
1.

We again use the Principle of Superposition to solve this
problem by separating the problem into forces acting along
the horizontal or Xaxis and those acting along the vertical
or Yaxis. Then, for each axis, all forces will be along a
line. We assign a positive or negative sign to the forces,
depending on their direction along the axis, and then add
them algebraically. There is no acceleration in the Ydirection,
a_{y }= 0. (F_{net})_{y }=
ma_{y }=m(0) = 0
(a) and (b),
F_{N}  F_{g} = 0 or
F_{N }= F_{g} = mg .
For m = 2.0 kg, F_{N }= F_{g} = (2.0 kg)(10
m/s^{2}) = 20 N.
(F_{net})_{x} = ma_{x}
F = ma_{x } or
a_{x} = F/m
For (c) a_{x }= 12 N/ 2.0 kg = 6.0 m/s^{2}.
For (d) a_{x }= 24 N/2.0 kg = 12.0 m/s^{2}.
The acceleration is directly proportional to the net force.
When you double the net force, you double the acceleration.
For (e) a_{x }= 12 N/1.0 kg = 12.0 m/s^{2}.
The acceleration is inversely proportional to the mass of
the object. When you halve the mass, you double the acceleration.


2.

Now the net force in the Xdirection depends on an applied
force to the right and a frictional force to the left.
(F_{net})_{x} = ma_{x}
F  f = ma_{x}
 12 N  4 N = 2.0 kg a_{x
} 4.0 m/s^{2} = a_{x
}
 24 N  4 N = 2.0 kg a_{x}
10 m/s^{2} = a_{x}
While the applied force F is doubled, the net force is
not doubled and the acceleration is not doubled. The second
net force = 20/8 x the first net force = 5/2 the first
net force. The second acceleration = 5/2 the first acceleration
= 5/2 x 4.0 m/s^{2} = 10 m/s^{2}.


3.

In Fig. 2a above, I have isolated (shown by dashed box) the
entire system in order to find the acceleration of the two blocks.
In Fig. 2b above, I have isolated A and B separately in order
to find F_{AB }and F_{BA}. F_{AB}
is the force of B on A and F_{BA} is the force of A
on B. F_{AB }is an external force to the system
which consists only of A and F_{BA} is an external force
to the system which consists only of B. F_{AB}
and F_{BA} are internal forces to the system of Fig.
2a. I do not use a Yaxis in the figures because I am
not interested in the normal forces and there is no friction
in the problem. For the entire system of Fig. 2a, I know the
external force that acts on it along the Xaxis and I know the
mass of the system so I can find its acceleration. Since I am
interested only in the Xdirection, I drop the subscript x on
F_{net }and a.
(a) F_{net external }= (mass of system)a
12
N = (2 + 4)kg (a) or
12
N/6 kg = 2 m/s^{2} = a
Now we isolate block A to find F_{AB }since we know
m_{A} and a and then isolate block B because we know
everything but F_{BA}, as shown in Fig. 2b. F
is only in contact with A and A pushes B to the right, while
B pushes A to the left.

For
block A 

For
block B 
(b) 
(F_{net external})_{on
A }= (m_{A})a 
(c) 
(F_{net external})_{on
B }= (m_{B})a 

F  F_{AB} =
(2 kg)(2 m/s^{2}) 

F_{BA} = (4
kg)(2 m/s^{2}) 

12 N  F_{AB }=
4 N 

F_{BA }= 8 N 

or
12 N  4 N = 8 N = F_{AB} 


F_{AB} and F_{BA
}are Newton third law of motion forces. They are equal
in magnitude, but opposite in direction. F_{AB},
the force of B on A, has a magnitude of 8 N and is to the left.
F_{BA}, the force of A on B, has a magnitude
of 8 N and is to the right.


4.

