Statics - Rutgers Physics & Astronomy · • The first condition of equilibrium is a statement of translational equilibrium • The net external force on the object must equal zero

Statics

Physics 116 2017

Tues. 2/7 and Thurs. 2/9

Conditions for Static Equilibrium • Equilibrium and static

equilibrium

• Static equilibrium conditions – Net external force must

equal zero

– Net external torque must equal zero

• Center of gravity or center of mass

• Solving static equilibrium problems

• The first condition of equilibrium is a statement of translational equilibrium

• The net external force on the object must equal zero

• It states that the translational acceleration of the object’s center of mass must be zero

Conditions for Equilibrium





0 amFF extnet





0 amFF extnet


• The second condition of equilibrium is a statement of rotational equilibrium

• The net external torque on the object must equal zero

• It states the angular acceleration of the object to be zero

• This must be true for any axis of rotation




• It states the angular acceleration of the object to be zero for any axis of rotation

0

Iextnet




• It states the angular acceleration of the object to be zero for any axis of rotation

0

Iextnet

A seesaw consisting of a uniform board of mass mpl and length L supports at rest a father and daughter with masses M and m, respectively. The support is under the center of gravity of the board, the father is a distance x from the center, and the daughter is a distance 2.00 m from the center. A) Find the magnitude of the upward force n exerted by the support on the board. B) Find where the father should sit to balance the system at rest.

A seesaw consisting of a uniform board of mass mpl and length L supports at rest a father and daughter with masses M and m, respectively. The support is under the center of gravity of the board, the father is a distance x from the center, and the daughter is a distance 2.00 m from the center. A) Find the magnitude of the upward force n exerted by the support on the board. B) Find where the father should sit to balance the system at rest.

A) Find the magnitude of the upward force n exerted by the support on the board.


0

0

0

,,

,,

,,

zextznet

yextynet

xextxnet

FF

FF


0

0

0

,,

,,

,,

zextznet

yextynet

xextxnet

FF

FF

gMgmgmn

gMgmgmnFF

FF

pl

plyextynet

xextxnet

0

0

,,

,,


0

0

0

,,

,,

,,

zextznet

yextynet

xextxnet

FF

FF

m00.22

000

0

,

,

M

m

M

mdx

Mgxmgd

Mgxmgdznet

nplfdextznet

B) Find where the father should sit to balance the system at rest.

Axis of Rotation

• The net torque is about an axis through any point in the xy plane

• Does it matter which axis you choose for calculating torques?

• NO. The choice of an axis is arbitrary

• If an object is in translational equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis

• We should be smart to choose a rotation axis to simplify problems

Axis of Rotation






Axis of Rotation






Axis of Rotation






Axis of Rotation






B) Find where the father should sit to balance the system at rest.

Rotation axis O Rotation axis P

O P

M

m

M

mdx

Mgxmgd

Mgxmgdznet

nplfdznet

2

000,

,

M

m

M

mdx

Mgxmgd

dgmmgMggdmMgxMgd

ndgdmdxMg

plplznet

plznet

nplfdznet

2

0)(

0)(0

,

,

,

Rank the magnitude of the vertical force F applied to the rod from the strongest to weakest. A. F = E > C > D = A > B B. F > E > C > D > A > B C. F > C > E = A > D = B D. E > F > C > A > D > B E. Same for all cases

Six identical 2-meter massless rods are supporting identical 12-Newton weights. In each case, a vertical force F is holding the rods and the weights at rest. The rods are marked at half-meter intervals.

Fr

Mgd

FrMgd

FrMgdznet

FMznet

0

0

,

,

i-Clicker!

Rank the magnitude of the vertical force F applied to the end of the rod from the strongest to the weakest. A. F > E > D > B > C > A B. D > E > F > A > C > B C. F > B > E > C > D > A D. F > E > B > D > C > A E. Same for all cases

Six identical massless rods are all supported by a fulcrum and are tilted at the same angle to the horizontal. Each rod is marked at 1-meter intervals.

