Transcript Examples and Hints for Chapter 5

Examples and Hints for
Chapter 5
Problem 5.17
A light rope is attached to a block with mass 4 kg
that rests on a frictionless, horizontal surface.
The horizontal rope passes over a
frictionless, massless pulley and a block with
mass m is suspended from the other end.
When the blocks are released, the tension in
the rope is 10 N.
a) Draw two free body diagrams, one for the 4
kg block and one for the block with mass m.
b) What is the acceleration of either block?
c) What is the mass m of the hanging block?
Free Body Diagrams
4 kg
T
T
mg
Use the sum of the forces
F or the horizontal block
ma 
F
4 a  10
a  2.5 m / s
2
F or the vertical block,
ma  T  mg
m (  2.5)  10  9.8 m
m  10 /(9.8  2.5)
m  1.37 kg
Problem 5.24
A box of bananas weighing 40 N rests on a horizontal surface. The
coefficient of static friction between the box and the surface is
0.40 and the coefficient of kinetic friction is 0.20.
a)
If no horizontal force is applied to the box and the box is at rest,
how large is the frictional force exerted on the box?
b)
What is the magnitude of the frictional force if a monkey applies
a horizontal force of 6 N to the box and the box is initially at
rest?
c)
What minimum horizontal force must the monkey apply to start
the box in motion?
d)
What minimum horizontal force must the monkey apply to keep
the box moving at a constant velocity?
e)
If the monkey applies a horizontal force of 18 N, what is the
magnitude of the friction force and what is the box’s
acceleration?
Free body diagrams
n-40=0
6-f=ma
n
f
6N
40 N
Part a) if there is no motion, is there
friction? No
n-40=0
6-f=ma
n
f
6N
40 N
Part b) f=ms*n for static situation
n-40=0
6-f=ma
n
f
6N
40 N
Therefore, n=40
and
ms=0.4 so
f=0.4*40=16 N
So 16 N is the
magnitude no
matter what
force applied!
Part c) f=ms*n for static situation
n-40=0
6-f=ma
n
f
6N
40 N
Therefore, n=40
and
ms=0.4 so
f=0.4*40=16 N
So 16 N is the
maximum static
friction so this is
what must be
applied
Part d) f=mk*n for moving situation
n-40=0
6-f=ma
n
f
6N
40 N
Therefore, n=40
and
mk=0.2 so
f=0.2*40=8 N
So 8 N is the
minimum force
which must be
applied to keep
the box moving
Part e) f=mk*n for moving situation
n-40=0
18-f=ma
n
f
18 N
40 N
Therefore, n=40
and
mk=0.2 so
f=0.2*40=8 N
m=40/9.8=4.08 kg
18-8=4.08*a
a=2.45 m/s2
Problem 5.61
Block A in the figure weighs 60 N. The coefficient of friction between
the block and the surface on which it rests is 0.25. The weight,
w, is 12 N and the system is in equilibrium.
a)
Find the frictional force exerted on block A
b)
Find the maximum weight for which the system will remain in
equilibrium
450
60 N
12 N
Free Body Diagrams
n
f
T
450
Th
T*sin(450)
T*cos(450)
60 N
,Note: at 450, Tsin(450)=Tcos(450)
Tv
450
60 N
12 N
12 N
From examination, T*cos(450)=12 N
=T*sin(450)
n
f
T
Th
450
T*sin(450)
T*cos(450)
60 N
,Note: at 450, Tsin(450)=Tcos(450)
Tv
So Th=Tv=12 N and since system in equilibrium,
f=Tv=12 N
12 N
From examination, T*cos(450)=12 N
=T*sin(450)
n
f
T
450
Th
T*sin(450)
T*cos(450)
60 N
,Note: at 450, Tsin(450)=Tcos(450)
Tv
n-60 N=0 so n=60 N
F=ms*n=0.25*60=15 N
Since TV=Th, then maximum weight
Is 15 N
12 N
Hints
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Problem 5.16– Draw it! Find the sum of the forces parallel to the
surface and set them equal to mass*acceleration
Problem 5.18—Sum the forces and calculate max acceleration.
Use my favorite kinematic formula (hint: it starts vf2=..) to solve for
distance. In part b) consider only the first glider
Problem 5.25– Remember constant speed implies that the
acceleration=0
Problem 5.50– Look at the nice example on page 187
Problem 5.60—Use the tension in the rope to connect between
the two blocks. In part a, see hint on 5.25. on part b) downward
motion implies friction is less.
Problem 5.62—Consider the blocks as a single unit of weight w
Problem 5.76– Find the maximum acceleration ams*g and use
my favorite formula (see problem 5.18). Back calculate to find v0