Dynamics | Conservation of Linear momentum | Friction | Numerical Problem Solutions | Class 11 (Physics) | Physics in Depth

Mass and Pulley system

Equation of motion for mass m is, $$\begin{array}{rl}F& =T-mg\\ ma& =T-mg...(i)\end{array}$$ Equation of motion for mass M is, $$\begin{array}{rl}F& =T-Mg\\ Ma& =T-Mg...(ii)\end{array}$$ Adding eqns (i) and (ii), $$\begin{array}{rl}ma+Ma& =Mg-mg\\ a& =\frac{(M-m)g}{(M+m)}...(iii)\end{array}$$ Substituting the vaue of a in eqn (i), $$\begin{array}{rl}m\left(\frac{(M-m)g}{M+m}\right)& =T-mg\\ \frac{m(M-m)g}{M+m}+mg& =T\\ T& =\frac{m(M-m)g+mg(M+m)}{M-m}\\ \therefore T& =\frac{2Mm}{M+m}g...(iv)\end{array}$$

Two masses 7 kg and 12 kg are connected at the two ends of a light inextensible string that passes over a frictionless pulley. Using free body diagram method, find the acceleration of masses and the tension in the string, when the masses are released.
Here, m = 7 kg and M = 12 kg. Prior to using the formula directly, you must deduce them at first.Now, using the relations (iii) and (iv),
$$a=\frac{(M-m)g}{(M+m)}=\frac{5\times 10}{19}=2.63m/s^2$$ and, $$T=\frac{2Mm}{M+m}g=\frac{2\times 7\times 12\times 10}{19}=88.42N$$

What is Tension? When a cord (or a rope, or a cable, or other such object) is attached to a body and pulled tightly, the cord pulls on the body with a force $\overrightarrow{T}$ directed away from the body and along the cord. This force is called the tension force because the cord is under the tension (i.e., it is being pulled).

Apparent weight of a person in a lift

Consider that you stand on a weighing machine inside the lift. Let 'R' be the reaction offered by the machine as your weight 'mg' acts downward on the machine. We will see here how the motion of the lift changes the reading of the weighing machine i.e., the reaction R which we will say the apparent weight .

• Lift at rest : In this case, R=mg (i.e., apparent weight will be equal to actual weight).


• Lift moving upward or downward with uniform speed : Consider when the lift is moving upward with uniform speed, $a=\frac{dv}{dt}=0\therefore a=0$, then, the equation of motion is, $$R-mg=0⟹R=mg$$ Since, $a=0$, even when the lift is moving downward with uniform speed,$R=mg$. So, apparent weight is equal to your actual weight.


• Lift accelerating upward : In this case, the equation of motion is, $$\begin{array}{rl}F& =R-mg\\ ma& =R-mg\\ R& =m(a+g)\end{array}$$ Thus, when the lift is accelerating upward, the apparent weight will be greater than the actual weight.


• Lift accelerating downward : Solve the equation of motion $F=mg-R$ to get $R=m(g-a)$. So, the apparent weight of the person will be less than the actual weight. In other words, we can say that the reaction of the floor on you is smaller than your real weight.


• Lift in free fall : In free fall, $a=g$, and on solving equation of motion $F=mg-R$, you will get $R=0$. This is the weightlessness condition. For more details on weightlessness jump to Gravitation .


• Lift accelerating downward with $a>g$ : On solving the equation of motion $F=mg-R$, you will get negative R which means the reaction of the floor on you will be in upward direction. You will stick on the ceiling of the lift.

A lift moves (i) up and (ii) down with an acceleration of 2$m/s^2$. In each case, calculate the reaction of the floor on a man of mass 5o kg standing in the lift.
Here, you need to draw the free body diagram of both the cases and solve it using the formulas deduced above. I have drawn free body digram for each of the cases in above diagram (indicating R, mg and a; do the same).
When the lift moves up, $$R=m(g+a)=50(10+2)=50\times 12=600N$$ When the lift moves down, $$R=m(g-a)=400N$$

What is normal force or normal reaction R? When a body presses against a surface, the surface deforms and pushes on the body with a force that is perpendicular to the surface. This force is called the normal force or normal reaction R. In the above examples, you have pressed the weighing machine or the floor with a force equal to mg and thus the weighing machine or flooe pushes up on you with a normal force R.