0

0

,

,

FrMgdznet

FMznet

i-Clicker!

Rank the magnitude of the vertical force F applied to the end of the rod from the strongest to the weakest. A. F > E > D > B > C > A B. D > E > F > A > C > B C. F > B > E > C > D > A D. F > E > B > D > C > A E. Same for all cases

Six identical massless rods are all supported by a fulcrum and are tilted at the same angle to the horizontal. Each rod is marked at 1-meter intervals.

Fr

Mgd

FrMgd

FrMgdznet

FMznet

0

0

,

,

i-Clicker!

Horizontal Beam Example A uniform horizontal beam with a length of l = 8.00 m and a weight of

Wb = 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of = 53 with the beam. A person of weight Wp = 600 N stands a distance d = 2.00 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.

Horizontal Beam Example

• The beam is uniform

– So the center of gravity is at the geometric center of the beam

• The person is standing on the beam

• What are the tension in the cable and the force exerted by the wall on the beam?

Horizontal Beam Example, 2

• Analyze

– Draw a free body diagram

– Use the pivot in the problem (at the wall) as the pivot point

• This will generally be easiest

– Note there are three unknowns (T, R, q)

• The forces can be resolved into components in the free body diagram

• Apply the two conditions of equilibrium to obtain three equations

• Solve for the unknowns



Nm

mNmN

l

lWdW

T

lWdWlT

bp

bpz

31353sin)8(

)4)(200()2)(600(

sin

)2

(

0)2

())(sin(

Choose pivot point about hinge near wall


0sinsin

0coscos

bpy

x

WWTRF

TRF

q

q

Nm

mNmN

l

lWdW

T

lWdWlT

bp

bpz

31353sin)8(

)4)(200()2)(600(

sin

)2

(

0)2

())(sin(



0sinsin

0coscos

bpy

x

WWTRF

TRF

q

q

Nm

mNmN

l

lWdW

T

lWdWlT

bp

bpz

31353sin)8(

)4)(200()2)(600(

sin

)2

(

0)2

())(sin(

NNT

R

T

TWW

T

TWW

R

R

bp

bp

5811.71cos

53cos)313(

cos

cos

1.71cos

sintan

cos

sintan

cos

sin

1

q

q

q

q

q


A beam is supported by a frictionless pin so that it is horizontal. A mass is suspended from the right end of the beam, and the left end is tethered by an anchored cable. Both cases below are identical, except for the location of the anchor point of the cable.

Is the magnitude of the torque due to the cable about the pivot pin greater in case A, greater in case B, or the same in both cases?

i-Clicker!

A. Greater in case A B. Greater in case B C. The same in both cases

A beam is supported by a frictionless pin so that it is horizontal. A mass is suspended from the right end of the beam, and the left end is tethered by an anchored cable. Both cases below are identical, except for the location of the anchor point of the cable.

Is the tension in the cable greater in case A, greater in case B, or the same in both cases? A. Greater in case A B. Greater in case B C. The same in both cases

i-Clicker!

Ladder Example

• A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is s = 0.40. Find the minimum angle q at which the ladder does not slip.

A Problem-Solving Strategy • Choose a convenient axis for calculating the net torque

on the object

– Remember the choice of the axis is arbitrary

• Choose an origin that simplifies the calculations as much as possible

– A force that acts along a line passing through the origin produces a zero torque

• Be careful of sign with respect to rotational axis

– I’ll choose positive if force tends to rotate object in CCW

– negative if force tends to rotate object in CW

– zero if force is on the rotational axis

• Apply the second condition for equilibrium

0,, zextznet

Choose an origin O that simplifies the calculations as much as possible ?

• A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is s = 0.40. Find the minimum angle.

mg mg mg mg

O

O

O

O

A) B) C) D)



mg mg mg mg

O

O

O

O

A) B) C) D)

Want to choose an origin that eliminates unknowns



mg mg mg mg

O

O

O

O

A) B) C) D)

Want to choose an origin that eliminates unknowns

Want to choose an origin that keeps values we know

Lever Arm

A lever arm is the shortest perpendicular distance between the line of action of a force and the axis of rotation. It provides an alternative way to determine the torque and is especially useful when the forces do not lie along coordinate axes. lever arm=rsin

Torque=F*r*sin

Torque=F*lever arm Sometimes it is simple to read the lever arm right off of the diagram.