It is a little difficult to solve (a) of this problem without
a sneak look at the figure for part (b). There are a lot of
forces here, but Fig. 34a below shows the only external
force in the horizontal direction acting on the system of
the two blocks. Again F_{net external} = ma, where
a is the acceleration of the blocks and m is the mass of the
system we have isolated. There are internal forces acting
in the system, for example, the force of the rope on A, F_{AR},
the force of the rope on B, F_{BR}. And we cannot
forget the force of A on the rope, F_{RA}, or the
force of B on the rope, F_{RB}. I have not drawn the
vectors for the gravitational forces on the blocks or the
normal force of the surface on the blocks because they were
not asked for and there is no frictional force.
While Figures 34b and c above show forces F_{AR},
F_{BR, } F_{RA} and F_{RB},
these are internal to the system shown in Fig. 34a
above. The only external force for the entire system
in the horizontal direction is F = 12 N.
 Thus for the system shown in Fig. 34a above,
(F_{net external})_{x } = (mass
of system)a
12 N = (2 + 4)kg (a) or
12 N/6 kg = 2 m/s^{2} = a
For part (b), let us stare at Fig. 34b and c, which I reproduce
below:
Let's cope with the massless rope first:
(F_{net external})_{on rope }
= (m_{R})a
Taking to the right to be positive,
F_{RB}  F_{RA }= (0)a, where
a = the acceleration of the rope
So,
F_{RB }= F_{RA}
It is important to notice that F_{RB }and
F_{RA}, while equal and in opposite directions,
are not Newton third law of motion forces because
they act on the SAME object. They are equal and in
the opposite direction because they must sum to zero because
a massless object has no net force acting on it. Then by
Newton's third law we know that:
F_{RA }=  F_{AR } and
F_{RB }= F_{BR}.
Thus since the magnitudes of these forces are equal, F_{RA
}= F_{RB}, F_{AR }= F_{BR}.
We throw all this careful notation to the wind and call
F_{AR}, F_{BR}, F_{RA} and F_{RB}
the tension T. With this information we redraw Fig. 34b
below:
 Now we can choose either object to find the tension T
in the rope. You can check the answer by isolating the other
block and finding T.
For block B, in
the Xdirection, 
For block A, in
the Xdirection, 
(F_{net external})_{x}
= (m_{B})a 
(F_{net external})_{x
}= (m_{A})a 
12 N  T = (4 kg)(2 m/s^{2})
= 8 N 
T = (2 kg)(2 m/s^{2})
= 4 N 
So
12 N  8 N = T = 4 N 



5.

The forces acting on the object are the applied force F and
the weight of the object mg. In all cases, F_{net} =
ma
(a) For constant velocity, a = 0 and F  mg = m(0) = 0
or
F = mg = 3.0 kg (10 m/s^{2}) = 30 N
(b) For a constant acceleration, a = 3.0 m/s^{2},^{
}F  30 N = 3.0 kg(3m/s^{2})
or
F = 9.0 N + 30 N = 39 N.


6.

 For entire system (Fig. 4a above),
F_{net } = Ma
F  3mg = (3m)a
9.0 N  (0.30kg)(10 m/s^{2}) = (0.30 kg)a
9.0 N  3.0 N = (0.30 kg)a
20 m/s^{2} = a
 To find the tension at the top of the rope T(b) first
use the lower figure in Fig. 4b above.
F_{net } = Ma
T(b)  2mg = (2m)a
T(b)  2.0 N = (0.20 kg)(20m/s^{2})
T(b)  2.0 N = 4.0 N.
T(b) = 6.0 N
Check with top figure in Fig. 4b above.
F_{net } = Ma
9.0 N  mg  T(b) = ma
9.0 N 1.0 N  T(b)=(0.10 kg)(20 m/s^{2})=2.0 N
6.0 N = T(b)
 The mass of 1/5 of the rope = m/5 = 0.02 kg and its weight=mg/5=
0.2N. Use the top figure in Fig. 4c above.
F_{net }= Ma
9.0 N  1.2 mg  T(c) = (0.12 kg)(20 m/s^{2}) =
2.4 N
T(c) = 5.4 N
Using the bottom figure in Fig. 4c above,
T(c)  1.8 mg = T(c)  1.8 N = (0.18 kg)(20 m/s^{2})
= 3.6 N
T(c) = 5.4 N


7.

(F_{net})_{x}
= ma_{x} 
(F_{net})_{y}
= ma_{y} 
26 N  f = 2.0 kg(a_{x}) 
F_{N } mg = m(0) 

F_{N }= mg = (2.0 kg)(10m/s^{2})=20N 
 f
= µ_{k}F_{N }= 1/5(20 N) = 4.0 N
26 N  4.0 N = 2.0 kg(a_{x})
 11 m/s^{2} = a_{x}


8.

