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Similar observations can be made using a meter stick held at different locations along its length.

A pole vaulter is standing on the ground holding a pole with his two hands. The center of gravity of the pole is between the hands of the pole vaulter and is near the right hand of the man. The weight W is shown as an arrow downward toward center of gravity. The reactions F sub R and F sub L of the hands of the man are shown with vectors in upward direction. A free body diagram of the situation is shown on the top right side of the figure.
A pole vaulter holds a pole horizontally with both hands.

A pole vaulter is standing on the ground holding a pole with his two hands. The center of gravity of the pole is between the hands of the pole vaulter and is near the right hand of the man. The weight W is shown as an arrow downward toward center of gravity. The reactions F sub R and F sub L of the hands of the man are shown with vectors in upward direction. A free body diagram of the situation is shown on the top right side of the figure.
A pole vaulter is holding a pole horizontally with both hands. The center of gravity is near his right hand.

A pole vaulter is standing on the ground holding a pole from one side with his two hands. The centre of gravity of the pole is to the left of the pole vaulter. The weight W is shown as an arrow downward at center of gravity. The reaction F sub R is shown with a vector pointing downward from the man’s right hand and F sub L is shown with a vector in upward direction at the location of the man’s left hand. A free body diagram of the situation is shown on the top right side of the figure.
A pole vaulter is holding a pole horizontally with both hands. The center of gravity is to the left side of the vaulter.

If the pole vaulter holds the pole as shown in [link] , the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If F L = F R size 12{F rSub { size 8{L} } =F rSub { size 8{R} } } {} , then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces F L size 12{F rSub { size 8{L} } } {} and F R size 12{F rSub { size 8{R} } } {} is straightforward, as the next example shows.

If the pole vaulter holds the pole from near the end of the pole ( [link] ), the direction of the force applied by the right hand of the vaulter reverses its direction.

What force is needed to support a weight held near its cg?

For the situation shown in [link] , calculate: (a) F R size 12{F rSub { size 8{R} } } {} , the force exerted by the right hand, and (b) F L size 12{F rSub { size 8{L} } } {} , the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand.

Strategy

[link] includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (net F = 0 size 12{F=0} {} ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (net τ = 0 ) size 12{ ital "net"`τ=0} {} if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand.

Solution for (a)

There are now only two nonzero torques, those from the gravitational force ( τ w size 12{τ rSub { size 8{W} } } {} ) and from the push or pull of the right hand ( τ R size 12{τ rSub { size 8{R} } } {} ). Stating the second condition in terms of clockwise and counterclockwise torques,

net τ cw = –net τ ccw . size 12{"net "τ rSub { size 8{"cw"} } ="net"τ rSub { size 8{"ccw"} } } {}

or the algebraic sum of the torques is zero.

Here this is

τ R = –τ w

since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise torque. Using the definition of torque, τ = rF sin θ size 12{τ= ital "rF""sin"θ} {} , noting that θ = 90º size 12{θ} {} , and substituting known values, we obtain

0 . 900 m F R = 0 .600 m mg . size 12{ left (0 "." "900"" m" right ) left (F rSub { size 8{R} } right )= left (0 "." "600"" m" right ) left ( ital "mg" right )} {}

Thus,

F R = 0.667 5.00 kg 9.80 m/s 2 = 32.7 N.

Solution for (b)

The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law:

F L + F R mg = 0 size 12{F rSub { size 8{L} } +F rSub { size 8{R} } - ital "mg"=0} {}

From this we can conclude:

F L + F R = w = mg size 12{F rSub { size 8{L} } +F rSub { size 8{R} } =w= ital "mg"} {}

Solving for F L size 12{F rSub { size 8{L} } } {} , we obtain

F L = mg F R = mg 32 . 7 N = 5.00 kg 9.80 m/s 2 32.7 N = 16.3 N alignl { stack { size 12{F rSub { size 8{L} } = ital "mg" - F rSub { size 8{R} } } {} #= ital "mg" - "32" "." 7 {} # =0 "." "333" ital "mg" {} #= left (0 "." "333" right ) left (5 "." "00"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right ) {} # ="16" "." 3" N" {}} } {}

Discussion

F L size 12{F rSub { size 8{L} } } {} is seen to be exactly half of F R size 12{F rSub { size 8{R} } } {} , as we might have guessed, since F L is applied twice as far from the cg as F R .

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If the pole vaulter holds the pole as he might at the start of a run, shown in [link] , the forces change again. Both are considerably greater, and one force reverses direction.

Take-home experiment

This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity!

Phet explorations: balancing act

Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game.

Balancing Act

Summary

  • Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies , still apply.

Conceptual questions

When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly above the person’s neck vertebrae.

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Problems&Exercises

To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?

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In [link] , the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in [link] , show that the second condition for equilibrium (net τ = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above.

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Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir
Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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