1. SPH4U: Practice Problems Today’s Agenda
Run and Hide

2. Review: Pegs & Pulleys Used to change the direction of forces
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: The tension is the same on both sides!
massless rope
F1 = -T i
| F1 | = | F2 |
ideal peg
or pulley
F2 = T j

3. Problem: Accelerometer
A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle  with respect to the vertical (y) axis. Solve for  in terms of a and g. a  i

4. Accelerometer... Draw a free body diagram for the mass:
What are all of the forces acting? m
T (string tension)
mg (gravitational force)  i

5. Accelerometer... Using components (recommended):
i: FX = TX = T sin  = ma
j: FY = TY - mg
= T cos - mg = 0 TX  TY T  j i m ma mg

6. Accelerometer... Using components : i: T sin  = ma
j: T cos - mg = 0
Eliminate T : TX TY T  j i m ma mg
T sin = ma
T cos = mg

7. Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
T (string tension) T  mg  m
FTOT
mg (gravitational force)

8. Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
Recall that FNET = ma: So
T (string tension)  T mg  m ma
mg (gravitational force)

9. Accelerometer... Let’s put in some numbers:
Say the car goes from 0 to 60 mph in 10 seconds:
60 mph = 60 x 0.45 m/s = 27 m/s.
Acceleration a = Δv/Δt = 2.7 m/s2.
So a/g = 2.7 / 9.8 =
 = arctan (a/g) = 15.6 deg a 

10. Understanding
A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces
Have equal magnitudes and form an action/reaction pair
Have equal magnitudes but do not form an action/reaction pair
Have unequal magnitudes and form an action/reaction pair
Have unequal magnitudes and do not form an action/reaction pair
None of the above
Because the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.

11. Angles of an Inclined plane
The triangles are similar, so the angles are the same! 
ma = mg sin  N  mg

12. Problem: Inclined plane
A block of mass m slides down a frictionless ramp that makes angle  with respect to the horizontal. What is its acceleration a ? m a 

13. Inclined plane...
Define convenient axes parallel and perpendicular to plane:
Acceleration a is in x direction only. i j m a 

14. Inclined plane... Consider x and y components separately:
i: mg sin  = ma a = g sin 
j: N - mg cos  = N = mg cos  ma i j
mg sin  N 
mg cos  mg

15. Problem: Two Blocks
Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m1, what is the force on the block of mass m2? F m1 m2

16. Problem: Two Blocks ( ) m2 Realize that F = (m1+ m2) a :
Draw FBD of block m2 and apply FNET = ma:
F / (m1+ m2) = a
F2,1 = m2 a
F2,1 m2 m2
2,1 ( ) ÷ ø ö ç è æ + = m1 F
Substitute for a :
(m1 + m2) m2
F2,1 F =

17. Problem: Tension and Angles
A box is suspended from the ceiling by two ropes making an angle  with the horizontal. What is the tension in each rope?   m

18. Problem: Tension and Angles
Draw a FBD: T1 T2 j i
T1sin 
T2sin   
T1cos 
T2cos  mg
Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
Fx,NET = T1cos  - T2cos  = T1 = T2
2 sin  mg
T1 = T2 =
Fy,NET = T1sin  + T2sin  - mg = 0

19. Problem: Motion in a Circle
A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v.
What is the tension T in the string at the top of the rock’s trajectory? v T R

20. Motion in a Circle...
Draw a Free Body Diagram (pick y-direction to be down):
We will use FNET = ma (surprise)
First find FNET in y direction:
FNET = mg +T y mg T

21. Motion in a Circle... FNET = mg +T Acceleration in y direction: v
ma = mv2 / R
mg + T = mv2 / R
T = mv2 / R - mg v y mg T
F = ma R

22. Motion in a Circle...
What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?
i.e. find v such that T = 0.
mv2 / R = mg + T
v2 / R = g
Notice that this does not depend on m. v mg
T= 0 R

23. Understanding Two-body dynamics
In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope. In case (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. m
10kg a a m
F = 98.1 N
Case (1)
Case (2)
(a) Case (1) (b) Case (2) (c) same

24. Solution For case (1) draw FBD and write FNET = ma for each block: (a)
T = ma (a)
mWg -T = mWa (b) m
10kg a
Add (a) and (b):
mWg = (m + mW)a
mW=10kg
(b)

25. Solution T = 98.1 N = ma For case (2) m 10kg a Case (1) m a F = 98.1 N
The answer is (b) Case (2). In this case the block experiences a larger acceleration

26. Problem: Two strings & Two Masses on horizontal frictionless floor:
Given T1, m1 and m2, what are a and T2?
T1 - T2 = m1a (a)
T2 = m2a (b)
Add (a) + (b): T1 = (m1 + m2)a a
Plugging solution into (b): i m2 m1 T2 T1 a

27. Understanding Two-body dynamics
Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? T3 T2 T1 3m 2m m a
(a) T1 > T2 > T (b) T3 > T2 > T (c) T1 = T2 = T3

28. Solution Draw free body diagrams!! T3 3m T3 = 3ma T3 T2 2m
T2 - T3 = 2ma
T2 = 2ma +T3 > T3 T2 T1 m
T1 - T2 = ma
T1 = ma + T2 > T2
T1 > T2 > T3

29. Alternative Solution T3 T2 T1 3m 2m m a
Consider T1 to be pulling all the boxes T3 T2 T1 3m 2m m a
T2 is pulling only the boxes of mass 3m and 2m T3 T2 T1 3m 2m m a
T3 is pulling only the box of mass 3m
T1 > T2 > T3

30. Problem: Rotating puck & weight.
A mass m1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m2.
What is the tension (T) in the string?
What is the speed (v) of the sliding mass? m1 m2 v R

31. Problem: Rotating puck & weight...
Draw FBD of hanging mass:
Since R is constant, a = 0.
so T = m2g m2
m2g m1 m2 v R T

32. Problem: Rotating puck & weight...
T = m2g N
Draw FBD of sliding mass:
T = m2g m1
Use F = T = m1a
where a = v2 / R
m1g
m2g = m1v2 / R m1 m2 v R T