 (F_{net})_{y} = ma_{y
}F_{N } + 26 N sin 22.6^{0}_{ }
mg = 0
F_{N }= + 20 N  26 N(5/13) = 20 N  10 N = 10 N.
f_{ }= µ_{k}F_{N }= (1/5)(10
N) = 2 N
 (F_{net})_{x} = ma_{x
}26 N cos 22.6^{0}  f = 2.0 kg(a_{x})
26 N(12/13)  2 N = 2.0 kg(a_{x})
24 N  2 N = 2.0 kg(a_{x})
11 m/s^{2} = a_{x
}


9.

(a) & (b) The xaxis is chosen in the direction of the
acceleration. The axes, forces and components of forces
are shown in the figure above.
(c) (F_{net})_{x} = ma_{x
}30 N sin 30^{0} = 3.0 kg(a_{x})
15 N = 3.0 kg(a_{x})
5.0 m/s^{2} = a_{x}


10.

 Now there is a frictional force up the plane.
_{ }(F_{net})_{y} = ma_{y
} F_{N }  mg cos 30^{o}_{ }=
0
F_{N }= 30 N (0.866) = 26 N
f = µ_{k }F_{N }= 0.154(26 N) = 4.0
N
 (F_{net})_{x} = ma_{x
}30 N sin 30^{o } f = 3.0 kg(a_{x})
15 N  4.0 N = 3.0 kg(a_{x})
a_{x} = 11/3m/s^{2}


11.

 First isolate m_{1}, taking the Xaxis in
the direction of the acceleration:
(F_{net})_{x}
= m_{1}a_{x} 
(F_{net})_{y}
= m_{1}a_{y} 
T  m_{1}g
sin 16.3^{o}  f = m_{1}a_{x } 
F_{N }
m_{1}g cos16.3^{o }= m_{1}(0) 

F_{N }=
25 N(24/25) = 24 N
f = µ_{k}F_{N }= 1/6(24 N) =
4 N 
T  25 N(7/25) 
4 N = 2.5 kg a 

T  11 N = 2.5 kg
a (Equation
1) 
Note:
I have dropped the subscript on a, the only acceleration. 
Now isolate m_{2},
taking down as positive:
F_{net }= m_{2}a
m_{2}g  T = m_{2}a
20 N  T = 2.0 kg a (Equation
2)
Adding Eq. 1 and Eq. 2:
20 N  11 N = 4.5
kg a or
a = 2.0 m/s^{2}
 Substituting a = 2.0 m/s^{2}
into Eq. 2:
20 N  T = 2.0 kg(2.0
m/s^{2}) and
T = 16 N.


12.

Figure for #12a above is a sketch of the object moving on the
table and all of the forces acting on it. As the object starts
to slip away from the circle, a frictional force acts into the
center of the circle to produce the centripetal acceleration.
I chose the Xaxis to the right because at the moment shown
in the picture, the centripetal acceleration is into the center
of the circle or to the right. The frictional force is the only
force into the circle or, as I have drawn it, along the positive
Xaxis. This is a three dimensional drawing showing that for
clockwise motion, at this instant of time, the velocity is along
the positive Yaxis. The weight of the object is always along
the negative Zaxis and the normal force of the table along
the positive Zaxis. Now our axis of interests are the Xaxis
and the Zaxis. The acceleration in the positive Xdirection
is the centripetal acceleration = v^{2}/r.
 (F_{net external})_{x} = ma_{x}
F_{friction }= mv^{2}/r = 2 kg (2 m/s)^{2}/(0.5
m) = 16 N
 There is no acceleration in the Zdirection.
(F_{net external})_{z} = ma_{z.
}F_{N}  mg = m(0) or
F_{N} = mg = (2 kg)(10 m/s^{2}) = 20 N
As shown in Fig. for #12b above,
Force of table on object = {(F_{friction})^{2
} + F_{N}^{2})^{1/2
}
= {(16)^{2} + (20)^{2}}^{1/2}
N
= {656}^{1/2 }N = 25.6 N
tan Θ = F_{N}/F_{friction
}= 20/16 = 1.25
Θ = 51.3^{o}


13.

