Force and Motion II: Group review
questions
Problem 1
Remembering the example used in class, the following pulley system was designed to
help lift a 100 kg weight:
T
T
2T
Pull
100
kg
Unfortunately, the pulley system is still not enough of an advantage for you to lift 100 kg
of weight. Can you design a better system to lift this weight?
Solution
The students can get creative with this one. The present system is able to reduce the
force needed to lift the block by ½. The following Pulley systems can reduce the force
by 1/3 or ¼:
T
T
T
2T
T
T
2T
100k
g
T T
T
Gravity:
100kg* 9.8 m/s2
100kg*9.8 m/s2 = 980 N =
3T
2T
2T
2T
2T
100k
g
T
Gravity:
100kg* 9.8 m/s2
100kg*9.8 m/s2 = 980 N =
4T
Problem 2
Variation of Serwat, Jewett, Physics for Scientists and Engineers, 6th Edition, page 135.
A hockey puck with mass m is placed on ice with an
initial speed V0=25 m/s as shown in Figure . After
traveling 150 m the puck is finally at rest. Determine the
coefficient of kinetic friction for the given
problem.
m
V0
Figure 2: sliding hockey puck
Solution
Which implies that
y
x
m
fk
n
V0
w
adopted frame of reference, the
negative sign in the acceleration
indicates that the puck is slowing down.
Because the acceleration is constant,
we can use:
Using
From Newton’s second law expressed in
component form we have:
and
Finally,
On the other hand, we know that
and
. Thus, we have:
. With the
we have
Problem 3
Two steel blocks with mass m1=2 kg and
m2=1kg are placed one over the other as
shown in Figure . The coefficient of static
friction between the blocks is μs=0.5 and
there is no friction between the first block
and the floor. What is the force F needed
to make the blocks slide with respect to
each other?
m
m1
F
Figure 3: when do the blocks slide?
Solution
.
Therefore,
For
the
x-
direction we have
y
m
x
n
fs
w2
Since
m1
n1
we can
compute the x-component of the upper
body acceleration is at this limiting case
n
fs
and
F
Newton’s second law expressed in
component form applied to the second
body gives
w1
In the lecture we learned that the
maximum friction force between two
bodies is given by
where
is
the static friction coefficient and
is the
normal force to the contact surface. In
order to solve the given problem we will
solve for the force F in this limiting
situation, and therefore any force larger
than F will produce the relative sliding of
the blocks. For the body on top we
have:
where
is the same as before due to
the action-reaction principle. Thus, we
have
Since the bodies are not sliding yet, their
accelerations
must
coincide,
i.e.
. Substituting the expression for
into the previous expression we have
Using the numerical values provided we
finally get
Any force larger than
sliding.
will produce
Problem 4
A block of mass m is hanging on two springs with
constants k1 and k2 as shown in Figure .
a) What is the displacement of the block at
equilibrium?
b) Is there any way to replace both springs
with an equivalent one? If so, what is the
equivalent spring constant?
k2
k1
m
Figure 4: mass hanging on two parallel
springs
Solution
y
f1
b) As it can be seen on the previous
expression, it can be rearranged to
have the form
f2
m
with
w
a) From Newton’s second law expressed
in component form we have:
Let’s suppose that the mass
equilibrium for a displacement
is in
from
its original position. Applying Hooke’s
we
law and using the fact that
have:
Therefore, the displacement of the
block at equilibrium is given by
and
.
Therefore, the springs can be replaced
by an equivalent spring with an
equivalent stiffness
.
Brain Teaser:
In case A,
Think carefully about this one: In both
Case a and Case b shown below a net
force of 100 N results in the acceleration
of Block A across the table toward the
pulley. Disregard friction altogether.
The acceleration of Block A is:
T
A
10kg
a) Greater in case A
T
b) Greater in case B
B
c) The same in both cases
10kg
Answer: Greater in Case b
Fg
mA = T
In case B,
mBa = Fg – T
the acceleration of the block:
mBa = Fg - mAa
mA * a = T => a = T/mA
a * (mA + mB) = Fg
T = 100 N
a = Fg/ (mA +mB) = 100 N / (10 kg + 10
kg)
a = 100N/10kg = 10 m/s2
a = 5 m/s2