Is the motion due to the net torque about the point A 1. clockwise 2. counterclockwise 3. zero What about point B? point C?

Three forces of equal magnitude are applied to a 3-m by 2-m rectangle. Forces F1 and F2 act at 45° angles to the vertical as shown, while F3 acts horizontally.

4 m

3 m

Another ladder example

A uniform ladder weighing 60 N leans against a frictionless vertical wall. What is the minimum coefficient of friction necessary between the ladder and the floor if the ladder is not to slip?

4 m

3 m

Fwall(normal)

Ffloor(normal)

Ffloor(friction)

Fearth(weight)

Another ladder example


What do we know?

4 m

3 m

Fwall(normal)

Ffloor(normal)

Ffloor(friction)

Fearth(weight)


Annotated solution We are trying to find the coefficient of friction,

How is it related to what we know (the weight)? Ffloor(friction)=Ffloor(normal)

Since SFvertical=0, by observation we can see that Ffloor(normal) Fearth(weight) 60 N

Since SFhorizontal=0, by observation we can see that Fwall(normal) Ffloor(friction) ??

There are two unknowns and only one equation – we need another equation. Sz=0 … to the rescue!

4 m

3 m

Fwall(normal)

Ffloor(normal)

Ffloor(friction)

Fearth(weight)

We know the forces have to balance… SF=0

4 m

3 m

Fwall(normal)

Ffloor(normal)

Ffloor(friction)

Fearth(weight)

In this case, what are the best choices for the axis of rotation?

A. I

B. II

C. III

D. I or II

E. II or III

I. At the top of the ladder

II. At the middle of the ladder

III. At the bottom of the ladder

i-Clicker!

Annotated solution We are trying to find the coefficient of friction,

How is the frictional torque related to the weight torque about the base? Ffloor(friction)=Ffloor(normal)

and Ffloor(friction)=Fwall(normal) Since Sbase=0, then Fearth(weight)*(its lever arm) must balance with Fwall(normal) *(its lever arm) How do you find the lever arm?

It is the shortest distance from the line of action of the force to the axis of rotation. The lever arm for the weight=1.5 m and the lever arm for the wall normal force is 4 m.

4 m

3 m

Fwall(normal)

Ffloor(normal)

Ffloor(friction)

Fearth(weight)

O

We know the torques about any

axis have to balance… S=0

Putting it all together

Sbase=0, so

Fearth(weight)*(its lever arm) must balance with Fwall(normal) *(its lever arm)

Putting it all together

Sbase=0, so

Fearth(weight)*(1.5m) = Fwall(normal)*(4 m)

Fwall(normal) = Fearth(weight)*(1.5m)/(4 m)

Ffloor(friction)=Fwall(normal) Fearth(weight)*0.375

Fwall(normal)/Ffloor(normal)

Since in this case Fearth(weight) = Ffloor(normal)

Fwall(normal)/Fearth(weight)

= [Fearth(weight)*0.375] /Fearth(weight)

= 0.375

How does this problem change if there is a person of known weight standing

on the ladder at a known position?

Old exam problem

A 50 N monkey climbs up a 120 N ladder and stops 1/3 of the way up. The upper and lower ends of the ladder rest on frictionless surfaces. The lower end is fastened to the wall by a massless horizontal rope. Find the tension in the rope when the monkey is 1/3 of the way up the ladder.

8 m

6 m

Old exam problem

A mass of M=225 kg hangs from the end of the uniform strut whose mass is mstrut = 45.0 kg. Find:

(a) the tension T in the cable

(b) horizontal and vertical force components exerted on the strut by the hinge.

Statics - Rutgers Physics & Astronomy · • The first condition of equilibrium is a statement of translational equilibrium • The net external force on the object must equal zero

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