Principle of conservation of linear momentum

It states that, " If no external force act on the system of two colliding objects then the vector sum of the linear momentum of each body remains constant and is not affected by their mutual interaction."
Consider two bodies of mass $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ respectively such that $u_2>u_1$. After collision for time 't', their respective velocities become $v_1$ and $v_2$. Due to collision, let $F_1$ be the force acting on $m_2$ due to mass $m_1$ and $F_2$ be the force acting on mass $m_1$ due to mass $m_2$.
Now, from Newton's second law of motion, $$F_1=\frac{\text{change in momentum}}{t}=\frac{m_1{v}_1-m_1{u}_1}{t}$$ Similarly, $$F_2=\frac{m_2{v}_2-m_2{u}_2}{t}$$ From Newton's third law of motion, $$\begin{array}{rl}{F}_1& =-F_2\\ \frac{m_1{v}_1-m_1{u}_1)}{t}& =-(\frac{m_2{v}_2-m_2{u}_2}{t})\\ m_1{v}_1-m_1{u}_1& =-m_2{v}_2+m_2{u}_2\\ m_1{u}_1+m_2{u}_2& =m_1{v}_1+m_2{v}_2\end{array}$$ This proves the principle of conservation of linear momentum.

An object A of mass 2 kg is moving with a velocity of 3 m/s and collides head on with an object B of mass 1 kg moving in the opposite direction with a velocity of 4 m/s. After collision both objects coalesce (becomes a single mass), so that they move with a common velocity v. Calculate v.
Initial momentum of A is, $$m_1{u}_1=2\times 3=6kgm/s$$ Initial momentum of B is, $$m_2{u}_2=1\times (-4)=-4kgm/s$$ Here, negative sign of velocity indicates the opposite direction of body B to that of A. Final momentum of A is, $$m_1{v}_1=2v$$ Final momentum of B is, $$m_2{v}_2=1v$$ From the principle of conservation of linear momentum, $$\begin{array}{rl}\text{total momentum before collision}& =\text{total momentum after collision}\\ 6-4& =2v+1v\\ v& =\frac{2}{3}m/s\end{array}$$

A bullet of mass 20 g travelling horizontally at 100 m/s embeds itself in the centre of a block of wood mass 1 kg which is suspended by light vertical strings 1 m in length. Calculate the maximum inclination of the strings to the vertical.
Suppose A is the bullet, B is the block suspended from a point O, and $\theta$ is the maximum inclination to the vertical as in figure. If v is the common velocity of block and bullet , then, from the principle of conservation of linear momentum,

$$\begin{array}{rl}0.02\times 100& =(1+0.02)v\because m_1{u}_1+m_2{u}_2=(m_1+m_2)v\text{and}{u}_2=0\\ v& =1.96m/s\end{array}$$ The vertical height risen by the block and the bullet is, $$h=l-lcos\theta =l(1-cos\theta )$$ From $v^2=u^2-2gh⟹u^2=2gh$ $\because v=0$ when the block and bullet comes to rest and remember! u=1.96 because the initial velocity of block and bullet is the common velocity of both, $$\begin{array}{rl}{u}^2& =2gl(1-cos\theta )\\ (1.96{)}^2& =2\times 9.8\times 1(1-cos\theta )\\ 1-cos\theta & =\frac{(1.96{)}^2}{2\times 9.8}\\ cos\theta & =0.8038\\ ⟹\theta & ={36.5}^{\circ }\end{array}$$ So, the maximum inclination of the string is ${36.5}^{\circ }$.

Friction

The resistance in the motion of a body when we either slide or attempt to slide a body over a surface, is called friction. This resistance is due to the bonding between the surface and the body in microscopic scale. When the two bodies are made in contact with each other, with their surfaces highly polished, they cold-weld together ( Cold welding or contact welding is a solid-state welding process in which joining takes place without fusion/heating at the interface of the two parts to be welded: Wikipedia ) and the force of friction increases. The force is required to break the welds and maintain the motion.

Static and Kinetic friction

Static friction is the force between two surfaces when the body is at rest. The maximum value of static friction is called the limiting friction. Kinetic friction is the force of friction when the body is sliding on the surface. For a limited speed, kinetic friction remains constant throughout the motion. Figure shows the graph of frictional force Vs applied force. Here, it can be noted that kinetic friction is less than static friction.

