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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

advantages of CRO over ordinary voltmeter
Dismas Reply
what is the difference between displacement and distance?!
Daniel Reply
what is equilibrium
Sade Reply
If a system is said to be under equilibrium whenever there is no force act upon it... And it remain in its initial stage..
soniya
What is conductivity
Saud Reply
It is the ease with which electrical charges or heat can be transmitted through a material or a solution.
Cffrrcvccgg
how to find magnitude and direction
Arjune Reply
how to caclculate for speed
Arjune
derivation of ohms law
Kazeem Reply
derivation of resistance
Kazeem
R=v/I where R=resistor, v=voltage, I=current
Kazeem
magnitude
Arjune
A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t 0 s, the x components of the puck’s ini￾tial velocity and acceleration are v0x 1.0 m/s and ax 2.0 m/s2 . The y components of the puck’s initial velocity and acceleration are v0y 2.0 m/s and ay 2.0
Arjune
Electric current is the flow of electrons
Kelly Reply
is there really flow of electrons exist?
babar
Yes It exists
Cffrrcvccgg
explain plz how electrons flow
babar
if electron flows from where first come and end the first one
babar
an electron will flow accross a conductor because or when it posseses kinectic energy
Cffrrcvccgg
electron can not flow jist trasmit electrical energy
ghulam
free electrons of conductor
ankita
electric means the flow heat current.
Serah Reply
electric means the flow of heat current in a circuit.
Serah
What is electric
Manasseh Reply
electric means?
ghulam
electric means the flow of heat current in a circuit.
Serah
electric means the flow of electric current through conductor
Sade
the continuos flow of electrons in a circuit is called electric
ANUBHA
electric means charge
ghulam
electric means current
Sade
a boy cycles continuously through a distance of 1.0km in 5minutes. calculate his average speed in ms-1(meter per second). how do I solve this
Jenny Reply
speed = distance/time be sure to convert the km to m and minutes to seconds check my utube video "mathwithmrv speed"
PhysicswithMrV
d=1.0km÷1000=0.001 t=5×60=300s s=d\t s=0.001/300=0.0000033m\s
Serah
A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t 0 s, the x components of the puck’s ini￾tial velocity and acceleration are v0x 1.0 m/s and ax 2.0 m/s2 . The y components of the puck’s initial velocity and acceleration are v0y 2.0 m/s and ay 2.0
Arjune
D=1km=1000m t=5mins×60secs=300sec s=d/t=3.333m/s
Daniel
I think Daniel Glorious is ryt
Amalia
why we cannot use DC instead of AC in a transformer
kusshaf Reply
becuse the d .c cannot travel for long distance trnsmission
ghulam
what is physics
Chiwetalu Reply
branch of science which deals with matter energy and their relationship between them
ghulam
Life science
the
what is heat and temperature
Kazeem Reply
how does sound affect temperature
Clement Reply
sound is directly proportional to the temperature.
juny
how to solve wave question
Wisdom Reply
I would like to know how I am not at all smart when it comes to math. please explain so I can understand. sincerly
Emma
Just know d relationship btw 1)wave length 2)frequency and velocity
Talhatu
First of all, you are smart and you will get it👍🏽... v = f × wavelength see my youtube channel: "mathwithmrv" if you want to know how to rearrange equations using the balance method
PhysicswithMrV
nice self promotion though xD
Beatrax
thanks dear
Chuks
hi pls help me with this question A ball is projected vertically upwards from the top of a tower 60m high with a velocity of 30ms1.what is the maximum height above the ground level?how long does it take to reach the ground level?
mahmoud
what is scalar quantities
babatunde
scaler quantity are quanties that have only direction and no magnitude
Natsu
ice Point
babatunde
scalar quantity are quantities that have magnitude but no direction
Ehigiamusoe
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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