Physics 2514Lecture 13

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

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Goals

We will discuss some examples that involve equilibrium.We then move on to a discussion of friction.

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Newton’s Laws of Motion

1) An object that is at rest will remain at rest, or an object thatis moving will continue to move (in a straight line) withconstant velocity, if and only if the net force acting on theobject is zero.

2) An object of mass m subjected to forces ~F1, ~F2, ~F3, . . . willundergo an acceleration ~a given by

~Fnet = m~a where ~Fnet =n

∑

i=1

~Fi

3) To every action there is always opposed an equal reaction;or, the mutual actions of two bodies upon each other arealways equal and directed to contrary parts.

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Equilibrium

Let’s first consider problems of static equilibrium

Equilibrium corresponds to the case that the object has zeronet force acting on it ~Fnet = 0

Two types of equilibrium conditionsStatic equilibrium corresponds to the case when theobject is at restDynamic equilibrium corresponds to the case when theobject moves at a constant velocityBoth correspond to ~Fnet = 0.

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Solving Problems

To solve problems using Newton’s laws of motion, we will followthe same problem solving procedure as for the case ofkinematics.

Rewrite the problem in a couple of sentences;1. State the situation;2. What variables are we solving for;

Draw a force diagram Free Body Diagram

State the known and unknown variables;Solve the problem algebraically;

Substitute numbers into the problem.

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Example 1

Consider a lead block of mass 5 Kg held against a frictionlesswall with a force applied at an angle of 30◦ as shown in thefigure. What is the magnitude of the force required to keep theblock from sliding?

PSfrag replacementsθ

~F

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Solution

Brief description A 5 Kg block is held against a frictionless wallwith a force acting at 30◦ relative to the horizontal. What is themagnitude of ~P required to keep the block in static equilibrium?

PSfrag replacements

~Pθ

Free body diagram

PSfrag replacementsx

y

~P

~n

m~g

θ

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Solution

Knownθ = 30◦

~w = −mg j = −49 j N

Unknown~P, ~n

PSfrag replacements

x

y

~P

~n

m~g

θ

Equations:∑

Fx = n − Px = n − P cos θ = 0∑

Fy = Py − mg = P sin θ − mg = 0

⇒

{

n = P cos θ

P = mg/ sin θ

}

⇒ P = mg/ sin(30) = 98 NPhysics 2514 – p. 8/18

Example 2

A 1000 Kg beam is supported by two ropes. Calculate thetension in each rope.

PSfrag replacements ~T1~T2

x

y

α β

m~g

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Example 3

A 1000 Kg beam is supported by two ropes. Calculate thetension in each rope.

Knownα = 20◦

β = 30◦

~w = −mg j = −9800 j N

Unknown~T1, ~T2

PSfrag replacements ~T1~T2

x

y

α β

m~g

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Example

Knownα = 20◦

β = 30◦

~w = −mg j = −9800 j N

Unknown~T1, ~T2

PSfrag replacements ~T1~T2

x

y

α β

m~g

T2x − T1x = T1 sin α − T2 sin β = 0

T1y + T2y − mg = 0 = T1 cosα + T2 cosβ − mg = 0

}

⇒

{

T1 = 6396 NT2 = 4375 N

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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.

1. 200 N2. 400 N3. 0 N4. 300 N5. None of the above

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Friction

Next to gravity, friction is the most common force that we interactwith. It allows us to walk, slows us down, and prevents us frommoving.

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.

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Clicker

For which situation is the frictional force the largest.

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Solution

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Example 3

A 50 kg steel box is in the back of a dump truck. The truck’s bed,also made of steel, is slowly tilted. At what angle will the filecabinet 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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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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Assignment

Read the sections on Newton’s second law and its applications:Sections 5.2, 5.3, 5.6

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