Why is kinetic friction always less than static friction? At first, it requires more force to loose the bonding between the surfaces in contact and when the force is sufficient enough to break the bonding between the surfaces, the body gains the inertia of motion. Due to this force of friction is reduced. In the static condition of body (in the former case), there is static friction between the surfaces and when on motion (in the latter case) kinetic friction comes to play. This means, kinetic friction is always less than static friction.

Force of friction is directly proportional to the normal reaction, R. The magnitude of static friction , $f_s$, has a maximum value called limiting friction $f_l$ that is given by, $$f_l={\mu }_{s}R$$ where ${\mu }_s$ is the coefficient of static friction. If the applied force parallel to the surface exceeds $f_l$, then the body begins to slide along the surface.
If the body begins to slide along the surface, the kinetic force of friction $f_k$ comes into play which is given by, $$f_k={\mu }_{k}R$$ where, ${\mu }_k$ is the coefficient of kinetic friction. ${\mu }_k$ and ${\mu }_s$ are dimensionless and depends on the nature of surfaces in contact.

Angle of friction

It is the angle which the resultant of limiting frictional force and the normal force on the body that just begins to slide along the horizontal surface makes with the normal force is called angle of friction $\alpha$ . From fig., In $\mathrm{\Delta }$ OYZ, $$\begin{array}{rl}tan\alpha & =\frac{YZ}{OY}\\ tan\alpha & =\frac{f_l}{R}\\ tan\alpha & ={\mu }_s\because f_l={\mu }_{s}R\end{array}$$ Thus, the coefficient of static friction (or limiting friction) is equal to the tangent of angle of friction.

Angle of repose

It is the minimum angle of inclination of a plane structure with the horizontal such that a body kept on it just begins to slide down along the plane.
In equilibrium condition, $$mgsin\theta =f_s$$ $$mgcos\theta =R$$ On dividing this two equations, we get, $$tan\theta =\frac{f_s}{R}$$ As, $$\begin{array}{rl}\frac{f_s}{R}& =\mu \\ tan\theta & =\mu \\ tan\theta & =tan\alpha \\ \theta & =\alpha \end{array}$$ So, the angle of repose is equal to the angle of friction.

What is the acceleration of a block sliding down a ${30}^{\circ }$ slope if the coefficient of sliding friction between two surfaces is 0.2.
The resulting downward force is, $$\begin{array}{rl}F& =mgsin\theta -f\\ ma& =mgsin\theta -\mu R\because f=\mu R\\ ma& =mgsin\theta -\mu mgcos\theta \\ a& =gsin\theta -\mu gcos\theta \\ & =10\times sin{30}^{\circ }-0.2\times 10\times cos{30}^{\circ }\\ \therefore a& =3.3m/s^2\end{array}$$

A 3.5 kg block is pushed along a horizontal floor by a force $\overrightarrow{F}$ of magnitude 15 N at an angle $\theta ={40}^{\circ }$ with the horizontal. The coefficient of kinetic friction between the block and the floor is 0.25. Calculate the magnitudes of a) the frictional force on the block from the floor and b) the block's acceleration.
Let us consider $f$ be the frictional force.

Here, $m=3.5$ kg, $F=15$ N, $\theta ={40}^{\circ }$ and ${\mu }_k=0.25$. Using free body diagram, Frictional force is given by, $$\begin{array}{rl}f& ={\mu }_{k}R\\ \because R& =mg+Fsin\theta \\ f& ={\mu }_k(mg+Fsin\theta )\\ & =0.25(3.5\times 9.8+15sin{40}^{\circ })\\ \therefore f& =10.99\text{N}\end{array}$$ Also, from the free body diagram, $$\begin{array}{rl}\text{Net force}& =Fcos\theta -f\\ ma& =15\times 0.76-10.99\\ a& =\frac{0.41}{3.5}\\ \therefore a=0.12m/s^2\end{array}$$ Thus, a)the frictional force is 10.99 N and b)block's acceleration is 0.12 m/s${}^2$.

Why is friction a necessary evil?
On the one hand, friction causes the tear and wear of objects and causes the energy loss. So, it could be considered as evil. However, on the other hand, it is very important in our daily life.

1. Friction has made possible to walk on the ground, drive the car, vehicles, etc.
2. If friction were totally absent, we could not write with pencil.
3. Woven clothes would fall apart in absence of friction.

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