Transcript SPH4U: Lecture 7 Notes

SPH4U: Practice Problems
Today’s Agenda
Run and Hide
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
ideal peg
or pulley
| F1 | = | F2 |
F2 = T j
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

Accelerometer...

Draw a free body diagram for the mass:
 What are all of the forces acting?
T (string tension)

m
i
mg (gravitational force)
Accelerometer...

Using components (recommended):
i: FX = TX = T sin  = ma
j: FY = TY - mg
= T cos - mg = 0

T

j
m
i
ma
mg
Accelerometer...

Using components :
i: T sin  = ma
T
j: T cos - mg = 0

j
m
i
ma

mg
Eliminate T :
T sin  = ma
T cos = mg
tan  
a
g
Accelerometer...

Alternative solution using vectors (elegant but not as systematic):

Find the total vector force FNET:
T
mg 
FTOT
T (string tension)

m
mg (gravitational force)
Accelerometer...



Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
Recall that FNET = ma:
T (string tension)
T
mg 

m
ma
mg (gravitational force)

So
ma a
tan  

mg g
tan  
a
g
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 = 0.28 .

tan  
 = arctan (a/g) = 15.6 deg
a

a
g
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
(A)
(B)
(C)
(D)
(E)
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.
Angles of an Inclined plane
The triangles are similar, so the angles are the same!
ma = mg sin 
N

mg

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

Inclined plane...

Define convenient axes parallel and perpendicular to plane:
 Acceleration a is in x direction only.
j
m
a

i
Inclined plane...

Consider x and y components separately:
i: mg sin  = ma.
a = g sin 

j: N - mg cos  = 0.

N = mg cos 
ma
j
mg sin 
N 
mg cos 
mg
i
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
Problem: Two Blocks

Realize that F = (m1+ m2) a :
F / (m1+ m2) = a

Draw FBD of block m2 and apply FNET = ma:
F2,1 = m2 a
F2,1
m2

Substitute for a :
F2,1


F

 m2 
 (m1 + m2 )
F2,1 
m2
F
(m1 + m2)
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
Problem: Tension and Angles

T1
Draw a FBD:
T1sin 

T1cos 
T2
j
T2sin 

T2cos 
i
mg

Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
Fx,NET = T1cos  - T2cos  = 0
T1 = T2
Fy,NET = T1sin  + T2sin  - mg = 0
mg
T1 = T2 =
2 sin 
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
Motion in a Circle...



Draw a Free Body Diagram (pick y-direction to be down):
We will use FNET = ma (surprise)
y
First find FNET in y direction:
FNET = mg +T
mg
T
Motion in a Circle...
FNET = mg +T

v
Acceleration in y direction:
ma = mv2 / R
y
mg
T
F = ma
mg + T = mv2 / R
R
T = mv2 / R - mg
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.
v
mv2 / R = mg + T
v2
/R=g
mg
T= 0
v  Rg

Notice that this does
not depend on m.
R
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
a
m
a
10kg
F = 98.1 N
Case (1)
(a)
Case (2)
Case (1)
(b) Case (2)
(c)
same
Solution

For case (1) draw FBD and write FNET = ma for each block:
(a)
T = ma
(a)
mWg -T = mWa

(b)
m
Add (a) and (b):
mWg = (m + mW)a
a
10kg
mW=10kg
(b)
mW g
a
m + mW
Solution

For case (2) T = 98.1 N = ma
a
a
98.1N
m + 10 kg
98.1N
m
a
98.1N
m
m
a
m
a
10kg
F = 98.1 N
Case (1)

Case (2)
The answer is (b) Case (2). In this case the block experiences a
larger acceleration
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
T1 = (m1 + m2)a
a 
m1 + m2
m2
Plugging solution into (b): T2  T1
m1 + m2
i
m2
T2
m1
T1
a
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?
a
3m
(a) T1 > T2 > T3
T3
2m
T2
(b) T3 > T2 > T1
m
T1
(c) T1 = T2 = T3
Solution

Draw free body diagrams!!
T3
3m
T3 = 3ma
T2 - T3 = 2ma
T3
T2 = 2ma +T3 > T3
T1 - T2 = ma
T2
T1 = ma + T2 > T2
T1 > T2 > T3
2m
m
T2
T1
Alternative Solution
a
Consider T1 to be pulling
all the boxes
3m
T3
T2
2m
m
T1
a
T2 is pulling only the
boxes of mass 3m
and 2m
3m
T3
T2
2m
m
T1
a
T3 is pulling only the
box of mass 3m
3m
T1 > T2 > T3
T3
2m
T2
m
T1
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?
v
m1
R
m2
Problem: Rotating puck & weight...

Draw FBD of hanging mass:
 Since R is constant, a = 0.
so
T = m2g
R
T
m2
m2
m2g
v
m1
T
T = m2g
Problem: Rotating puck & weight...
N

T = m2g
Draw FBD of sliding mass:
m1
Use F = T = m1a
m1g
where a = v2 / R
v
m2g = m1v2 / R
v
m1
R
T
m2
gR
m2
m1