1. Forces and Motion beyond 1 D
PHYSICS 220
Lecture 05
Forces and Motion beyond 1 D
Textbook Sections 3.7, 4.1
Lecture 5
Purdue University, Physics 220

2. Purdue University, Physics 220
Apparent Weight
Recall: F = m a
Consider person accelerating up in an elevator
Draw FBD
Apply Newton’s second law
N – mg = ma
N = m(g+a)
Apparent weight is equal to normal
force from scale or floor
Note: in free fall a=-g so N=0 y x N mg
Lecture 5
Purdue University, Physics 220

3. Purdue University, Physics 220
iClicker
You are traveling up on an elevator to the 30th floor of the Sears tower. As it nears the 30th floor, your weight appears to be
A) heavier B) the same C) lighter N
SF = ma
N – mg = ma
N = m(g+a)
a < 0 so N < mg mg
Lecture 5
Purdue University, Physics 220

4. Purdue University, Physics 220
Exercise
A person has a mass of 50 kg. What is their apparent weight when they are riding on an elevator
Going up with constant speed 9.8 m/s
Going down with constant speed 9.8 m/s
Accelerating up at a rate of 9.8 m/s2
Accelerating down at a rate of 9.8 m/s2
N = m(g+a)
a = 0 so N= mg = 490 Newtons
a = 0 so N= mg = 490 Newtons
a = +9.8 m/s2 so N= 2 mg = 980 Newtons
a = -9.8 m/s2 so N= 0 mg = 0 Newtons
Lecture 5
Purdue University, Physics 220

5. Purdue University, Physics 220
Exercise
You are standing on a scale inside an elevator. You weigh 125 pounds, but the scale reads 140 pounds.
The elevator is going (up down can’t tell)
The elevator is accelerating (up down can’t tell) y x
A B C
A B C
N = m(g+a)
Weight increases when accelerating up
Weight decreases when accelerating down.
Lecture 5
Purdue University, Physics 220

6. Purdue University, Physics 220
Exercise
Fred throws a ball 30 m/s vertically upward. How long does it take to hit the ground 2 meters below where he released it?
y = y0 + vy0t g t2
y-y0 - vy0t g t2 = 0
-2 – 30 t + 0.5*9.8 t2 = 0
y = y0 + vy0t - 1/2 gt2
vy = vy0 - gt
vy2 = vy02 - 2g(y-y0)
t = 6.19 s or -.06 s
Lecture 5
Purdue University, Physics 220

7. Purdue University, Physics 220
Exercise
Fred throws a ball 30 m/s vertically upward. What is the maximum height the ball reaches? How long does it take to reach this height?
v2-vo2 = 2 a Dy
Dy = (v2-vo2 )/ (2 a)
= -302 / (2 * (-9.8))
= 46 m
v = v0 + a t
t = (v-v0) / a
= (0 – 30 m/s )/ (-9.8 m/s2)
= 3.1 seconds
Have students make plot?
Lecture 5
Purdue University, Physics 220

8. Drag and Terminal Velocity
Drag force = air resistance
It is proportional to the square of the speed Fd W
It opposes the direction of motion
At the terminal velocity the drag force is equal to weight
Lecture 5
Purdue University, Physics 220

9. Drag and Terminal Velocity
Terminal velocity depends on size, mass, and shape of the object
Lecture 5
Purdue University, Physics 220

10. Purdue University, Physics 220
Equilibrium
Net Force: If are all the forces acting on an object
When an object is in equilibrium, the net force acting on it is zero
Lecture 5
Purdue University, Physics 220

11. Purdue University, Physics 220
Question
Compare the net force on the two books
A) Fphysics > Fbiology B) Fphysics = Fbiology C) Fphysics < Fbiology
Net force is zero on both books! The normal force from table exactly cancels downward forces.
Physics
Biology
Lecture 5
Purdue University, Physics 220

12. Book Pushed Across Table
Calculate force of hand to keep the book sliding at a constant velocity, if the mass of the book is 1 kg, ms = .84 and mk=.75.
Constant velocity  SF=0
x-direction: SF=0
Fhand-Ffriction = 0
Fhand=Ffriction
Fhand=mk FNormal
Combine:
Fhand=mk FNormal Fhand=0.759.8 N
Fhand=7.3 N
y-direction: SF=0
FNormal-FGravity = 0
FNormal = FGravity
FNormal =19.8=9.8 N
hand
Gravity
Normal
friction y x
Key idea here is we can do each direction independently!
Physics
Lecture 5
Purdue University, Physics 220

13. Forces in 2 Dimensions: Ramp
Calculate tension in the rope necessary to keep the 5 kg block from sliding down a frictionless incline of 20 degrees.
1) Draw free-body diagram
2) Label axes
Weight is not in x or y direction! Need to decompose it! N
Tell them any choice is OK, but this way only have 1 force to decompose T y x W q
Lecture 5
Purdue University, Physics 220

14. Forces in 2 Dimensions: Ramp
Calculate force necessary to keep the 5 kg block from sliding down a frictionless incline of 20 degrees.
x- direction:
S F = 0
W sinq – T = 0
W sin q W q
W cos q y x q
Lecture 5
Purdue University, Physics 220

15. Forces in 2 Dimensions: Ramp
Calculate force necessary to keep the 5 kg block from sliding down a frictionless incline of 20 degrees.
x- direction:
W sinq – T = 0
T = W sinq
= m g sinq = 16.8 Newtons N
Tell them any choice is OK, but this way only have 1 force to decompose T y x W q
Lecture 5
Purdue University, Physics 220

16. Purdue University, Physics 220
iClicker
What is the normal force of ramp on block?
A) N > mg B) N = mg C) N < mg
In “y” direction:
SF = ma
N – W cos q = 0
N = W cos q T N W W q
W sin q
W cos q y x
N = m g cos q q
Lecture 5
Purdue University, Physics 220

17. Purdue University, Physics 220
Inclined Plane Ff N
N = m g cos
m g sin - Ff = 0
Ff = N
m g sin -  m g cos = 0  mg
Special case 1: Start with  at zero and slowly increase . Just before it slides
s m g cos = m g sin tan = s
Special case 2: Object is sliding down at constant velocity, that is a = 0
k m g cos = m g sin tan = k
Lecture 5
Purdue University, Physics 220

18. Purdue University, Physics 220
Force at Angle Example
A person is pushing a 15 kg block across a floor with mk= 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block?
x- direction: SFx = 0
Fpush cosq – Ffriction = 0
Fpush cosq – m FNormal = 0
FNormal = Fpush cosq / m
Combine:
( Fpush cosq / m)–mg – FPush sinq = 0
Fpush ( cosq / m - sinq) = mg
Fpush = m g / ( cosq/m – sinq)
y- direction: SFy = 0
FNormal –Fweight – FPush sinq = 0
FNormal –mg – FPush sinq = 0
Fpush = 80 N
Normal
Could do this on over head to help people work through it.
Pushing y x q q
Friction
Weight
Lecture 5
Purdue University, Physics 220