MedAdvance Internship Program MedAdvance is an internship program based out of the Scientific Education and Research Institute in Thornton, CO. As a member.

MedAdvance Internship Program• MedAdvance is an internship program based out of the Scientific Education and

Research Institute in Thornton, CO. • As a member you may lead and participate in Anatomy/Suture labs and Shadow

Doctors. In addition to these opportunities, the Scientific Education and Research Institute is a medical device training facility that has doctors visit from all around the world to learn about new medical technology. This gives interns the opportunity to, not only meet health professionals from around the nation, but also to become familiar with the latest medical technology!!!

• MedAdvance is having an Open House on Thursday September 29th at 6-8:30pm at the Scientific Education and Research Institute. The address is 9005 Grant st. Thornton, CO 80229.

• If you have any more questions about the internship or would like to RSVP for the Open House, please feel free to contact [email protected].

Announcements

• CAPA Set #5 due Friday at 10 pm

• This week in Section Assignment #3: Forces, Newton’s Laws, Free-Body Diagrams Print out and bring the assignment to your section…

• Complete reading all of Chapter 4

Step 1: Draw a free-body diagram

Fg = mg

Note that “Normal Force” is always Perpendicular

(i.e. normal) to the surface.

In this case, it is not in the opposite direction to the

gravitational force.

Fg = mg

x

y We could choose our “usual” axes. However, we know that we will have motion

in both the x and y directions.

Step 2: Choose a coordinate system

x

yIf we pick these rotated axes, we know that

the acceleration along y must be zero.

Step 3: Write down the equations Σ Fx = m ax , Σ Fy = m ay

Problem: Fg doesn’t point in a coordinate direction.Must break it into its x- and y-components.

Fg = mg

|F(gravity)| = mg

F(gravity)y

F(gravity)x

Clicker Question Room Frequency BA

What is the angle a labeled above?)A a = 90 degrees B) a = q

C) a = 90 – q D) Cannot be determined

a90 – q90

Fg = mg

Step 3: Write down the equations Σ Fx = m ax , Σ Fy = m ay

N Fx = mg sin(q)

Fy = N – mg cos(q)

We also know that ay = 0, there is no motion in that

direction.

Fy =may=0= N – mg cos(q)

Step 4: Solve the Equations

)sin(mgmaF xx

)cos(mgNmaF yy

)sin(gax

0)cos( gm

Na y

)cos(mgN

As expected, maximum

acceleration if straight down

(q=90 degrees), ax = g.

Recall x-y definition

A mass m accelerates down a frictionless inclined plane.

Which statement is true?A) N < mgB) N > mgC) N = mg

+x

+y

θ

θ-mg cos θ

-

0 = Fy = N – mg cos θN = mg cos θ < mg

Clicker Question Room Frequency BA

The driver of a car parked on a hill releases the brake and

puts the car in neutral

Let = normal force = force of gravity = acceleration of the car.Which of the following is true?

N

Fg

a

A) A a,E N

,B F

g

B) E N

,D a,C F

g

C) E N

,D Fg

,C a

D) A a,B Fg

,E N

Clicker Question Room Frequency BA

+x

F fc

m

Two boxes with masses M and m are glued together. Force F is applied to Box 1.

Force fc is the “contact force” that Box 1 exerts on Box 2.

Box 1 Box 2A)F = fc

B) F > fc

C) F < fc

D)Indeterminate from information given

Clicker Question Room Frequency BA

M

* Frictionless surface

M m

+x

F fc

m

Box 1Box 2

Draw the free-body diagram for each box!

M mF fc

What other forces are acting on either box?Gravitational force down on both boxes, and

Normal force up on both boxes.However, since there is no vertical motion

we sometimes do not bother to write them down.

What about Newton’s Third Law?

fc

Apply Σ Fx = ma. (ignore vertical forces, no vertical motion)

M mF fcfc

F-fc = MaM fc = mam

Since the boxes are fixed (glued) together aM = am

aM=(F-fc)/M am =fc/m =

FmM

mf

FmmMf

MfmfFm

c

c

cc

)(

Ffc

Assume Box 1 has a mass of 2 kg and Box 2 has a mass of 1 kg. If a force of 1 N is applied to Box 1,

what is the contact force between the boxes?

A) 1 NB) ½ NC) 1/3 ND) 2 NE) 3 N

M

+x

F fc

m

Box 1Box 2

fc m

M m

F

1kg

2kg 1kg

1N 1

3N

fc m

M m

F

Clicker Question Room Frequency BA

Assume a force F is applied to Box 1 at an angle θ from the horizontal. What will be the equation for the contact force fc?

M m

+x

fc

m

Box 1Box 2

A) fc m Mm

F tan

B) fc m

M m

F sin

C) fc m

M m

F cos

D) fc m(M M )

F cos

θfc

m

M m

F

Previous equation:

The horizontal component of produces horizontal

motion.

F

F

F cosθ

F sinθ

Clicker Question Room Frequency BA