Download Chapter 4 P36 THE PROBLEM STATEMENT

Chapter 4 P36 Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 4 P36. An object with mass m1 =5.00 kg rests on a frictionless horizontal
table and is connected to a cable that passes over a pulley and is then
fastened to a hanging object with mass m2 = 10.0 kg, as shown in Figure
P4.36. Find the acceleration of each object and the tension in the cable.
Ch4 P40. In Figure P4.36, m1 = 10 kg and m2 = 4.0 kg. The coefficient of
static friction between m1 and the horizontal surface is 0.50, and the
coefficient of kinetic friction is 0.30. (a) If the system- is released from rest, what will its
acceleration be? (b) If the system is set in motion with m2 moving downward, what will be the
acceleration of the system?
Ch4 P45. Objects with masses m1 = 10.0 kg and m2 = 5.00 kg are connected by a light string that
passes over a frictionless pulley as in Figure P4.36. If, when the system starts from rest, m2 falls
1.00 m in 1. 20 s, determine the coefficient of kinetic friction between m1 and the table.
Page 1 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 4 P36. An object with mass m1 =5.00 kg rests on a frictionless horizontal table and is connected to a
cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 10.0 kg, as shown
in Figure P4.36. Find the acceleration of each object and the tension in the cable
Ch4 P40. In Figure P4.36, m1 = 10 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and
the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system- is released
from rest, what will its acceleration be? (b) If the system is set in motion with m2 moving downward, what
will be the acceleration of the system?
Ch4 P45. Objects with masses m1 = 10.0 kg and m2 = 5.00 kg are connected by a light string that passes
over a frictionless pulley as in Figure P4.36. If, when the system starts from rest, m2 falls 1.00 m in 1. 20 s,
determine the coefficient of kinetic friction between m1 and the table.
BRAINSTORMING-Definitions, concepts , principles and
Discussion
Discussion
A. m1 has forces in both the horizontal and vertical directions. It is in
equilibrium in the vertical direction (Fv = 0). So, Newton’s 2nd Law must be
applied in both directions.
B. m2 has forces only in the vertical direction. So, Newton’s 2nd Law must be
applied in vertical directions.
C. F = k N , 0 in Prob. 36; static s = 0.5 and kinetic k = 0.3 in Prob 40; and find k in Prob. 45
Problem 36
Replace all physical contacts
with action-reaction pairs by
rd
Newton’s 3 Law of Motion.
pulley pulls m1
to the right.
m1
T
m1
m1 pulls pulley
Tto the left.
Floor pushes
N
m 1 up.
T
Newton’s 2nd Law must be
applied to Free Body Diagram.
FDB of m1
T T
a
N W1=m1g
T
m 2 pulls
pulley down.
m2
N
T Pulley pulls
a
m2 up.
m1 pushes N
down on floor..
W2=m2g
FDB of m2
m2
Problems 40 &45
Replace all physical contacts
with action-reaction pairs by
rd
Newton’s 3 Law of Motion.
m1
F
a
T
F
m2
N
a
T
F
m 1 pulls floor to the
right in reaction to
floor’s action on m1 .
N
T
T
is pulling m1 to right.
N
T T
N W1=m1g
m2
FDB of m2
Vertical direction (up Positive)
In equilibrium
Fv = 0; N – m1g = 0
For m2
Only vertical .( down ,direction
of a, Positive.)
Fv = m2a ; m2g -T = m2a
FDB of m1
Everything is the same as in Prob 36
except friction is added.
Floor by way of friction
T T
m1
pulls m1 to left because T
F
For m1
Horizontal direction.( ->
,direction of a, Positive.)
Fh = m1a ; T = m1a .
T
W2=m2g
Problems 40&45 same as
Problem 36 except add friction
for m1.
For m1
Horizontal direction.( ->
,direction of a, Positive.)
Fh = m1a ; T - F = m1a .
The diagrams for Problem 36 and Problems 40 & 45 are identical except
for the friction force F .
Page 2 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 4 P36. An object with mass m1 =5.00 kg rests on a frictionless horizontal table and is
connected to a cable that passes over a pulley and is then fastened to a hanging object with mass
m2 = 10.0 kg, as shown in Figure P4.36. Find the acceleration of each object and the tension in
the cable
Ch4 P40. In Figure P4.36, m1 = 10 kg and m2 = 4.0 kg. The coefficient of static friction between
m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the
system- is released from rest, what will its acceleration be? (b) If the system is set in motion with
m2 moving downward, what will be the acceleration of the system?
Ch4 P45. Objects with masses m1 = 10.0 kg and m2 = 5.00 kg are connected by a light string that
passes over a frictionless pulley as in Figure P4.36. If, when the system starts from rest, m2 falls
1.00 m in 1. 20 s, determine the coefficient of kinetic friction between m1 and the table.
Basic Solution - Strategy : Solve Problem 40 symbolically then use this to solve
all three by substituting the appropriate numbers.
General Symbolic Solution
For m1
Vertical direction (up Positive)
In equilibrium
Fv = 0; N – m1g = 0. So N = m1g
Horizontal direction.( -> ,direction of a, Positive.)
Fh = m1a ; T - F = m1a . F= N = m1g. So, T = m1a + F = m1a + m1g = m1(a + g) (1)
For m2
Only vertical .( down ,direction of a, Positive.)
Fv = m2a ; m2g -T = m2a So, from (1) m2g - m1a - m1g = m2a, gives a= (m2 - m1)g/(m1 + m2). (2)
Problem 39: m1 =5.00 kg, m2 = 10.0 kg , coefficient  =0, so F = 0. Find a and T.
a= (m2 - m1)g/(m1 + m2) = (10.0kg-0)9.8m/s2/(10.0 +5.00)kg = 10.0*9.8m/s2/15 = 6.53 m/s2.
T = m1(a + g) = 5.00kg(6.53-0)m/s2 = 5.00*6.53 kg m/s2 = 32.7 N.
Problem 40. m1 =10 kg, m2 = 4.0 kg ; coefficient of friction static s = 0.5 , kinetic k = 0.3.
(a) Max static tension with a=0 m/s2 Tstatic = m1(a + sg) = 10kg (0+0.5*9.8)m/s2 = 10*5*9.8 m/s2
= 49 N
This exceeds the pulling force of W2 = m2g= 4.0 kg*9.8m/s2 = 39N So, there in no movement against
the static friction.
(b) So, we must solve for a using the kinetic friction, , kinetic k = 0.3.
akinetic = (m2 - km1)g/(m1 + m2) = (4.0kg-0.3*10kg)9.8m/s2/(10.0 +4.0)kg = 9.8/14 m/s2 = 0.70 m/s2 .
Problem 45. m1 =10 kg, m2 = 5.0 kg . m2 falls y = -1.00m in t = 1.20 s. Find k .
With v0 = 0m/s , released from rest, y = v0 t – ½ at2 = - ½ at2
So, a = - 2y/ t2 = -(-2*1.00m)/1.2s2 = 1.39 m/s2/.
Solving for k from a = (m2 - km1)g/(m1 + m2) .
a*(m1 + m2)= (m2 - km1)g , a* (m1 + m2)/g= (m2 - km1), a/g *(m1 + m2) - m2 = - km1 .
So, k = (m2 - a/g *(m1 + m2))/m1 = (5.0kg - 1.39/9.8 *(10+5.0)kg)/10kg = (5.0kg -2.128)/10
=0.287.
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