Force 

Newton's law of motion

•First law: “A body continues in its state of rest or uniform motion in a straight line unless acted by external force”

-It gives the idea of inertia and gives the qualitative definition of force. It is the condition for equillibirium of body. eg A book lying on table, moving fan etc.

•Second law: “The rate of change of momentum is directly proportional to the force applied and change takes place on direction of force.”

Momentum: It is the measure of quantity of motion contained on body. Let p be the momentum and v be the velocity then,

$\vec{p}=m\vec{v}$

i.e, $F=\frac{d\vec{P}}{dt}=\frac{d(m\vec{v})}{dt}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}$

-It gives the quantitative definition of force.

•Third law: “To every action there is equal and opposite reaction.”

$\overrightarrow{F_{\text{12}}}=-\overrightarrow{F_{\text{21}}}$

It states that force exists on pair.

Motion of lift and apparent weight

Let, M be mas of lift ,m=mass of body ,T=tension on rope and R=apparent weight.R

(i) When lift moves up or down with constant Velocity(a=0)

mg-R=ma

so R=mg

∴T=(M+m)g

(ii)When lift moves upward with constant acceleration(a)

R-mg=ma

so, R=mg+ma

∴T=(M+m)(g+a)

$\therefore$net acceleration =(g+a)

(iii)When lift moves downward with constant acceleration(a)

mg-R=ma

so, R=mg-ma

∴T=(M+m)(g-a)

$\therefore$net acceleration=(g-a)

(iv) If a=g(free fall) R=0 weightlessness.                                  

(v)If a>g, R is negative i.e acts upward person will rise from floor.   

System of variable mass

Example:

What is the force needed to hold pipe in which water is flowing with velocity v?

We have,

$F=v.\frac{dm}{dt}$

or, $F=v.\frac{d(\rho Al)}{dt}=v\ast \rho A\ast \frac{dl}{dt}=\rho A{v}^2$ (away from himself)

Case I: When water rebounds with equal speed

F= ρAv(v-(-v))=2ρAv²

Case II:When directed at angle θ with horizontal and rebounding is neglected,

F= ρAv(vcosθ -0)= ρAv²cosθ

Case III:When directed at angle θ with horizontal and rebounds at same angle.

F= ρAv[vcosθ –(-vcosθ)] = 2ρAv²cosθ

Mass pulley system

(i)If m₂>m₁ Net acceleration=$\frac{\text{net force}}{\text{Total mass}}$

                             =$\frac{m_{2}g-m_{1}g}{m_2+m_1}=\frac{(m_2-m_1)g}{(m_2+m_1)}$

Also T-m₁g=m₁a and m₂g-T=m₂a

Solving, $T=\frac{2m_2{m}_1}{m_1+m_2}g$

Note, the net thrust on pulley is T+T=2T

(ii) T=m₂a

M₁g-T=m₁a

Also net acceleration=$\frac{\text{net force}}{\text{Total mass}}$

                                     =$\frac{m_{1}g}{m_2+m_1}$

$T=\frac{2m_2{m}_1}{m_1+m_2}g$

Note, the net thrust on pulley is , $\sqrt{T^2+T^2}=\sqrt{2}T$

If m₂gsinβ1gsinα , 

Net acceleration=$\frac{\text{net force}}{\text{Total mass}}$

  =$\frac{-m_{2}gsin\beta +m_{1}gsin\alpha }{m_2+m_1}$

Also, m₁gsinα-T=m₁a and T -m₂gsinβ =m₂a

Solving,

 $T=\frac{m_2{m}_1}{m_2+m_1}g(sin\alpha +sin\beta )$

Note, the net thrust on pulley can be found  vectorically.

When moving carriage is given horizontal force

•If a mass suspended on moving carriage is given horizontal force:

Tcosθ=mg and Tsinθ=ma

So, Tanθ=$\frac{a}{g}$

And Tension on rope is

$T=m\sqrt{a^2+g^2}$

Rocket Motion

•It based on conservation of linear momentum states that, total momentum remain constant.

∑mv=constant

For two bodies,

$m_1{v}_1+m_2{v}_2=m_1{u}_1+m_2{u}_2$

Rocket motion,

V=u*$ln\frac{m_o}{m}$

Where, V=velocity, u=exhaust speed of gas

m₀=initial mass, m=residual mass

Real weight on pan

If an object is placed on one pan its weight is ‘w1’, in another pan its weight is w2, then true weight of object:

•If balance has equal arms, true weight is $w=\frac{w1+w2}{2}$

•If balance has unequal arms, true weight is $w=\sqrt{w_1.w_2}$

Contact forces on smooth surfaces

The forces developed on surface of contact between two bodies is called Contact force.

•Acceleration of system is

$a=\frac{F}{m_1+m_2}$

And F-F’=m₁a, F’=m₂a

Solving, contact force is

$F=\frac{m_{2}F}{m_1+m_2}$

If force was applied on block m₂ ,

 contact force is $F=\frac{m_{1}F}{m_1+m_2}$

Tension on connected systems

Acceleration of system is $a=\frac{F}{(m_1+m_2+m_3)}$

$T_1=\frac{F(m_2+m_3)}{(m_1+m_2+m_3)}$

$T_2=\frac{F{m}_3}{(m_1+m_2+m_3)}$  

Generally, tension on any part of string is = $\frac{force\ast (\text{sum of masses behind it})}{\text{total sum of masses}}$

Reaction on different surfaces

Rxn always lies perpendicular to contact plane.

(i) On simple plane:

Reaction =weight(R=mg)

(ii)On inclined plane:

R=mgcosθ

(iii)Pulled by force F

R=mg-Fsinθ

And ma=Fcosθ

(iv)Pushed by force F

R=mg+Fsinθ>mg

ma=Fcosθ

Hence, pulling is easier than pushing because It reduces normal reaction.

Problem on chain or rope

•At a distance x from force applied end. (M is total mass of rope)

-Mass of remaining portion is ,$m=\frac{L-x}{L}\ast M$

Acceleration of each part is, $a=\frac{F}{M}$

So, Tension at C is $T_c=m\ast a$

 or, $T_c=\frac{L-x}{L}\ast F$

Similarly, for suspended vertically tension at x distance from free end (force applied end) is,

$T_c=\frac{L-x}{L}\ast F+\frac{xg}{L}\ast M$

Example

•If a stone of mass m is dropped from height h  and it penetrates s distance on ground, then,

average retardation is $a=\frac{gh}{s}$ (as,energy conservation, mgh=mas)

-Time of penetration is$t=\sqrt{\frac{2s^2}{gh}}$ $(0=\sqrt{2gh}-at)$

-Average resistance offered by ground is,

R=mg+ma=$mg(1+\frac{a}{g})$= $mg(1+\frac{h}{s})$

•Note that, if a man takes a quick step on platform his weight first decreases then increases.

Impulse and Momentum

•It is the product of force and time.If I denotes impulse, F is force and dt be small interval of time then∫$\vec{F}.dt$

$\vec{I}=\vec{F}.dt$=$\int \frac{d\vec{P}}{dt}$. dt =$dp=p_f-p_i$

Momentum: It is the product of mass and velocity.

It is vector and denoted by $\vec{p}$ and given by

$\vec{p}=m\vec{v}$ where, m is mass of body and v is velocity.