Goals Equilibrium Weight The End

Physics 2514Lecture 13

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

P. Gutierrez (University of Oklahoma) Physics 2514 February 23, 2011 1 / 14

Goals Equilibrium Weight The End

Goal

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

Equilibrium Static and Dynamic.

Gravity and weight.

Continue with the application of friction Static and Kinetic

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Goals Equilibrium Weight The End

Clicker

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

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Goals Equilibrium Weight 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

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Goals Equilibrium Weight 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 Weight 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 Weight 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?

P. Gutierrez (University of Oklahoma) Physics 2514 February 23, 2011 6 / 14

Goals Equilibrium Weight The End

Solution

Knownµs = 0.80, m = 50 kg

Unknownangle θ of inclineNormal force n

Newton’s second law:

mg sin θ − µsn = 0

n − mg cos θ = 0

}

⇒ µs = tan θ ⇒ θ = tan−1 µs = 38.7◦

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Goals Equilibrium Weight The End

Solution

Knownµs = 0.80, m = 50 kg

Unknownangle θ of inclineNormal force n

Newton’s second law:

mg sin θ − µsn = 0

n − mg cos θ = 0

}

⇒ µs = tan θ ⇒ θ = tan−1 µs = 38.7◦

P. Gutierrez (University of Oklahoma) Physics 2514 February 23, 2011 7 / 14

Goals Equilibrium Weight The End

Solution

Knownµs = 0.80, m = 50 kg

Unknownangle θ of inclineNormal force n

Newton’s second law:

mg sin θ − µsn = 0

n − mg cos θ = 0

}

⇒ µs = tan θ ⇒ θ = tan−1 µs = 38.7◦

P. Gutierrez (University of Oklahoma) Physics 2514 February 23, 2011 7 / 14

Goals Equilibrium Weight The End Example

Gravity and Weight

Mass is an intrinsic property of an object (objects inertia).

Gravity is a long range force acting between any twomasses. (Newton’s law of gravity.)

~FG =Gm1m2

r2r̂

Gravity near Earth’s surface (Newton’s second law.)

m~a = ~FG =GmMe

R2e

r̂ ⇒ g ≡ a =GMe

R2e

= 9.8 m/s2

G = 6.67 × 10−11 N · m2/kg2

Me = 5.98 × 1024 kg Re = 6.37 × 106 m

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Goals Equilibrium Weight The End Example

Gravity and Earth

m~a =GmMe

(Re + y)2r̂ ⇒ g ≡ a =

GMe

(Re + y)2

y ⇒ distance above the earth’s surface

y (km) g (m/s2)

0 9.830 Earth’s surface

10 9.799 Cruising altitude of a commercial aircraft

100 9.528 60 miles up

400 8.703 Height of space shuttle

400,000 0.003 Height of the moon

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Goals Equilibrium Weight The End Example

Weight

Weight is a measurement

An objects weight results from the force exerted on aspring scale could be a pulling or pushing contact force.

If the system is in static equilibrium (rest frame) w = mg.

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Goals Equilibrium Weight The End Example

Weight

Weight is a measurement

An objects weight results from the force exerted on aspring scale could be a pulling or pushing contact force.

If the system is in static equilibrium (rest frame) w = mg.

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Goals Equilibrium Weight The End Example

Example

You are standing on a scale inside an elevator. When theelevator is stationary, you find your weight is 680 N. What isyour weight if the elevator is moving downward with a velocityof 10 m/s, but the accleration is upward at 3 m/s2?

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Goals Equilibrium Weight The End Example

Example

You are standing on a scale inside an elevator. When theelevator is stationary, you find your weight is 680 N. What isyour weight if the elevator is moving downward with a velocityof 10 m/s, but the accleration is upward at 3 m/s2?

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Goals Equilibrium Weight The End Example

Solution

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Goals Equilibrium Weight The End Example

Solution

∑

i

Fyi = Fsp − mg = ma

w = N = Fsp

⇒ w = ma+mg = m(a+g) = 888 N

P. Gutierrez (University of Oklahoma) Physics 2514 February 23, 2011 12 / 14

Goals Equilibrium Weight The End Example

Solution

∑

i

Fyi = Fsp − mg = ma

w = N = Fsp

⇒ w = ma+mg = m(a+g) = 888 N

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Goals Equilibrium Weight The End Example

Clicker

In the example the elevator is moving downwards at 10 m/s,accelerating upwards at 3 m/s2 and the weight is found to be888 N. Now assume that the elevator is moving upwards at10 m/s and accelerating upwards at 3 m/s2, what is the weight?(Recall that for the stationary elevator the weight was 680 N.)

A) 888 N

B) 680 N

C) 471 N

D) 0 N

C) -888 N

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Announcements

Read chapter 6.

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

New homework available.

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