Physics 1ALecture 5B

"And in knowing that you know nothing, that makes you the smartest of all.”

--Socrates

FrictionRecall that you can have either static friction (when the object is motionless) or kinetic friction (when the object is moving).

Kinetic friction is always given by:

Whereas static friction can be any value in a range given by:

Obviously when calculating any type of friction, it becomes helpful to calculate the normal force.

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0 ≤ fs ≤ µsFN

Normal ForceExampleAn applied 12N horizontal force (Fapplied) pushes a block weighing 5.0N against a vertical wall. The coefficient of static friction between the wall and the block is 0.60, and the coefficient of kinetic friction is 0.40. Assume that the block is not moving initially. Will the block ever move?AnswerFirst, you must define a coordinate system.Let’s choose up as positive y and the direction of the applied force as positive x.

blockFapplied

Normal ForceNext, draw a force diagram for the block.

block

Fgravity, Earth on blockFapplied, you on block

Fnormal, wall on block

Ffriction, wall on block

No need to break the forces into components, so we can turn to Newton’s Laws.

Which of Newton’s Laws do we apply?

We know it doesn’t accelerate in the x-direction, but we are not sure about the y-direction.

Answer

Normal ForceAnswerWe should apply Newton’s 1st Law in the x-direction.

ΣFx = Fapplied - Fnormal = 0Fnormal = Fapplied = 12N

If the block is not going to move then we can say that the maximum static friction force must be greater than or equal to the force of gravity.

Ffriction,max = (μs) Fnormal = (0.60)12NFfriction,max = 7.2N

ΣFx = 0

Ffriction,max > Fgravity = 5.0N

So, the block will not move, since Ffriction can be 5N!

ExampleExampleA box lies on an inclined plane at an angle of 30.0o to the horizontal plane. The mass of the box is 5.00kg. What is value of the coefficient of static friction if it is noted that the static friction is at a maximum?

AnswerGo through the guidelines.First, you must define a coordinate system.

ExampleAnswerWe will make a clever choice of coordinate systems:

Next we should draw a force diagram for the box:

boxFgravity, Earth on box

Fnormal, plane on box Ffriction, ground on box

The angled coordinate system will allow us to only break one force into components.

ExampleAnswerThe normal force only points in the y-direction and the friction force only points in the x-direction. We do not need to break them into components.

But we do need to break the gravitational force into x and y components.

We get:

Fgx = mg sinθFgy = mg cosθ

ExampleAnswerNext we should apply the appropriate Newton’s Laws.

Since the box is at rest we shall apply Newton’s First Law in both independent directions (x and y).

ΣFx = 0 and ΣFy = 0

Now we can perform the math, let’s start with the x-direction:

ExampleAnswerNext we can turn our attention to the y-direction:

Turning to the equation for maximum static friction:

<- unitless

PulleysPulleys will change the direction of the tension force in ropes.

This could mean that the tension for two masses may be in the same direction.

Or this also could mean that a tension force on one mass in the horizontal direction will have a Third Law Pair with another mass in the vertical direction.

PulleysExampleTwo objects with masses 2.00kg (m1) and 6.00kg (m2) are connected by a light string that passes over a frictionless pulley. Determine the acceleration of each mass and the tension in the string.

AnswerFirst, you must define a coordinate system.Let’s choose up as positive y.

PulleysAnswer

Next we should draw a force diagram for each mass:

Forces are already broken up into components.

So we should apply Newton’s 2nd Law separately to each object.

6kg massFgravity, Earth on 6kg mass

Ftension, string on 6kg mass

2kg massFgravity, Earth on 2kg mass

Ftension, string on 2kg mass

PulleysAnswer

Since the string cannot be stretched, a = a1 = -a2.

Also the tensions will be the same magnitude: T

For the 2kg mass (m1) we have:

ΣFy = T - m1g = m1aT = m1a + m1g

For the 6kg mass (m2) we have:

ΣFy = T - m2g = m2(-a)T = m2g - m2a

ΣFy = m1a1

ΣFy = m2a2

PulleysAnswer

T = T

Plugging in the values we get:

m1a + m1g = m2g - m2a m1a + m2a = m2g - m1g a(m2 + m1) = g(m2 - m1)

a = g(m2 - m1)/(m2 + m1)

a = (9.80m/s2)(6kg - 2kg)/(6kg + 2kg)

a = (9.80m/s2)(4kg)/(8kg) = 4.90m/s2

Set T = T and solve for the resulting acceleration.

PulleysAnswer

Recall the equation that we have for tension from Newton’s Second Law (either one will work):

T = m2g - m2a

T = m2(g - a) T = 6kg(9.80m/s2 - 4.90m/s2)

T = 6kg(4.90m/s2) = 29.4N

Conceptual QuestionLet’s say that in the previous example the masses were changed to m1 = 4kg and m2 = 8kg. How would this affect the resulting motion of the Atwood’s Machine?

A) Acceleration of the two masses would decrease.

B) Acceleration of the two masses would increase.

C) Acceleration of the two masses would remain the same.

EquilibriumAn object either at rest or moving with a constant velocity is said to be in equilibrium.

The net force acting on the object is zero (since the acceleration is zero):

Though we usually work with components when dealing with equilibrium.

ΣFx = 0 and ΣFy = 0

For Next Time (FNT)

Keep working on Chapter 5 HW