 The centripetal acceleration a = v^{2}/r = (3m/s)^{2}/0.5
m = 18 m/s^{2}. The object travels in a counterclockwise
direction so the directions of the velocities are as shown
in Fig. 7a above. Remember the velocity is always tangent
to the path. Thus,
 at B the velocity is to the right,
 at T the velocity is to the left, and
 at S the velocity is down.
 At B the acceleration is into the center of the circle
or up. At B we take the Xaxis up.
 At T the acceleration is into the circle or down. At T
the Xaxis is down.
 At S the acceleration is to the right and the Xaxis is
to the right. All of the accelerations are in the direction
of the net force.
For all cases F_{net }= ma = 2 kg(18 m/s^{2})
= 36 N.
Looking at Fig. 7b below, we see that at B and T, there
are no components of the force in any direction except the
Xdirection so we drop the subscripts on F_{net}.
 At B,
F_{net }= ma
F_{or at B } mg = ma
F_{or at B } 20 N = 36 N
F_{or at B }= 56 N
 At T, I have drawn F_{or at T} down, but I must
use the equation for a final decision, so I write it with
± to find its sign and, therefore, its direction.
F_{net }= ma
±F_{or at T} + 20 N = 36 N so
F_{or at T }= +16 N or it is down.
 At S, F_{or at S } must have
a component into the center or along the +Xaxis to produce
the centripetal acceleration and a component along the +Yaxis
so the net force along the Yaxis is zero keeping the magnitude
of the velocity for the uniform circular motion constant.
Thus, we must look at F_{net} along the X and Yaxes.
^{}
(F_{net})_{x
at S }= ma_{x} 
(F_{net})_{y
at S }= ma_{y} 
(F_{or})_{x
at S } = 36 N 
(F_{or})_{y
at S } mg = m(0)
(F_{or})_{y at S }= mg = 20 N 
F_{or
at S }= {(F_{or})^{2}_{x at
S} + (F_{or})^{2}_{x at S}}^{1/2} 
F_{or
at S }= {(36)^{2} + (20)^{2}}^{1/2
}N = 41 N 

tan
Θ = (F_{or})_{y}/(F_{or})_{x}
= 20 N/ 36 N = 0.55.
Θ = 29^{o} 
_{
}
^{
}


14.

 Forces shown in Fig. 8 below:
 (F_{net})_{x} = ma_{x}
(F_{net})_{y} = ma_{y
} mg sin Θ= ma_{x
}T  mg cos Θ=
mv^{2}/L
Acceleration in the Xdirection = g sin Θ
Acceleration in the Ydirection = v^{2}/L, where
v is speed of the object at that instant and L is the radius
of the circular arc.
a_{y }= v^{2}/L = T/m  g cos Θ
 In general, a_{x }= g sin Θ
(i) at maximum displacement, a_{x}
= g sin Θ_{max}
(ii) at the equilibrium position, Θ
= 0, sin Θ= 0, a_{x}
= 0.
In general, _{ } T mg cos Θ=mv^{2}/L.
 At its maximum displacement, the bob comes momentarily
to rest,
v = 0 and T = mg cos Θ_{max}.
 At the equilibrium position, v = v_{max},
T = mg cos 0^{0} + mv_{max}^{2}/L
= mg + mv_{max}^{2}/L.


15.

The forces acting on the object are its weight mg
and the tension T in the string. The acceleration
is into the center or, at this instant, to the right. We
take the Xaxis in the direction of the acceleration and
the Yaxis perpendicular to it. The weight mg is
in the negative Ydirection, but T is in neither
the X nor Y direction so we take components, as is shown
in Figure 9 above. The acceleration in the Xdirection
is the centripetal acceleration v^{2}/r. There is
no acceleration in the Ydirection.
(F_{net})_{x}
= ma_{x} 
(F_{net})_{y }=
ma_{y} 
T sin Θ
= mv^{2}/r (Equation
1) 
T cos Θ
 mg = 0 or


T cos Θ
= mg (Equation
2) 
Dividing Eq. 1 by Eq. 2:
T sin Θ/T
cos Θ = tan Θ
= (mv^{2}/r)/(mg) = v^{2}/rg (Equation
3)
From geometry, we see that r/L
= sin Θ or
r = L sin Θ
(Equation
4)
 Substituting Eq. 4 into Eq.
3:
tan Θ= v^{2}/(L
sin Θ)g or
(tan Θ) (L sin
Θ)g= v^{2} and
v =(Lg sin Θtan Θ)^{1/2
}
 v = (distance traveled/time)
= 2πr/T or
T= 2πr/v = 2 π(L
sin Θ)/v =
T = 2 π(L sin Θ)/(Lg
sin Θtan Θ)^{1/2
} = 2 π(L
sin Θ)/(Lg sin^{2}Θ/cos
Θ)^{1/2} =
T = 2 π(L cos Θ/g)^{1/2}


16.

