Download Exercise_4a_Solutions

1.) The NET force required to keep a 2kg object accelerating at
10m/s2 is:
F  ma
net
Fnet
Fnet
 m 
 2kg10 2 
 s 
 20N
2.) The magnitude of the weight of a 2kg object is:
w  mg
Fnet
Fnet


m
 2kg9.8 2 

s 
 19.6N
Fnet  19.6N
3.) The NET force required to keep a 2kg object moving with a
constant velocity of 10m/s is:
Fnet  ma
constant velocity  a = 0
Fnet  0
Questions 4-6
A 30kg object is at rest on a horizontal surface. The coefficients of
static and kinetic friction are; ms = 0.55, mk = 0.25. A force of
F=125N at 20° above the horizontal is applied to the object.
F=125N
30kg
20°
Forces Acting on Object:
N
F
Fy  F sin
30kg
Fx  Fcos 
f  mN
mg
Newton’s Second Law for Object:
x-direction
F cos   mN  ma
y-direction
N  F sin  mg  0
N  mg  Fsin 
F cos   mmg  Fsin   ma
Fcos   mmg  F sin
a
m
Will the object accelerate?
Fcos   m s mg  Fsin 
a

0
m
F(cos   m s sin)  m s mg
a
0
m
125Ncos 20  .55(sin 20)  .55(30kg)(9.8 m2 )
s
a
30kg
a  0.69 m2  0
s
Object will NOT accelerate.
What is the minimum force required to start the object moving?
F(cos   ms sin )  m s mg
0
m
F(cos   ms sin )  m s mg  0
m s mg
F
(cos   m s sin)
.55(30kg)(9.8 m2 )
F  cos 20  .55(sins20)
F  143.3N
Questions 7-10
In the diagram below object #1 has a mass of 10kg and object #2
has a mass of 8kg. Both inclines are frictionless.
y
x
x
y
#1
#2
1.5m
.75m
30°
55°
Forces Acting on m1 and m2
N1
T
#1
T
N2
#2
m1gsin1
30°
m 2 g cos2
m1 gcos1
55°
m2g sin 2
Newton' s Second Law for m 1
x  direction
T  m 1g sin1  m 1a 1
T  m 1 a1  m1 g sin 1
y  direction
N1  m1 gcos 1  0
N1  m1 gcos 1
Newton' s Second Law for m2
x  direction
T  m 2 gsin 2  m 2 a 2
T  m 2 a 2  m 2 gsin 2
y  direction
N 2  m 2 g cos 2  0
N 2  m 2 g cos 2
m1 a 1  m1 gsin 1  m2 a 2  m2 gsin 2
a 2  a 1
m1 a1  m1 gsin 1  m2 a1  m2 g sin 2
m1 a1  m2 a1  m2 gsin  2  m1g sin1
a 1 m1  m2   m2 gsin 2  m1 gsin 1
m 2 sin 2  m1 sin1 
a 1  g 

 m1  m 2 


 m 8kgsin 55  10kgsin 30 
a1  9.8 2 

10kg  8kg

 s 
a1  .85 m2
s
a2  a1  -.85 m2
s
T  m1 a1  m1 gsin 1
 m 
 m 
T  10kg.85 2   10kg9.8 2 sin 30
 s 
 s 
T  57.5N
T  m 2 a 2  m2 gsin 2


 m 
m
T  8kg.85 2  8kg9.8 2 sin 55

 s 
s 
T  57.4N
Questions 11-13
Three objects are connected as shown below. Object #1 has a mass
of 18kg and moves on a horizontal surface. Object #2 has a mass of
15kg and moves on an incline of 50° . Object #3 has a mass of 30kg.
There is friction on all surfaces with mk= 0.3. When the system is
released object #2 is seen accelerating up the incline.
30kg
Forces and Newton’s Second Law for Object #1
N1
f1
#1
T1
a1
m1 g
T1  f1  m1 a 1
N1  m 1 g  0  N1  m 1g
T1  mm 1 g  m1 a 1
T1  m1a1  mm1 g
Forces and Newton’s Second Law for Object #2
N2
f2
T1
a2
#2
T2
m 2 g sin
m 2 g cos
T2  T1  f2  m 2 gsin   m 2 a 2
N 2  m 2 g cos   0  N 2  m 2 g cos 
T2  T1  mm 2 gcos   m2 gsin   m 2 a2
T2  m2 a 2  T1  mm2 g cos   m2 gsin
Forces and Newton’s Second Law for Object #3
T2
a3
#3
m3g
T2  m3 g  m3 a 3
T2  m3 a 3  m3 g
T1  m1a1  mm1 g
T2  m2 a 2  T1  mm2 g cos   m2 gsin
T2  m3 a 3  m3 g
a 1  a 2  a 3
T1  T2  m 2 a 2  mm 2 gcos   m2 gsin 
m1a1  mm1g  T2  m2a 2  mm2g cos  m2g sin
m1a1  mm1g  m3a 3  m 3g  m2a 2  mm2g cos  m2 gsin
m1a1  mm1g  m3a1  m 3g  m2a1  mm 2gcos  m2gsin
m1 a1  m3 a1  m2 a1 m3 g  mm2 gcos   m 2 gsin   mm1 g
g m3  mm2 cos   m 2 sin  mm1 
a1 
m1  m3  m2
a1 
9.8 30  .55(15)(cos 50)  (15)(sin 50)  .55(18)
18  15  30
a1  1.6 m2
s
a2  1.6 m2
s
a3  1.6 m2
s
T1  m1a1  mm1 g
 m 
 m 
T1  18kg1.6 2  .318kg9.8 2 
 s 
 s 
T1  81.7N
T2  m3 a 3  m3 g


 m 
m
T2  30kg1.6 2  30kg9.8 2 

 s 
s 
T2  246N