By now you are already familiar with Newton’s Laws of motion. These laws will help you in solving problems in mechanics. Typically, a problem in mechanics doesn’t deal with multiple forces acting on a single object. On the contrary, it deals with an assembly of different bodies exerting forces on each other apart from experiencing the gravitational force. In this article, we will look at some tips for solving problems in mechanics.
When you are trying to solve a problem in mechanics, it is important to remember that you can choose any part of the assembly and apply the laws of motion to that part. All you need to ensure is that you account for all forces acting on the ‘chosen part’ due to the remaining parts of the assembly. For the sake of simplicity, we call the chosen part of an assembly as the ‘system’ and the remaining part as the ‘environment’.
Steps for Solving Problems in Mechanics
These steps will help you solve the problems in a systematic manner:
- A diagram speaks a thousand words. Ensure that you draw a diagram which represents the environment in its entirety.
- Look at the diagram and choose a convenient part of the environment as the system.
- Draw a separate diagram for the system including the forces exerted on it by the environment. Remember to NOT include forces on the environment by the system in this diagram. We will call this diagram a ‘free-body’ diagram. In this diagram:
- Include information about the forces that are either given to you or you are sure of
- The information should contain the magnitude and direction of these forces
- Anything unknown should be left as it is and determined with the help of the laws of motion.
- If required, follow the same procedure for another choice of the system.
Let’s look at an example to understand these steps:
A wooden block of mass 2 kg rests on a soft horizontal floor as shown in the figure below. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m/s2. What is the action of the block on the floor (a) before and (b) after the floor yields? Take g = 10 m/s2. Identify the action-reaction pairs in the problem.
a. The block is at rest on the floor.
If the block is at rest on the floor, there are forces on the block; the gravitational force and the normal force R of the floor on the block. Since the block weighs 2 kg, the gravitational force is 2 x 10 = 20N.
According to Newton’s First Law, the normal force R = – 20N. Also, as per the Third Law, the action of the block or the force exerted by the block on the floor is 20N and is directed downwards.
b. The system (block + cylinder) accelerates downwards with 0.1 m/s2.
When the system of the block and cylinder accelerates downwards, there are two forces on the system; the gravitational force [ (25 + 2) x 10 = 270N] and the normal force R’ by the floor. Internal forces between the block and the cylinder are not taken into consideration. Applying per the Second Law of motion, we get
270 N – R′ = 27 × 0.1 N
R′ = 267.3 N
Hence, according to the third law, the action of the system on the floor is equal to 267.3 N vertically downward.
Before the floor yields –
- The gravitational force on the block of 20N (say, action) and the force on the earth by the block equal to 20N but directed upwards (reaction)
- The force on the floor by the block (action) and the force on the block by the floor (reaction)
After the floor yields –
- The gravitational force on the system of 270N (say, action) and the force on the earth by the system equal to 270N but directed upwards (reaction).
- The force on the floor by the system (action) and the force on the system by the floor (reaction)
The force on the block by the cylinder and the force on the cylinder by the block is also an action-reaction pair.
Action-reaction pairs consist of mutual forces which are always equal and opposite between two bodies. There can be two forces on the same body which are equal and opposite, but they don’t constitute an action-reaction pair. For example, the force of gravity on an object and the normal force on the object by the floor is NOT an action-reaction pair.
Since the object is at rest, these forces happen to be equal and opposite. In the example above, you can observe that while the weight of the system is 270N, the normal force is only 267.3N. Practice drawing free-body diagrams as they can really help in solving problems in mechanics.
Solving Problems in Mechanics
Q1. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in the figure below. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding.
Solution: We know that,
Mass of the block = m = 25kg
Mass of the man = M = 50kg
Acceleration due to gravity = g = 10 m/s2
Therefore, Force applied on the block by gravity = F = 25 x 10 = 250N
Weight of the man = W = 50 x 10 = 500N
Case a: The man lifts the block directly
In this case, since the man applies force in the upward direction, it increases his apparent weight on the floor. Therefore,
Action on the floor by the man = 250 + 500 = 750N
Case b: The man uses a pulley to lift the block
In this case, since the man applies the force in the downward direction, his apparent weight on the floor decreases. Therefore,
Action on the floor by the man = 500 – 250 = 250N
If the floor yields to a normal force of 700N, then the man should use the pulley to lift the block to ensure that the floor does not yield.