 For motion in a horizontal circle, from the solution
to #15, we see that:
T sin Θ= mv^{2}/L
sinΘ or
v^{2} = TL sin^{2} Θ/m
(1)
Also, T cos Θ= mg
or
cos Θ= mg/T
From Fig. a, we see that:
sin Θ= (T^{2}
 m^{2}g^{2})^{1/2}/T (2)
Substituting Eq. 2 into Eq. (1):
v^{2} = TL/m (T^{2}  m^{2}g^{2})/T^{2}
= TL/m {1  (mg/T)^{2}}
or
v = (TL/m)^{1/2} {1  (mg/T)^{2}}^{1/2
}
 For a vertical circle, maximum tension occurs when the
object is at the lowest point and the tension and weight
of the object are in opposite directions. For this case,
T  mg = mv^{2}/L or
v ={(T  mg)L/m}^{1/2}


17.

 Plots of the position in the x and y direction as a function
of time are straight lines. The slopes of these lines are
the constant velocities, v_{x} and v_{y},
respectively. For constant velocity, there is no acceleration
nor force. F_{x} = 0. F_{y }=
0.
 For the projectile, the object travels equal horizontal
distances in equal time with a constant velocity. F_{x}
= 0. There is a constant acceleration down. F_{y
}=  constant.
 The speed of the particle is constant. The velocity is
always tangent to the circle. This is uniform circular motion
with the acceleration in toward the center. It is constant
in magnitude, but changes direction.
F_{x} =  (constant)x and
F_{y }=  (constant)y.
F = F_{x}i + F_{y}j =  (constant)(xi
+ yj) or
F =  (constant)r, where r is
a radial vector.
 The slope of v_{x }versus t = k = a_{x}.
The slope of v_{y }versus t = k’ = a_{y},
where k and k’ are constants.
F_{x} = c and F_{y} =
c’, where c and c’ are new constants.


18.

(b)

(a) First isolate the entire system as shown in Fig. 11a
above. For this system, T_{2}, T_{3},
the force F_{mp }of the platform on the man,
and the force F_{pm} of the man on the platform are
internal forces. The external forces are T_{1}, the
weight of the man (mg)_{man}, and the weight of the
platform (mg)_{platform}.
F_{net external }= Ma
T_{1} 1000 N  600 N = (100 + 60)kg (5.0 m/s^{2})
T_{1} = 2400 N
T_{2} = T_{3} = T_{1}/2 = 1200 N
(c) Now isolate the man as shown in Fig. 11b above:
F_{net external }= m_{man }a
T_{3 } (mg)_{man} + F_{mp} = 100
kg (5 m/s^{2})
1200 N  1000 N + F_{mp }= 500 N
The force of the platform on the man F_{mp} = 300
N.
As a check, let us isolate the platform, as in Fig. 11c above:
F_{net external }= m_{platform }a
T_{2 } (mg)_{platform}  F_{pm} =
60 kg (5 m/s^{2})
1200 N  600 N  F_{mp }= 300 N
Force of the man on the platform F_{pm} = 300 N.
F_{mp} and F_{pm} are Newton third law
of motion forces. They are equal in magnitude and opposite
in direction.


19.

