Question 1195 of 3158 | MHT-CET (Maharashtra Common Entrance Test) Physics

• Block ${\displaystyle {}^{ʹ}Aʹ}$ cannot be in equilibrium. Due to Fsin6 component (upwards) on block ${\displaystyle ʹBʹ}$, Block ${\displaystyle ʹAʹ}$ moves downwards relative to block ${\displaystyle ʹBʹ}$.
• So, friction on block ${\displaystyle ʹBʹ}$ downwords.

Newton՚s Third Law

To every action, there is always an equal (in magnitude) and opposite (in direction) reaction.

• When a body exerts a force on any other body, the second body also exerts an equal and opposite force on the first.
• Forces in nature always occurs in pairs. A single isolated force is not possible.
• Any agent, applying a force also experiences a force of equal magnitude but in opposite direction. The force applied by the agent is called ‘Action’ and the counter force experienced by it is called ‘Reaction’.
• Action and reaction never act on the same body. If it were so, the total force on a body would have always been zero i.e.. the body will always remain in equilibrium.
• If ${\displaystyle F_{AB}}$ = force exerted on body A by body B (Action) and FBA = force exerted on body B by body A (Reaction) Then according to Newton՚s third law of motion ${\displaystyle F_{AB}=F_{BA}}$
• Example: (i) A book lying on a table exerts a force on the table which is equal to the weight of the book. This is the force of action.

• The table supports the book, by exerting an equal force on the book. This is the force of reaction.
• As the system is at rest, net force on it is zero. Therefore, force of action and reaction must be equal and opposite.
• Swimming is possible due to third law of motion.
• When a gun is fired, the bullet moves forward (action). The gun recoils backward (reaction).
• Rebounding of rubber ball takes place due to third law of motion.

• While walking a person presses the ground in the backward direction (action) by his feet. The ground pushes the person in forward direction with an equal force (reaction). The component of reaction in horizontal direction makes the person move forward.
• It is difficult to walk on sand or ice.
• Driving a nail into a wooden block without holding the block is difficult.

Static Friction

• When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.
• Static friction is friction between two or more solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface.
• The coefficient of limiting static friction, typically denoted as vs, is usually higher than the coefficient of kinetic friction.
• The static friction force must be overcome by an applied force before an object can move. The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: ${\displaystyle F_{\max }={\mu }_s{F}_n}$.
• When there is no sliding occurring, the friction force can have any value from zero up to ${\displaystyle F_{\max }}$ ax force smaller than ${\displaystyle F_{\max }}$ attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction. Any force larger than F_max overcomes the force of static friction and causes sliding to occur.
• The instant sliding occurs, static friction is no longer applicable the friction between the two surfaces is then called kinetic friction.
• An example of static friction is the force that prevents a car wheel from slipping as it rolls on the ground.
• Even though the wheel is in motion, the patch of the tire in contact with the ground is stationary relative to the ground, so it is static rather than kinetic friction.

Equilibrium of Forces

Triangle law of vector addition states that when two vectors are represented by two sides of a triangle in magnitude and direction taken in same order then third side of that triangle in opposite order represents in magnitude and direction the resultant of the vectors.

Parallelogram Law

Parallelogram Law of Vector Addition states that when two vectors are represented by two adjacent sides of a parallelogram by direction and magnitude then the resultant of these vectors is represented in magnitude and direction by the diagonal of the parallelogram starting from the same point.

If all the forces working on a body are acting on the same point, then they are said to be concurrent.

A body, under the action of concurrent forces, is said to be in equilibrium, when there is no change in the state of rest or of uniform motion along a straight line.

The necessary condition for the equilibrium of a body under the action of concurrent forces is that the vector sum of all the forces acting on the body must be zero.

Mathematically for equilibrium

${\displaystyle \sum \stackrel{⇾}{F}=0\text{or}\sum \stackrel{⇾}{F}x=0;}$

${\displaystyle \sum \stackrel{⇾}{F}y=0;\sum \stackrel{⇾}{F}z=0}$

Three concurrent forces will be in equilibrium, if they can be represented completely by three sides of a triangle taken in order.

Lami՚s Theorem

For three concurrent forces in equilibrium

${\displaystyle \frac{F_1}{\sin \alpha }=\frac{F_1}{\sin \alpha \beta }=\frac{F_1}{\sin \gamma }}$

Solved Examples:

Question:

If a system in mechanical equilibrium and is subjected with

three forces ${\displaystyle \stackrel{⇾}{F}1,\stackrel{⇾}{F}2,\stackrel{⇾}{F}3}$ in vector form …

${\displaystyle \stackrel{⇾}{F}1=5\widehat{𝚤}-6\widehat{𝚥};\stackrel{⇾}{F}2=4\widehat{𝚥}+2\widehat{𝚤}+3\widehat{k}}$

Then find ${\displaystyle \stackrel{⇾}{F}3}$?

Solutions:

As the forces are in mechanical equilibrium therefore

${\displaystyle \stackrel{⇾}{F}1,\stackrel{⇾}{F}2,\stackrel{⇾}{F}3=0;i\mathrm{.}e,5\widehat{𝚤}-6\widehat{𝚥}+4\widehat{𝚥}+2\widehat{𝚤}+3\widehat{k}+F_3=0;7\widehat{𝚤}-2\widehat{𝚥}+3\widehat{k}+F_3=0;F_3=2\widehat{𝚥}-7\widehat{𝚤}-3\widehat{k}}$

${\displaystyle \frac{T}{\sin 90}=\frac{mg}{\sin (180-\theta )}=\frac{F}{\sin (90+\theta )}}$