Transcript Exercise Class

Exercise Class
For College Physics
俞颉翔(Jiexiang Yu)
2010-10-20
Email: [email protected]
Office: 2401, East Guanghua Building
Problem 4.58 on P113

Consider the 52kg mountain climber.
a)
b)
Find the tension in the rope (T) and the force
that the climber must exert with her feet on
the vertical rock face to remain stationary (Fl).
What is the minimum coefficient of friction
between her shoes and the cliff?

For x direction:
T sin 31o  Fl cos15o
So
sin31o
Fl  T
 0.533T
o
cos15

For y direction:
31o
T cos31o  Fl sin 15o  m g
T
0
sin 31
0
T (cos31 
sin
15
)  mg
0
cos15
T hen wehave:
T  1.005m g  512.1N
y
Fl
0
Fl  0.533T  273.0 N
x
15o
W=mg
Problem 4.58 on P113
——Sine Law
a
b
c


 2R
sin A sin B sin C
R refers to the radius of circumscribed
circle in a triangle.
Another solution

Since the system remains stationary, the
additions of the force vectors is zero.
Fl
mg
T


sin 74o sin 75o sin 31o
sin 75o
T  mg
 512.1N
o
sin 74
sin 31o
Fl  m g
 273.0 N
o
sin 74
31o
T
W=mg
75o
15o
74o
Fl
74o  75o  31o  180o
Problem 4.58 on P113

Coefficient of friction
N  Fl cos15o
f  Fl sin 15o
T hen t he s is given by
f
 s   t an15o  0.268
N
f
15o
Fl
N
Something about friction



Coefficient of static friction, μs
Coefficient of kinetic friction,μk.
Usually, μs > μk in the same situation
Problem 1

A horizontal force F = 12N pushes a block
weighing 5N against a vertical wall. μs = 0.60
and μk = 0.40. Assume the block is not
moving initially. Will the block start moving?
F
W



N = F =12N
The maximum of static
friction is:
f s max  s N  7.2N  W
So the block remains
stationary.
f
N
W
W
F
Problem 2

F
Someone exerts a force F directly up
on the axle of the pulley. Consider the
pulley and string to be mass-less and
the bearing frictionless. Two object,
m1=1.2kg and m2=1.9kg, are attached
to the opposite ends of the string,
which passes over the pulley. The m2
is in contact with the floor.
a)
b)
c)
Find the largest value of F may have so
that m2 will remain at rest on the floor.
what is the tension in the string if the
upward force F is 110N?
With the tension determined in b), what
is the acceleration of m1?
m1
m2
a)
Since the pulley is mass-less, the
net force exerted on it is zero.
Thus the tension in the string:
1
T F
2
Then we get the maximum of the tension:
F
T
T
Tmax  m2 g
and the maximum of F:
Fmax  2Tmax  2m2 g  2 1.9  9.8  37.24N
T
m2
W2 = m2g
b)
The tension in the string:
1
T  F  55 N
2
c)
T
m2
W2=m2g
The acceleration of m1:
T  m1 g 55
a1 

 9.8  36.03m/s2
m1
1.2
T
m1
W 1= m1g
Problem 3

The two blocks, m = 16kg and M = 88kg are
free to move. The coefficient of static
friction between the blocks is μs = 0.38, but
the surface beneath M is frictionless. Find
the minimum horizontal force F required to
hold m against M.
F
m
M
No friction

Acceleration of the two blocks:
F
m
F
a
mM

No friction
Acceleration of the m block:
FN
a
m

M
N
f
F
m
W=mg
The friction :
f  mg  s N
N’

Acceleration of the M block:
N
a
M
M
No friction
f
N
W=Mg

From the same acceleration
of the two blocks, we have:
F
m
No friction
M
N
F
mM

Put this into the function of
the friction f:
f  m g  s N  s
M
N
f
F
m
W=mg
M
F
mM
So
N’
mM
16  88
F  mg
 16 9.8 
 487.66N
M s
0.38 88
No friction
M
f
N
W=Mg
Problem 4

You throw a ball with a speed of 25.3m/s at an angle
of 42.0o above the horizontal directly toward a wall.
The wall is 21.8m from the release point of the ball.
a)
b)
c)
d)
how long is the ball in the air before it hits the wall?
how far above the release point does the ball hit the
wall?
What are the horizontal and vertical components of its
velocity as it hits the wall?
Has it passed the highest point on its trajectory when
it hits?
25.3m/s
42.0o
21.8m
Solution
a)
The time taken for the ball to hit the wall:
l
21.8
t

 1.16s
o
v0 cos 25.3  cos 42
b)
The Vertical distance above the release
point as the ball hits the wall:
1 2
y  v0 sin t  gt  13.04 m
2
25.3m/s
42.0o
21.8m
Solution
c)
The vector of velocity as it hits the wall:
vx  vx 0  v0 cos  18,8m/s
v y  v y 0  gt  v0 sin   gt  5.57m/s
d)
Since vy > 0, the ball hasn’t passed the peak
point of its trajectory.
25.3m/s
42.0o
21.8m
Problem 5

A chain consisting of five
links, each with mass 100g, is
lifted vertically with a
constant acceleration of
2.5m/s2. Find:
a)
b)
c)
The forces acting between
adjacent links,
The force F exerted on the top
link by the agent lifting the
chain,
The net force on each link.
F
5
4
3
2
1
For link 1:

a
t 21  w
, t 21  m(a  g )  0.1 (2.5  9.8)  1.23 N
m

For link 2:
a
t32  t12  w
, t32  m(a  g ) t 12  0.1 (2.5  9.8)  2  2.46 N
m

For link 3:

For link 4:
a
t54  t34  w
, t54  m(a  g ) t 34  0.1 (2.5  9.8)  4  4.92 N
m

For link 5:
t t  w
a  43 23
, t 43  m(a  g ) t 23  0.1 (2.5  9.8)  3  3.69 N
m
F  t 45  w
a
, F  m(a  g ) t 45  0.1 (2.5  9.8)  55  6.15 N
m
F
5
4
3
2
1


The net force on each link:
Fn  ma  0.1 2.5  0.25N
F
Question：

A vertical force F is exerted on a rope
of which mass is M and length L with
a constant acceleration of a. Find the
tension as a function of the distance x
from the bottom of the rope.
x