First isolate the entire system (Fig. 12a above). For the entire
system, the external forces are the applied force F, the frictional
force f_{2 }of the surface on m_{2}, the
normal force F_{N on 1 & 2} of the surface on the two
blocks and the attraction of the earth for the objects or the
weight of the objects, (m_{1} + m_{2})g.
As the objects move to the right, f_{2 }acts
to the left.
(F_{net})_{x}
= Ma_{x} 
(F_{net})_{y}
= Ma_{y} 
45 N  f_{2}
= 6.0 kg a 
F_{N on 1 & 2 }
60 N = 0
F_{N on 1 & 2 }= 60 N 
f_{2} = µ_{k} F_{N on 1 & 2 }
= 0.25(60 N) = 15 N 
45 N  15 N = 6.0 kg
a 

5.0 m/s^{2}
= a 

Now isolate m_{1 }(Fig. 12b
above). There must be a frictional force f_{1 } on
m_{1} due to m_{2} to
the right to produce the acceleration of m_{1}.
(F_{net})_{x}
= m_{1}a_{x} 
(F_{net})_{y}
= m_{1}a_{y} 
f_{1
}= 2.0 kg(5 m/s^{2}) 
F_{N1}
 20 N = 0
F_{N1} = 20 N and
f_{1} = µ_{s}(20 N) 
µ_{s}(20
N) = 10 N
µ_{s }= 1/2 

As a check isolate m_{2}
(Fig. 12c above). By Newton's third law, m_{1} must
exert a force f_{1} on m_{2}
to the left.
(F_{net})_{x} = m_{2}a_{x}
45 N  f_{2}  f_{1} = 4.0 kg(5 m/s^{2})
or
45 N  15 N  f_{1} = 20 N
and again,
f_{1} =10 N with
µ_{s}(20 N) = 10 N and
µ_{s }= 1/2.


20.

As the block starts to slip down, a frictional force f opposes
the motion.

(F_{net})_{x}
= ma_{x} 
(F_{net})_{y}
= ma_{y} 

N
= mv^{2}/R 
f  mg = 0 or
f = mg = µ_{s}N 
Since 
v
= 2 πR/T_{o}
= 2 πRf_{o} 
N = mg/µ_{s
} (Equation
1) 

N
= m(4 π^{2}R^{2}f_{o}^{2})/R 
_{ (Equation
2)} 
Substituting Eq. 1 into Eq. 2 gives:
f_{o } = (1/2 π)(g/µ_{s}R)^{1/2}


21.

Take the Xaxis in the direction of the acceleration or up the
plane and the
Yaxis perpendicular to it (Fig. 14 above).
(F_{net})_{x} ma_{x
}F cos 37^{o}  mg sin 37^{o}  f = ma
60 N(4/5)  20 N(3/5)  f = 2.0 kg(a)
48 N  12 N  f = 0
(Equation
1)
(F_{net})_{y} = ma_{y
} F_{N}  F sin 37^{o}  mg cos 37^{o
}= m(0)
F_{N }= 60 N(3/5) + 20 N(4/5) = 52 N
^{ } f = µ_{k}F_{N} = 1/2(52 N)
= 26 N (Equation
2)
Substituting Eq. 2 into Eq. 1:
48 N  12 N  26 N = 2.0 kg(a) or
a = 5.0 m/s^{2}


22.

 Net force on sled = m_{sled}a_{sled }=
10 kg(2.0 m/s^{2}) = 20 N = force of man on sled.
 By Newton's third law, force of man on sled = force of
sled on man
= 20 N.
 Net force on man = m_{man}a_{man} = 60
kg(2.0 m/s^{2}) = 120 N = frictional force of snow
on man  force of sled on man. 120 N = frictional force
of snow on man  20 N. Frictional force of snow
on man = 140 N.


23.

 Let F_{os} = the force of the scale on the object
when the elevator moves with a constant velocity. For
constant velocity, F_{net }= ma = m(0).
F_{os}  mg = 0, and F_{os }= mg =
(10 kg)(10 m/s^{2}) = 100 N. By Newton's third
law, the force of the scale on the object = the force of
the object on the scale. The reading of the scale is 100
N.
 Let F'_{os} = the force of the scale on the object
when the elevator moves upward with a constant acceleration
of 5 m/s^{2}.
F_{net }= ma = (10 kg)(5 m/s^{2}).
F'_{os}  mg = 50 N, or F'_{os} =
50 N + 100 N = 150 N.
F'_{os} = F'_{so} = reading of scale = 150
N.


24.

F_{net }= mg  f = ma or a =
g  f/m. The lead sphere has a greater density than the wood
sphere. The volume of the spheres = 4/3 πr^{3}
and the mass m of the spheres = density x volume. The lead sphere
has the same volume as the wood sphere since their radii are
identical, but it has a larger mass because it has a greater
density. Since a = g  f/m, the sphere of the larger mass has
the greater acceleration.


