Lect 17

Physics 211 Lecture 17, Slide Physics 211 Lecture 17, Slide 11

Physics 211Physics 211Lecture 17Lecture 17

Today's Concepts:Today's Concepts:

Torque due to gravityTorque due to gravityStatic EquilibriumStatic Equilibrium

I would like to know how torque and statics are going to be combined with I would like to know how torque and statics are going to be combined with rotation problems on the future exams.rotation problems on the future exams.

How does friction provide torque but do no work on rolling objects? Also How does friction provide torque but do no work on rolling objects? Also why is the shortest distance to the force applied the lever arm in torque.why is the shortest distance to the force applied the lever arm in torque.

can the preflights and prelectures go up earlier?can the preflights and prelectures go up earlier?

Can we please discuss everything static equilibrium? I am so lost and scared Can we please discuss everything static equilibrium? I am so lost and scared for exam 3. (seriously).for exam 3. (seriously).

If I am in space and traveling at the speed of light, so the light from me does not If I am in space and traveling at the speed of light, so the light from me does not travel faster than my body, and I hold a mirror in front of me, am I invisible?travel faster than my body, and I hold a mirror in front of me, am I invisible?

is the person who does the prelectures from Cal Poly? when i refreshed that is the person who does the prelectures from Cal Poly? when i refreshed that page, it asked me to log onto a Cal Poly account.page, it asked me to log onto a Cal Poly account.

Maybe it's just because none of the concepts are really new, but I prefer to Maybe it's just because none of the concepts are really new, but I prefer to think it's the masterful prelecture. Either way, these concepts seem very think it's the masterful prelecture. Either way, these concepts seem very clear.clear.

Can we just do a lot of these? They don't seem too bad, but we need to get Can we just do a lot of these? They don't seem too bad, but we need to get used to using them.used to using them.

I really like torque. It's my physics happy place.I really like torque. It's my physics happy place.

If the Yankees win we should have a pizza party…If the Yankees win we should have a pizza party…

I don't have any questions, but I'd like to point out that I don't have any questions, but I'd like to point out that physics is the one thing stopping me from pursuing my dream physics is the one thing stopping me from pursuing my dream to be a ninja.to be a ninja.

New Topic, Old Physics:New Topic, Old Physics: Statics:Statics:

As the name implies, “statics” is the study of systems that don’t move. Ladders, sign-posts, balanced beams, buildings, bridges...


The key equations are familiar to us: I F ma

If an object doesn’t move: If an object doesn’t move: 0a

0

0F

0

The net force on the object is zeroThe net force on the object is zero

The net torque on the object is zero The net torque on the object is zero ((for any axis)for any axis)

Statics:Statics:

Example:Example:

What are all of the forces acting on a car parked on a hill?What are all of the forces acting on a car parked on a hill?

xy

N f


mg

Car on Hill:Car on Hill:• Use Newton’s 2nd Law: Use Newton’s 2nd Law: FFNET NET = M= MAACM CM = 0= 0• Resolve this into Resolve this into xx and and yy components: components:

N

mg

f


xy

x: f - mg sin = 0

f = mg sin

y: N - mg cos = 0

N = mg cos

Torque due to GravityTorque due to Gravity


sin( ) cmR Mg

Magnitude:

MgR

Lever arm

R

CMR

Example:Example:

Now consider a plank of mass Now consider a plank of mass MM suspended by two strings as suspended by two strings as shown. We want to find the tension in each string:shown. We want to find the tension in each string:

L/2 L/4

Mx cm

T1 T2

Mgy

x


Balance forces:Balance forces:

L/2 L/4

Mx cm

T1 T2

Mgy

x

T1 + T2 = Mg

0F


Choose the rotation axis to be out of the Choose the rotation axis to be out of the page through the CM:page through the CM:

2 2 4T

L

The torque due to the string on the right about this axis is:

1 1 2T

L

The torque due to the string on the left about this axis is:

