# 12.2 Examples of static equilibrium  (Page 8/9)

 Page 8 / 9

We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P :

$\text{pivot at}\phantom{\rule{0.2em}{0ex}}P\text{:}\phantom{\rule{0.2em}{0ex}}{\tau }_{w}+{\tau }_{Bx}+{\tau }_{By}=0.$

We use the free-body diagram to find all the terms in this equation:

$\begin{array}{ccc}\hfill {\tau }_{w}& =\hfill & dw\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\text{−}\beta \right)=\text{−}dw\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\beta =\text{−}dw\frac{b\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}2}{d}=\text{−}w\frac{b}{2}\hfill \\ \hfill {\tau }_{Bx}& =\hfill & a{B}_{x}\text{sin}\phantom{\rule{0.2em}{0ex}}90\text{°}=+a{B}_{x}\hfill \\ \hfill {\tau }_{By}& =\hfill & a{B}_{y}\text{sin}\phantom{\rule{0.2em}{0ex}}180\text{°}=0.\hfill \end{array}$

In evaluating $\text{sin}\phantom{\rule{0.2em}{0ex}}\beta ,$ we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into [link] and compute ${B}_{x}:$

$\text{pivot at}\phantom{\rule{0.2em}{0ex}}P\text{:}\phantom{\rule{0.2em}{0ex}}\text{−}w\phantom{\rule{0.1em}{0ex}}\frac{b}{2}+a{B}_{x}=0\phantom{\rule{0.5em}{0ex}}⇒\phantom{\rule{0.5em}{0ex}}{B}_{x}=w\phantom{\rule{0.1em}{0ex}}\frac{b}{2a}=\left(400.0\phantom{\rule{0.2em}{0ex}}\text{N}\right)\phantom{\rule{0.1em}{0ex}}\frac{1}{2·2}=100.0\phantom{\rule{0.2em}{0ex}}\text{N.}$

Therefore the magnitudes of the horizontal component forces are ${A}_{x}={B}_{x}=100.0\phantom{\rule{0.2em}{0ex}}\text{N}.$ The forces on the door are

$\begin{array}{}\\ \text{at the upper hinge:}\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{F}}_{A\phantom{\rule{0.2em}{0ex}}\text{on door}}=-100.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i}+200.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j}\hfill \\ \text{at the lower hinge:}{\stackrel{\to }{F}}_{B\phantom{\rule{0.2em}{0ex}}\text{on door}}=\text{+}100.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i}+200.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j}.\hfill \end{array}$

The forces on the hinges are found from Newton’s third law as

$\begin{array}{}\\ \\ \text{on the upper hinge:}\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{F}}_{\text{door on}\phantom{\rule{0.2em}{0ex}}A}=100.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i}-200.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j}\hfill \\ \text{on the lower hinge:}\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{F}}_{\text{door on}\phantom{\rule{0.2em}{0ex}}B}=-100.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i}-200.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j}.\hfill \end{array}$

## Significance

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.

Check Your Understanding Solve the problem in [link] by taking the pivot position at the center of mass.

${\stackrel{\to }{F}}_{\text{door on}\phantom{\rule{0.2em}{0ex}}A}=100.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i}-200.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{;}\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{F}}_{\text{door on}\phantom{\rule{0.2em}{0ex}}B}=-100.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i}-200.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j}$

Check Your Understanding A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

711.0 N; 466.0 N

Check Your Understanding A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

1167 N; 980 N directed upward at $18\text{°}$ above the horizontal

## Summary

• A variety of engineering problems can be solved by applying equilibrium conditions for rigid bodies.
• In applications, identify all forces that act on a rigid body and note their lever arms in rotation about a chosen rotation axis. Construct a free-body diagram for the body. Net external forces and torques can be clearly identified from a correctly constructed free-body diagram. In this way, you can set up the first equilibrium condition for forces and the second equilibrium condition for torques.
• In setting up equilibrium conditions, we are free to adopt any inertial frame of reference and any position of the pivot point. All choices lead to one answer. However, some choices can make the process of finding the solution unduly complicated. We reach the same answer no matter what choices we make. The only way to master this skill is to practice.

## Conceptual questions

Is it possible to rest a ladder against a rough wall when the floor is frictionless?

Show how a spring scale and a simple fulcrum can be used to weigh an object whose weight is larger than the maximum reading on the scale.

