Page 1 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! F 1 = -T i ideal peg or pulley F 2 = T j | F 1 | = | F 2 | massless rope 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? m T (string tension) mg (gravitational force) i
Transcript
Page 1
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!
F1 = -T i
ideal peg
or pulley
F2 = T j
| F1 | = | F2 |
massless rope
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?
m
T (string tension)
mg (gravitational force)
i
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Accelerometer...
Using components (recommended):
i: FX = TX = T sin = ma
j: FY = TY - mg
= T cos - mg = 0
T
mg
m
ma
j
i
Accelerometer...
Using components :
i: T sin = ma
j: T cos - mg = 0
Eliminate T : mg
m
ma
T sin = ma
T cos = mgtan
a
g
j
i
T
Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
T
mg
FTOT
m
T (string tension)
mg (gravitational force)
Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
Recall that FNET = ma:
So
ma
tanma a
mg g
T
mg
tana
g
m
T (string tension)
mg (gravitational force)
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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 .
= arctan (a/g) = 15.6 deg
tana
g
a
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) Have equal magnitudes and form an action/reaction pair
(B) Have equal magnitudes but do not form an action/reaction pair
(C) Have unequal magnitudes and form an action/reaction pair
(D) Have unequal magnitudes and do not form an action/reaction pair
(E) 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
ma = mg sin
mg
N
The triangles are similar, so the angles are the same!
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 ?
ma
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Inclined plane...
Define convenient axes parallel and perpendicular to plane:
Acceleration a is in x direction only.
ma
i
j
Inclined plane...
Consider x and y components separately:
i: mg sin = ma. a = g sin
j: N - mg cos = 0. N = mg cos
mg
N
mg sin
mg cos
ma
i
j
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?
m1 m2
F
Problem: Two Blocks Realize that F = (m1+ m2) a :
Draw FBD of block m2 and apply FNET = ma:
F2,1F2,1 = m2 a
F / (m1+ m2) = a
m22,1 ( )
+
m2m1
FF
Substitute for a :
m2
(m1 + m2)
m2F2,1 F
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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
Draw a FBD:
T1 T2
mg
Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
T1sin T2sin
T2cos T1cos
j
i
Fx,NET = T1cos - T2cos = 0 T1 = T2
2 sin
mgT1 = T2 =Fy,NET = T1sin + T2sin - mg = 0
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?
R
v
T
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 T
mg
y
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Motion in a Circle...
FNET = mg +T
Acceleration in y direction:
ma = mv2 / R
mg + T = mv2 / R
T = mv2 / R - mg
R
T
v
mg
y
F = ma
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 doesnot depend on m.
R
mg
v
T= 0
v Rg
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.
(a) Case (1) (b) Case (2) (c) same
m
10kg
a a
m
F = 98.1 N
Case (1) Case (2)
Solution
m
10kg
a
Add (a) and (b):
mWg = (m + mW)a
W
W
m ga
m m
+
(a)
(b)
T = ma (a)
mWg -T = mWa (b)
For case (1) draw FBD and write FNET = ma for each block:
mW=10kg
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Solution
The answer is (b) Case (2). In this case the block experiences a larger acceleration
98.1Na
mT = 98.1 N = ma For case (2)
m
10kg
a
Case (1)
kg10m
N198a
+
.
m
a
F = 98.1 N
Case (2)
m
N198a
.
Problem: Two strings & Two Masses on
horizontal frictionless floor:
m2 m1T2 T1
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
a
i
1
1 2
T
m m
+
Plugging solution into (b): 22 1
1 2
mT T
m m
+
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) T1 > T2 > T3 (b) T3 > T2 > T1 (c) T1 = T2 = T3
T3 T2 T13m2m m
a
Solution
Draw free body diagrams!!
T33mT3 = 3ma
T3 T22m
T2 - T3 = 2ma
T2 = 2ma +T3 > T3
T2 T1m
T1 - T2 = ma
T1 = ma + T2 > T2
T1 > T2 > T3
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Alternative Solution
T3 T2 T13m 2m m
a
Consider T1 to be pulling all the boxes
T3 T2 T13m 2m m
a
T2 is pulling only the boxes of mass 3m and 2m
T3 T2 T13m 2m m
a
T3 is pulling only the box of mass 3m
T1 > T2 > T3
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
