Grade 11 Physics Note

Equilibrium

State principle of moments:

It states that, “If an object is equilibrium under the action of many forces acting on it. Then, the sum of moments of all forces acting about a point is zero.”

I.e. clockwise moment = anticlockwise moment

Let’s take a rod in which it is pivoted at point O such that it is in equilibrium at point O under the action of force f1, f2, f3, f4. Then, at equilibrium,

${{\rm{f}}_1}{\rm{*}}{{\rm{A}}_0} + {{\rm{f}}_2}{\rm{*}}{{\rm{B}}_0} - {{\rm{f}}_3}{\rm{*}}{{\rm{O}}_{\rm{C}}} - {{\rm{f}}_4}{\rm{*OD}} = 0$

${\rm{or}},{\rm{\: }}{{\rm{f}}_1}{\rm{*}}{{\rm{A}}_0} + {{\rm{f}}_2}{\rm{*}}{{\rm{B}}_0} = {{\rm{f}}_3}{\rm{*}}{{\rm{O}}_{\rm{c}}} + {{\rm{f}}_4}{\rm{*OD}}$

Or, anticlockwise moment = clockwise moment               

 

Couple and momentum of a couple:

A couple is a pair of forces, equal in magnitude, oppositely directed, and displaced by perpendicular distance or moment.

The product of force and perpendicular distance from the axis of relation is called moment of force and is also called torque. So,

Moment of force = f ${\rm{*}}$ r

The SI unit of moment of force is Nm.

Dimensions: $\left[ {{\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}} \right]$

The torque on a body is measured by the product of the force applied and the torque arm(r), i.e.,

${\rm{torque\: }}\left( {\rm{T}} \right) = {\rm{force}}\left( {\rm{F}} \right){\rm{*Torque\: arm}}\left( {\rm{r}} \right)$

Torque Due to a Couple

http://www.ustudy.in/imagebrowser/view/image/10190/_original

 

Two equal unlike parallel forces acting on a body at different points constitute a couple.

The moment of the couple is called torque.

Hence, torque due to a couple Γ = force * couple arm=F * l

 

Center of mass (C.M):

Center of mass is the point on the object in which the applied force produces acceleration without rotation. Generally, the center of mass lies within the object but in some object no mass lies at the center of mass.

In a ring or hollow sphere, no mass lies at the center of mass because the center of mass in such object lies at the center which is empty.

Generally, the center of gravity and center of mass coincide. If two object rotate about center of mass, then we have,

${{\rm{m}}_1}{{\rm{x}}_1} = {{\rm{m}}_2}{{\rm{x}}_2}$

But, if a rigid body is rotating about the center of mass then, the co-ordinate of center of mass is

${\rm{X}} = \frac{{{{\rm{m}}_1}{{\rm{x}}_1} + {{\rm{m}}_2}{{\rm{x}}_2} +  \ldots  + {{\rm{m}}_{\rm{n}}}{{\rm{x}}_{\rm{n}}}}}{{{{\rm{m}}_1} + {{\rm{m}}_2} +  \ldots  + {{\rm{m}}_{\rm{n}}}}}$

$ = \frac{{ \in {\rm{mx}}}}{{ \in {\rm{m}}}} = \frac{{ \in {\rm{mx}}}}{{ \in {\rm{M}}}}$

And ${\rm{Y}} = \frac{{ \in {\rm{my}}}}{{\rm{M}}}$

Centre of mass of two bodies;

Suppose two bodies of masses m1 and m2.

Let masses be connected by rod and their centre of mass be C.

Let x be the distance

Then,

m1 x1 = m2x2

or   x1 = x2$\frac{{{{\rm{m}}_2}}}{{{{\rm{m}}_1}}}$

x2 = x1$\frac{{{{\rm{m}}_1}}}{{{{\rm{m}}_2}}}$

So greater the mass of the body, nearer is their C.G. and so on.