Gravity exerts notorque about the CM

L/2 L/4

Mx cm

T1 T2

Mgy

x


Balance torquesBalance torques

0

The sum of all torques must be The sum of all torques must be 00::

We already found that

T1 + T2 = Mg

T Mg11

3

T Mg22

3

L/2 L/4

Mx cm

T1 T2

Mgy

x


Finish the problemFinish the problem

1 2 0 1 2 02 4

L LT T

T T2 12

Choose the rotation axis to be out of the Choose the rotation axis to be out of the page at the left end of the beam:page at the left end of the beam:

2 2

3

4

LT

The torque due to the string on the right about this axis is:

1 0

The torque due to the string on the left is zero L/2 L/4

Mx cm

T1 T2

Mgy

x


What if you choose a different axis?What if you choose a different axis?

0

Torque due to gravity:2g

LMg

The sum of all torques must be The sum of all torques must be 00::

We already found that

T1 + T2 = Mg

T Mg11

3

T Mg22

3

L/2 L/4

Mx cm

T1 T2

Mgy

x


You end up with the same You end up with the same answer!answer!

1 2 0g 2

30

4 2

L LT Mg

T Mg22

3

Approach to Statics: SummaryApproach to Statics: Summary

In general, we can use the two equationsIn general, we can use the two equations

to solve any statics problem.to solve any statics problem.

When choosing axes about which to calculate torque, we can sometimes be clever and make the problem easier....When choosing axes about which to calculate torque, we can sometimes be clever and make the problem easier....

0F

0


Hanging Lamp AgainHanging Lamp AgainA lamp of mass m hangs from the end of plank of mass M and length L. One end of the plank is held to a wall by a hinge, and the other end is supported by a massless string that makes an angle with the plank.

a) What is the tension in the string?

b) What are the forces supplied by the hinge on the plank?

hinge

m

M

L


First use the fact that in both x and y directions:

F 0

x: T cos - Fx = 0

y: T sin + Fy - Mg - mg = 0

0Now use

Just to be different, choose the rotation axis to be through the end of the rod.


L/2Fx

Fy

TT

L/2

mg

Mgm

M

y

x

02

Y

LF L Mg

Now mg and T and Fx will not enter into the torque equation:

So we have three equations and So we have three equations and three unknowns:three unknowns:

T cos - Fx = 0

T sin + Fy - Mg - mg = 0

which we can solve to find:which we can solve to find:

F mgy 1

2

02

Y

LF L Mg

L/2Fx

Fy

TT

L/2

mg

Mgm

M


y

x

( / 2 )

tan( )

x

M m gF

( / 2 )

sin( )

M m g

T

L/2Fx

Fy

TT

L/2

Mg

mgM

m


y

x

F mgy 1

2

( / 2 )

tan( )

x

M m gF

( / 2 )

sin( )

M m g

T

Same as in the pre-lectureSame as in the pre-lecture

In In case 1case 1 one end of a horizontal massless rod of length one end of a horizontal massless rod of length LL is attached to a vertical wall by a hinge, and the other end holds a is attached to a vertical wall by a hinge, and the other end holds a ball of mass ball of mass MM. . In In case 2case 2 the massless rod holds the same ball but is twice as the massless rod holds the same ball but is twice as long and makes an angle of long and makes an angle of 3030oo with the wall as shown. with the wall as shown.

In which case is the total torque about the hinge biggest?In which case is the total torque about the hinge biggest?A) Case 1A) Case 1 B) Case 2B) Case 2 C) Both are the sameC) Both are the same

LL

2L2L

9090oo

3030oo

Case 1Case 1 Case 2Case 2

MM

MM

gravitygravity


PreflightPreflight

67% got this right67% got this right

In which case is the total torque about the hinge biggest?In which case is the total torque about the hinge biggest?A) Case 1A) Case 1 B) Case 2B) Case 2 C) Both are the sameC) Both are the same

LL

2L2L

9090oo

3030oo

Case 1Case 1 Case 2Case 2

MM

MM


C) The torque is force times its lever arm. In this question, since gravity C) The torque is force times its lever arm. In this question, since gravity always point down, the lever will be the horizontal distance between the always point down, the lever will be the horizontal distance between the ball and the wall, which in both case is equal to L. ball and the wall, which in both case is equal to L.

