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net F y = 0 size 12{"net "F rSub { size 8{y} } =0} {}

where we again call the vertical axis the y -axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that

F p w 1 w 2 = 0 . size 12{F rSub { size 8{p} } - w rSub { size 8{1} } - w rSub { size 8{2} } =0} {}

This equation yields what might have been guessed at the beginning:

F p = w 1 + w 2 . size 12{F rSub { size 8{p} } =w rSub { size 8{1} } +w rSub { size 8{2} } } {}

So, the pivot supplies a supporting force equal to the total weight of the system:

F p = m 1 g + m 2 g . size 12{F rSub { size 8{p} } =m rSub { size 8{1} } g+m rSub { size 8{2} } g} {}

Entering known values gives

F p = 26.0 kg 9.80 m/s 2 + 32.0 kg 9.80 m/s 2 = 568 N.

Discussion

The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the location of the seesaw’s actual pivot!

Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since F p is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force F p is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem.

Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case . Always enter the correct forces—do not jump ahead to enter some ratio of masses.

Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation—the distances r 1 and r 2 are the distances to points directly below the center of gravity    of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point.

Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter.

Take-home experiment

Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies?

Section summary

  • The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be
    τ = rF sin θ size 12{τ= ital "rF""sin"θ} {}

    where τ size 12{τ} {} is torque, r size 12{r} {} is the distance from the pivot point to the point where the force is applied, F size 12{F} {} is the magnitude of the force, and θ size 12{θ} {} is the angle between F size 12{F} {} and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm r size 12{r rSub { size 8{ ortho } } } {} is defined to be

    r = r sin θ size 12{r rSub { size 8{ ortho } } =r"sin"θ} {}

    so that

    τ = r F . size 12{τ=r rSub { size 8{ ortho } } F} {}
  • The perpendicular lever arm r size 12{r rSub { size 8{ ortho } } } {} is the shortest distance from the pivot point to the line along which F size 12{F} {} acts. The SI unit for torque is newton-meter (N·m) . The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero:
    net τ = 0 size 12{"net "τ=0} {}

    By convention, counterclockwise torques are positive, and clockwise torques are negative.

Conceptual questions

What three factors affect the torque created by a force relative to a specific pivot point?

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A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base.

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Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help? (It is also hazardous since it can break the bolt.)

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Problems&Exercises

(a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges?

a) 46.8 N·m

b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges. (Children don’t have a tougher time opening a door because they push lower than adults, they have a tougher time because they don’t push far enough from the hinges.)

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When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. (a) How much torque are you exerting in newton × meters (relative to the center of the bolt)? (b) Convert this torque to footpounds.

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Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.

23.3 N

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Use the second condition for equilibrium (net τ = 0) to calculate F p in [link] , employing any data given or solved for in part (a) of the example.

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Repeat the seesaw problem in [link] with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium.

Given:

m 1 = 26.0 kg, m 2 = 32.0 kg, m s = 12.0 kg, r 1 = 1.60 m, r s = 0.160 m, find (a) r 2 , (b) F p

a) Since children are balancing:

net τ cw = net τ ccw w 1 r 1 + m s gr s = w 2 r 2 alignl { stack { size 12{"net "τ rSub { size 8{"cw"} } =" net "τ rSub { size 8{"ccw"} } } {} #drarrow w rSub { size 8{1} } r rSub { size 8{1} } +m rSub { size 8{s} } ital "gr" rSub { size 8{s} } =w rSub { size 8{2} } r rSub { size 8{2} } {} # {}} } {}

So, solving for r 2 size 12{r rSub { size 8{2} } } {} gives:

r 2 = w 1 r 1 + m s gr s w 2 = m 1 gr 1 + m s gr s m 2 g = m 1 r 1 + m s r s m 2 = ( 26.0 kg ) ( 1.60 m ) + ( 12.0 kg ) ( 0.160 m ) 32.0 kg = 1.36 m

b) Since the children are not moving:

net F = 0 = F p w 1 w 2 w s F p = w 1 + w 2 + w s alignl { stack { size 12{"net "F=0=F rSub { size 8{p} } - w rSub { size 8{1} } - w rSub { size 8{2} } - w rSub { size 8{s} } } {} #drarrow F rSub { size 8{p} } =w rSub { size 8{1} } +w rSub { size 8{2} } +w rSub { size 8{3} } {} } } {}

