Goals Equilibrium Friction The End

Physics 2514Lecture 12

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 1 / 13

Goals Equilibrium Friction The End

Goal

Goals for today’s lecture:Application of Newton’s second law.

Equilibrium Static and Dynamic.

Define weight.

Introduce friction Static and Dynamic

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Goals Equilibrium Friction The End

Clicker

An elevator suspended by a cable is moving upward andslowing to a stop. Which free-body diagram is correct?

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Goals Equilibrium Friction The End

Equilibrium

Newton’s Second Law.

An object of mass m subjected to forces ~F1, ~F2, . . . will undergoan acceleration ~a given by

~a =~Fnet

m

where ~Fnet =∑n

i=1~Fi is the vector sum of all forces acting on

the object. The acceleration vector ~a points in the samedirection as the force vector ~Fnet.

Condition for equilibrium.∑

i

~Fi = 0 ⇒ ~a = 0

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 4 / 13

Goals Equilibrium Friction The End

Equilibrium

Newton’s Second Law.

An object of mass m subjected to forces ~F1, ~F2, . . . will undergoan acceleration ~a given by

~a =~Fnet

m

where ~Fnet =∑n

i=1~Fi is the vector sum of all forces acting on

the object. The acceleration vector ~a points in the samedirection as the force vector ~Fnet.

Condition for equilibrium.∑

i

~Fi = 0 ⇒ ~a = 0

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 4 / 13

Goals Equilibrium Friction The End

Equilibrium

Newton’s Second Law.

An object of mass m subjected to forces ~F1, ~F2, . . . will undergoan acceleration ~a given by

~a =~Fnet

m

where ~Fnet =∑n

i=1~Fi is the vector sum of all forces acting on

the object. The acceleration vector ~a points in the samedirection as the force vector ~Fnet.

Condition for equilibrium.∑

i

~Fi = 0 ⇒ ~a = 0

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 4 / 13

Goals Equilibrium Friction The End

Equilibrium

Equilibrium∑

i~Fi = 0.

Static equilibrium system is at rest.

Dynamic equilibrium system moves with a constantvelocity; ~a = 0.

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Goals Equilibrium Friction The End

Example

An accident victim with a broken leg is placed in traction. Thepatient wears a special boot with a pulley attached to the sole.The foot and boot together have a mass of 4.0 kg and thedoctor decides to hang a 6.0 kg mass from the rope. The bootis held suspended by the ropes and does not touch the bed.

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Goals Equilibrium Friction The End

Example

What is the tension in the rope?

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Goals Equilibrium Friction The End

Example

What is the tension in the rope?

T

FG

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 7 / 13

Goals Equilibrium Friction The End

Example

What is the tension in the rope?

T

FG

The force due to gravity FG = ma ⇒ FG = mg

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 7 / 13

Goals Equilibrium Friction The End

Example

What is the tension in the rope?

T

FG

The force due to gravity FG = ma ⇒ FG = mg

∑

i

Fi = T−m6g = 0 ⇒ T = m6g = 6.0 kg×9.8 m/s2 = 58.8 N

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Goals Equilibrium Friction The End

Example

The net traction force needs to pull straight out on the leg.What is the proper angle θ for the upper rope?

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13

Goals Equilibrium Friction The End

Example

The net traction force needs to pull straight out on the leg.What is the proper angle θ for the upper rope?

T1

T2

F

m4g

x

y

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13

Goals Equilibrium Friction The End

Example

The net traction force needs to pull straight out on the leg.What is the proper angle θ for the upper rope?

T1

T2

F

m4g

x

y

T = T1 = T2 = 58.8 N∑

Fy = T sin θ − m4g − T sin(15◦) = 0

⇒ sin θ = (m4g + T sin(15◦))/T ⇒ θ = 67.5◦

∑

Fx = T cos(15◦) + T cos θ − F = 0 ⇒ F = 79.0 N

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13

Goals Equilibrium Friction The End

Example

The net traction force needs to pull straight out on the leg.What is the proper angle θ for the upper rope?

T1

T2

F

m4g

x

y

T = T1 = T2 = 58.8 N∑

Fy = T sin θ − m4g − T sin(15◦) = 0

⇒ sin θ = (m4g + T sin(15◦))/T ⇒ θ = 67.5◦

∑

Fx = T cos(15◦) + T cos θ − F = 0 ⇒ F = 79.0 N

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13

Goals Equilibrium Friction The End

Example

The net traction force needs to pull straight out on the leg.What is the proper angle θ for the upper rope?

T1

T2

F

m4g

x

y

T = T1 = T2 = 58.8 N∑

Fy = T sin θ − m4g − T sin(15◦) = 0

⇒ sin θ = (m4g + T sin(15◦))/T ⇒ θ = 67.5◦

∑

Fx = T cos(15◦) + T cos θ − F = 0 ⇒ F = 79.0 N

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13

Goals Equilibrium Friction The End

Friction

Next to gravity, friction is the most common force that weinteract with. It allows us to walk, slows us down, and preventsus from moving.

There are 3 types of friction (all act to oppose motion)Static fraction fs ≤ µsN when no motion occurs;

Kinetic friction fk = µkN , when object is moving;

Rolling friction fr = µrN , when objects rolls.

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 9 / 13

Goals Equilibrium Friction The End

Clicker

Joe and Bill are playing tug-o-war. Joe is pulling with a force of200 N. Bill is simply hanging on, but skidding towards Joe at aconstant velocity. What is the magnitude of the force of frictionbetween Bill’s feet and the ground.

A) 200 N

B) 400 N

C) 0 N

D) 300 N

D) None of the above

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 10 / 13

Goals Equilibrium Friction The End

Example

A 50 kg steel box is in the back of a dump truck. The truck’sbed, also made of steel, is slowly tilted. At what angle will thefile cabinet begin to slide?

Brief description: A 50 kg steel box is on a steel incline plane.What is the maximum angle the incline can have for the box toremain in static equilibrium?

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Goals Equilibrium Friction The End

Solution

Knownµs = 0.80, m = 50 kg

Unknownangle θ of inclineNormal force n

Equations of motion:

mg sin θ − µsn = 0

−mg cos θ + n = 0

}

⇒ µs = tan θ ⇒ θ = tan−1 µs = 38.7◦

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Goals Equilibrium Friction The End

Announcements

Read chapter 6.

Reading quiz on chapter 5 and 6; due by 11:59 PM today.

New homework available.

P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 13 / 13