B) f is the same in both cases, and r is 2L in case 2, so the torque is bigger B) f is the same in both cases, and r is 2L in case 2, so the torque is bigger in case 2. in case 2.

A) the force of gravity is perpendicular to the rod, which will cause more A) the force of gravity is perpendicular to the rod, which will cause more torque. torque.

An object is made by hanging a ball of mass M from one end An object is made by hanging a ball of mass M from one end of a plank having the same mass and length L. The object is of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below.plank supporting the ball, as shown below.

Is the object balanced?Is the object balanced?A) Yes B) No, it will fall left C) No, it will fall rightA) Yes B) No, it will fall left C) No, it will fall right

MM

LL/4gravitygravity


PreflightPreflight


How far to the right of the pivot is the center of mass of just How far to the right of the pivot is the center of mass of just the plank.the plank.

A) L/4 B) L/2 C) 3L/4A) L/4 B) L/2 C) 3L/4

M


ActAct

xx

Is the object balanced?Is the object balanced?A)A) Yes Yes B)B) B) No, it will fall left B) No, it will fall left C)C) C) No, it will fall rightC) No, it will fall right

MM

L/4


xxL/4


C) The torque is greater on the right because the distance from the center C) The torque is greater on the right because the distance from the center of mass is greater on the right. . of mass is greater on the right. .

B) because part of the plank is on the left side of the pivot so there is more B) because part of the plank is on the left side of the pivot so there is more mass on the left side mass on the left side

A) Ball of mass M balanced at L/4 left of pivot; beam of mass M balanced A) Ball of mass M balanced at L/4 left of pivot; beam of mass M balanced L/4 right of pivot. So system balanced. L/4 right of pivot. So system balanced.

In case 1, one end of a horizontal plank of mass M and In case 1, one end of a horizontal plank of mass M and length L is attached to a wall by a hinge and the other end is length L is attached to a wall by a hinge and the other end is held up by a wire attached to the wall. In case 2 the plank is held up by a wire attached to the wall. In case 2 the plank is half the length but has the half the length but has the same mass same mass as in case 1, and the as in case 1, and the wire makes the wire makes the same angle same angle with the plank. with the plank.

In which case is the tension in the wire biggest?In which case is the tension in the wire biggest?A) Case 1A) Case 1 B) Case 2B) Case 2 C) Both are the sameC) Both are the same

gravitygravity

L/2L/2

MM

LL

MM


PreflightPreflight


Which equation correctly expresses the fact that the total Which equation correctly expresses the fact that the total torque about the hinge is zero?torque about the hinge is zero?

A) A)

B) B)

C)C)LL

MgMg


ActAct

TT

Lsin

(Ls

in(

))

sin( ) 02

L

TL Mg

02

L

TL Mg

sin( ) 0 TL MgL

sin( )2

Mg

TThe The LL cancels out: cancels out:

In which case is the tension in the wire biggest?In which case is the tension in the wire biggest?A) Case 1A) Case 1 B) Case 2 B) Case 2 C) Same C) Same

L/2L/2

MM

LL

MM


C) The tension will be the same in both cases because T=Mg/2sin(theta), C) The tension will be the same in both cases because T=Mg/2sin(theta), and theta and M are equal in both cases. and theta and M are equal in both cases.

B) they both have a net force and torque of zero case two has a shorter B) they both have a net force and torque of zero case two has a shorter lever arm so it needs a bigger force to produce the same torque to cancel lever arm so it needs a bigger force to produce the same torque to cancel out their equal weights. out their equal weights.

A) the force is at a point that is twice the length away so the torque is twice A) the force is at a point that is twice the length away so the torque is twice as big as big

Lect 17

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Transcript of Lect 17