(Proof)

#### Questions & Answers

y oxygen is shift to first place
What is coriolis firce
Angular accelaration force,as the result of the rotatation of earth
Arzoodan
Due to the rotation of the earth the winds at the equator get deflected in opposite direction and therefore cause some currents in the Northern and Southern hemisphere which are opposite in there spin.
Varsha
Thats correct.
Arzoodan
if cos = x/y then sec = y/x if sin = y/x then cosec = x/y
if cosβ=x/y then what is cosecβ
. if cosβ=x/y what is cosecβ
Abubakar
if cosΦ=x/y what is cosecΦ
Abubakar
What are the unknown symbols?
Jan
Î²
Jan
Î¦
Jan
It is defined that cosec = 1/sin and sec = 1/cos
Jan
Do you understand this?
Jan
of cours
Arzoodan
∇(f/g) = (g∇f − f∇g)/g^2 , at points x where g(x) 6= 0 please help me to solve the problem ....
sir please add answer sheet of spigel vector analysis ..
In a pulley system 2 boxes r hanging in both sides of pulley. An other box was joined to the left box through a rope & get accelerated downward. If all the boxes have same mass ,then what will be the acceleration of that system ?
it may be mg
Amalesh
∇(f/g) = (g∇f − f∇g)/g2 , at points x where g(x) 6= 0 plus solve it
11
What is momentum
is the product of mass to it's velocity (mv)
Ahmed
momentum
quatity of motion present in a body or product of mass and velocity
Shakeel
show that the kE of a uniform ring of mass m rolling along a smooth horizontal surface so that its centre of mass has a velocity v is mv×v
folder
what is hydration energy
the energy................................................
Rika
what is momentum
dont know want to know the answer
Rika
it's the product of mass multiplied by velocity
harsha
the quantity of motion possessed by a body is called its momentum.by virtue of which a body can exert a force in the agency which tend to stop it .it is a common experience that stronger force is required to stop more massive body.also faster the body moves harder it is to stop it .this is why
Manoj
momentum is product of mass and velocity and it is denoted by "P".
Manoj
show that the cross product of vector axb=-bxa
Nature of the physical low
1/2 mv2 =mgh prove that
Rashika
study work energy theorem properly
Rohan
v2=u2+2gh...u=0...so 1/2 mv2=1/2m(2gh)=mgh
Talha
how can I become confident in physics
practice more and more question ,,,,if u get trouble in any one question infact do that again and again.
jyotirmayee
is theory important or numericals
arshdeep
both ,,,,because without understanding theory ,,,,u can't able to solve problemsss or numericals
jyotirmayee
which books I can follow for IIT jee prep for mcq and which for mcq
arshdeep
what can make me good in physics
lovet
how can I solve qsns in physics cuz I understand most of the theories but facing problem during solving the qsns.
Rohan
what can make me good in physics
Rika
Can someone please tell what really happened to planet Pluto
Pluto was consider as a dwarf planet and not a planet a planet should satisfy the three criteria viz.it should have a orbit around the sun,it should be spherical in shap,and it should clear its orbit . since,Pluto satisfy the first two criteria so it was not considered as a planet .
Manoj
but it was seen that earth also not cleared its orbit but earth is a planet so another criteria was made that a planet should have grater gravitational force other than its orbit and earth follow this but Pluto wasn't satisfied this criteria .so Pluto wasn't consider as a planet.since it follow the
Manoj
first two criteria so it gets a another name called dwarf planet
Manoj
what are the first 20 elements in periodic table.
what happened to the rest now, or have you forgotten?
Ejiba
Hydrozen helium lithium berrilium boron carbon nitrogen oxygen flourine neon sodium magnesium aluminium silicon phosphorous sulphur chlorine argon potassium calcium
Rika
you said 20 elements and thats here
Rika
correct
osobase
yes i know
Rika
excuse me any one here
Rika
yes
harsha
what is momentum
Rika
it is the product of mass and velocity
harsha
thanks
Rika
momentum p = mass X velocity
harsha
Hydration energy?
osobase
p= m*v
harsha
Hydration enery?
Rika
Hydration energy?
Rika
The heat energy released when new bonds are made between the ions and water molecules is known as the hydration enthalpy of the ion
harsha
hydration energy is also called as hydration enthalpy.
harsha
yes what is hydration Energy?
osobase
correct
osobase
Tripple point of water?
osobase
yes it is.
harsha
where water exists in all the three states
harsha
at that point it exists as water (liquid) , ice (solid) and vapours (gas)
harsha
yeap
Rika
What is d tripple point pressure of water?
osobase
it's 6 milli bar
harsha
in mmhg?
osobase
mmHg?
osobase
I can't remember the exact number
harsha
76 cm of Hg = 1 bar
harsha
calculate it for 6millibar
harsha
wrong
osobase
okay.
harsha
4.6mmHg
osobase
what is hydrodynamics
Manoj
the branch of physics that deal with the motion of fluids and the force acting on a solid body immersed in it . fluids means a substance that has no fixed shape and yields easily to external pressure viz.gas or liquid in other words a substance that able to flow easily.
Manoj