0

+Tcos θ-W=0

Tcos θ=W

Now,​​ considering the forces acting on the ball horizontally.​​ 

Fx=0

-Tsin θ+P=0

Tsin θ=P

We may write the above values as​​ 

Tv=Tcos θ=W

Th=Tsin θ=P

Since​​ Tv & Th​​ are the components of tension in the string, by applying the resultant vector formula, we may find another expression.​​ 

Resultant Vector=horizintal component of  vector2+vertical component of vector2

T=Th2+Tv2

T=P2+W2

OR

T2=P2+W2

  • Correct Option:

By​​ comparing the values of​​ Th​​ and​​ Tv​​ to the given options, none fits in any option. However, the final expression​​ T2=P2+W2​​ is the correct​​ answer is option C.​​ 

 

Q3: 9702 MJ 22/ P11/ Q7

An object is moving along the ground in a straight line at a constant speed.

Which statement about the resultant force acting on the object is correct?

A The resultant force acting on the object is equal to its weight.

B The resultant force acting on the object is equal to the product of its mass and its velocity.

C The resultant force acting on the object is equal to the resistive force.

D The resultant force acting on the object is equal to zero.

QUESTION BREAKDOWN:

Context:​​ An object is moving along the ground in a straight line at a constant speed.

Focus:​​ Determine the correct statement​​ about the resultant force acting on the object.

SOLUTION​​ 

  • Analysing the Options​​ 

Option A:​​ The resultant force acting on the object is equal to its weight.

Evaluation: This statement does not necessarily hold true when the object is moving at a constant speed. Weight is the force due to gravity, but if the object is not accelerating vertically, the vertical forces are balanced.

Conclusion: Incorrect.

Option B:​​ The resultant force acting on the object is equal to the product of its mass and its velocity.

Evaluation: This statement describes the formula for calculating momentum (F=m×v), not the resultant force. It does not account for the constant speed scenario.

Conclusion: Incorrect.

Option C:​​ The resultant force acting on the object is equal to the resistive​​ force.

Evaluation: This option is closer to the correct concept. When an object moves at a constant speed, the resistive force (like friction) is equal in magnitude but opposite in direction to the applied force, resulting in a balanced situation.

Conclusion: Partially correct but not the best choice.

Option D:​​ The resultant force acting on the object is equal to zero.

Evaluation: This option accurately represents the scenario. When an object moves at a constant speed, it experiences no acceleration.​​ According to Newton's first law of motion, the object is in a state of equilibrium, with the net force being zero.

Conclusion: Correct choice.

  • Correct Answer:

The​​ correct option is D. The resultant force acting on the object is equal to zero.

Q4: MJ 22/ P13/ Q13

A street lamp is fixed to a wall by a metal rod and a cable.

Which vector triangle could represent the forces acting on the end of the rod at point P?

QUESTION BREAKDOWN:

Context:​​ The street lamp is held stable/fixed by a metal rod and cable.​​ 

Focus:​​ Determine the correct vector diagram of the given options.​​ 

SOLUTION​​ 

  • How to know this question is about the application of 1st​​ Law of Newton?

The question says that the lamp is fixed, a condition which aligns with the principles of Newton's First Law.

  • Understanding the Forces

Tension in the string as it supports the lamp.

On the other hand, the weight acts in downwards direction as the earth attracts it to the centre of the earth.​​ 

Remember, to form a vector triangle, it is necessary that the​​ vectors, forces in this case, must be combined with each head’s coinciding with other vector’s tail. In other words, there should be no resultant vector. If the bottom vector is drawn with its direction leftward, tension in the cable would become​​ resultant​​ force, which is not the case in this question. Recall that it is must for the body to be equilibrium that the polygon of forces is closed, otherwise the resultant force would exist, acting on a body. ​​ 

Hence, the correct vector diagram/triangle is shown below.​​ 

  • Correct Answer​​ 

The correct option is D,​​ as the weight acts downwards, not upwards as shown in option A. other than options A and D, there has been a resultant vector forming therefore, they cant be the correct options. ​​ 

 

Q5: Edexcel IAL U1/Oct/2021/Q6

A block of wood is stationary on a frictionless ramp as shown. The block is held in place by a string. The weight of the block is W. The force applied to the block by the string is F.

A triangle of forces can be used to determine the magnitude and​​ direction of the normal contact force N acting on the block.​​ 

Which of the following triangles is correct?

 

QUESTION BREAKDOWN

Context:​​ A block is held by the string on an inclined surface.​​ 

Focus:​​ Identify the vector diagram showing the polygon of forces closed.​​ 

SOLUTION

  • How to know this question is about the application of 1st​​ Law of Newton?

The question says that the block is stationary, a condition which aligns with the principles of Newton's First Law.​​ 

  • Understanding the Forces

For an object to be​​ stationary, it is necessary that the polygon of forces acting on the object is closed that is the​​ starting point of the first vector should meet the ending point of the last vector, forming a closed geometric shape, typically a polygon.​​ 

  • Correct option​​ 

The correct option is A that the normal force mut act in upward direction.​​