# 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,

i.e,

-It gives the quantitative definition of force.

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

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)

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)

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,

or, (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=

=

Also T-m1g=m1a and m2g-T=m2a

Solving,

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

(ii) T=m2a

M1g-T=m1a

Also net acceleration=

=

Note, the net thrust on pulley is ,

If m2gsinβ1gsinα ,

Net acceleration=

=

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

Solving,

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=

And Tension on rope is

Rocket Motion

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

∑mv=constant

For two bodies,

Rocket motion,

V=u*

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

•If balance has unequal arms, true weight is

Contact forces on smooth surfaces

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

•Acceleration of system is

And F-F’=m1a, F’=m2a

Solving, contact force is

If force was applied on block m2 ,

contact force is

Tension on connected systems

Acceleration of system is

Generally, tension on any part of string is =

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 ,

Acceleration of each part is,

So, Tension at C is

or,

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

Example

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

average retardation is (as,energy conservation, mgh=mas)

-Time of penetration is

-Average resistance offered by ground is,

R=mg+ma==

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∫

=. dt =

Momentum: It is the product of mass and velocity.

It is vector and denoted by and given by

where, m is mass of body and v is velocity.