 

Lami’s theorem:

If a body is in equilibrium under the action of three forces, then each of the forces proportional to the site of the angle between the other two forces.

http://img.bhs4.com/9e/c/9ec49a51565272a51739660a331f37e6eadfaa6d_large.jpg

 

Consider the three forces P, Q, R are acting on the particles A such that particles is in equilibrium.

${\rm{if\: }}\alpha ,\beta ,\gamma $  are the angles between the three forces as shown in the figure.

From Lami’s theorem, mathematically, we have

$\frac{{\rm{P}}}{{{\rm{sin}}\alpha }} = \frac{{\rm{Q}}}{{{\rm{sin}}\beta }} = \frac{{\rm{R}}}{{{\rm{sin}}\gamma }}{\rm{\: }}$

if the body is in equilibrium.

$\frac{{{\rm{AB}}}}{{{\rm{sin}}\alpha }} = \frac{{{\rm{BC}}}}{{{\rm{sin}}\beta }} = \frac{{{\rm{CA}}}}{{{\rm{sin}}\gamma }}$

$\frac{{\rm{P}}}{{{\rm{sin}}180 - \alpha }} = \frac{{\rm{Q}}}{{{\rm{sin}}180 - \beta }} = \frac{{\rm{R}}}{{{\rm{sin}}180 - \gamma }}$

$\frac{{\rm{P}}}{{{\rm{sin}}\alpha }} = \frac{{\rm{Q}}}{{{\rm{sin}}\beta }} = \frac{{\rm{R}}}{{{\rm{sin}}\gamma }}{\rm{\: }}$

Hence lami’s theorem proved

 

The conditions under which a rigid body remains in equilibrium under action of a set of coplanar forces:

Generally, the base of the object should lie in lowest position so that the object becomes more stable. Similarly, the area of base should also be large. Also, the vertical line passing through C.G should lie on the large area of base. Under such three conditions, the objects become stable. 

A rigid body will be in equilibrium if the following two conditions are met.

1. The vector sum of the forces acting on the body must be zero

2. The net torque acting on the body must be zero.

Forces acting in a single plane or in a same plane are called coplanar forces. If only two forces act through a point, they must be coplanar. However none parallel forces that do not act through a point cannot be coplanar. Three or more non-parallel forces acting through a point may not be coplanar.

Co-planner force:

If the line of action of all forces lies on a plane then, these forces are called co-planer forces. If the sum of two forces is equal and opposite to third force then, their resultant will be zero, but these all forces should be co-planner.

Let${\rm{\vec P}}$, ${\rm{\vec Q}}$ and ${\rm{\vec R}}$ be three co-planner forces such that

${\rm{\vec P}}$ +${\rm{\vec Q}}$ = $\overrightarrow { - {\rm{R}}} $ , then their resultant or sum is zero.

i.e${\rm{\vec P}} + {\rm{\vec Q}} + {\rm{\vec R}} = 0$

 

It easier to open or close a door by pulling from a point nearer its outermost edge than pulling nearer the hinge:

It easier to open or close a door by pulling from a point nearer its outermost edge than pulling nearer the hinge because the centre of mass of a body is point where all the mass of the body can be considered to be concentrated. Therefore we can replace whole body by a single particle located at the centre of mass.

 

 

It is easier to stand on two legs than on one leg:

It is easier to stand on two legs than on one leg. A body will be stable if its base area is more and C.G. of the body lies within the base area as low as possible. The base area of two legs on the ground is much more than that of the one leg.

 

Two unequal coplanar forces acting together produce condition of equilibrium:

Forces acting in a single plane or in a same plane are called coplanar forces. If only two forces act through a point, they must be coplanar. However none parallel forces that do not act through a point cannot be coplanar. Three or more non-parallel forces acting through a point may not be coplanar.

 

It easier to hold a load with arm folded than outstretched:

The centre of mass of a body is point where all the mass of the body can be considered to be concentrated thus the force acts at this point, it easier to hold a load with arm folded than outstretched. 


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