So that

F p = ( 26.0 kg + 32.0 kg + 12.0 kg ) ( 9.80 m / s 2 ) = 686 N alignl { stack { size 12{F rSub { size 8{p} } = left ("26" cdot 0" kg "+" 32" cdot "0 kg "+"12" cdot "0 kg" right ) left (9 cdot "8 " {m} slash {s rSup { size 8{2} } } right )} {} #= {underline {"686"" N"}} {} } } {}
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Questions & Answers

What does mean ohms law imply
Victoria Reply
what is matter
folajin Reply
Anything that occupies space
Kevin
Any thing that has weight and occupies space
Victoria
Anything which we can feel by any of our 5 sense organs
Suraj
Right
Roben
thanks
Suraj
what is a sulphate
Alo
any answers
Alo
the time rate of increase in velocity is called
Blessing Reply
acceleration
Emma
What is uniform velocity
Victoria
Greetings,users of that wonderful app.
Frank Reply
how to solve pressure?
Cruz Reply
how do we calculate weight and eara eg an elefant that weight 2000kg has four fits or legs search of surface eara is 0.1m2(1metre square) incontact with the ground=10m2(g =10m2)
Cruz
P=F/A
Mira
can someone derive the formula a little bit deeper?
Bern
what is coplanar force?
OLADITI Reply
what is accuracy and precision
Peace Reply
How does a current follow?
Vineeta Reply
follow?
akif
which one dc or ac current.
akif
how does a current following?
Vineeta
?
akif
AC current
Vineeta
AC current follows due to changing electric field and magnetic field.
akif
you guys are just saying follow is flow not follow please
Abubakar
ok bro thanks
akif
flows
Abubakar
but i wanted to understand him/her in his own language
akif
but I think the statement is written in English not any other language
Abubakar
my mean that in which form he/she written this,will understand better in this form, i write.
akif
ok
Abubakar
ok thanks bro. my mistake
Vineeta
u are welcome
Abubakar
what is a semiconductor
Vineeta Reply
substances having lower forbidden gap between valence band and conduction band
akif
what is a conductor?
Vineeta
replace lower by higher only
akif
convert 56°c to kelvin
Abubakar
How does a current follow?
Vineeta
A semiconductor is any material whose conduction lies between that of a conductor and an insulator.
AKOWUAH
what is Atom? what is molecules? what is ions?
Abubakar Reply
What is a molecule
Samuel Reply
Is a unit of a compound that has two or more atoms either of the same or different atoms
Justice
A molecule is the smallest indivisible unit of a compound, Just like the atom is the smallest indivisible unit of an element.
Rachel
what is a molecule?
Vineeta
what is a vector
smith Reply
A quantity that has both a magnitude AND a direction. E.g velocity, acceleration, force are all vector quantities. Hope this helps :)
deage
what is the difference between velocity and relative velocity?
Mackson
Velocity is the rate of change of displacement with time. Relative velocity on the other hand is the velocity observed by an observer with respect to a reference point.
Chuks
what do u understand by Ultraviolet catastrophe?
Rufai
A certain freely falling object, released from rest, requires 1.5seconds to travel the last 30metres before it hits the ground. (a) Find the velocity of the object when it is 30metres above the ground.
Mackson
A vector is a quantity that has both magnitude and direction
Rufus
the velocity Is 20m/s-2
Rufus
derivation of electric potential
Rugunda Reply
V = Er = (kq/r^2)×r V = kq/r Where V: electric potential.
Chuks
what is the difference between simple motion and simple harmonic motion ?
syed
hi
Peace
hi
Rufus
hi
Chip
simple harmonic motion is a motion of tro and fro of simple pendulum and the likes while simple motion is a linear motion on a straight line.
Muinat
a body acceleration uniform from rest a 6m/s -2 for 8sec and decelerate uniformly to rest in the next 5sec,the magnitude of the deceleration is ?
Patricia Reply
The wording not very clear kindly
Moses
6
Leo
9.6m/s2
Jolly
the magnitude of deceleration =-9.8ms-2. first find the final velocity using the known acceleration and time. next use the calculated velocity to find the size of deceleration.
Mackson
wrong
Peace
-3.4m/s-2
Justice
Hi
Abj
Firstly, calculate final velocity of the body and then the deceleration. The final ans is,-9.6ms-2
Muinat
8x6= 48m/-2 use v=u + at 48÷5=9.6
Lawrence
can i define motion like this motion can be define as the continuous change of an object or position
Shuaib Reply
Any object in motion will come to rest after a time duration. Different objects may cover equal distance in different time duration. Therefore, motion is defined as a change in position depending on time.
Chuks
Practice Key Terms 4

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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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