Preparation problems
6-2. Two forces act on a 3.1-kg mass, which undergoes
acceleration a = 0.91 i - 0.27 j m/s2.
If one of the forces is
F1 = -1.2 i -2.5 j N, what is the other force?
Solution:
We know that the net force is what causes an object to
accelerate. This net force is given by
We also know Newton's Second Law,
We can find the second force by noting that
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(3.1 kg) |
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0.91 |
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- 0.27 |
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- |
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-1.2 |
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- 2.5 |
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N |
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2.821 |
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- 0.837 |
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+ |
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1.2 |
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+ 2.5 |
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N |
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6-13. A camper hangs a 26-kg pack between two trees,
using two separate pieces of rope of different lengths, as shown in Fig.
6-54 (on page 145). What is the tension in each rope?
Solution:
We begin by drawing a free-body diagram for the backpack.
where W represents the weight of the backpack, T1
is the tension in the left-hand rope, and T2 is the tension
in the right-hand rope. Since the backpack is not accelerating, we know
that the net force acting on the backpack must be zero. As is usually
the case with vector equations, we will make it easier on ourselves if
we write each vector in its component form.
where q = 71° and f = 28°. We see we have two
unknowns, T1 and T2. We expect then that we must solve two
equations in order to find both values.
We will start by writing down Newton's Second Law in the
x-direction.
Substituting in from above, we get an equation that looks like
Since we have two unknowns in this equation, we must use another to
solve the problem. So we must write down Newton's Second Law in the
y-direction:
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Fnet,y = 0 = T1,y + T2,y + W |
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Substituting in for the weight and the y-components of the tensions
from above, we get an equation that looks like
Now we can set Equations (1) and (2) equal to
each other and solve for T2.
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(254.8 N)
cos(28°) tan(71°) + sin(28°)
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Finally, we can substute this back into Equation (1) to get
6-42. Eight 80-kg rugby players climb on a 70-kg
"scrum machine," and their teammates proceed to push them with
constant velocity across a field. If the coefficient of kinetic
friction between the scrum machine and the field is 0.78, with what
force must they push?
Solution:
We begin by drawing a free-body diagram for the scrum machine.
where Wmachine represents the weight of the
machine, Nplayers is the normal force exerted on
the machine by the eight rugby players standing atop it,
Nground is the normal force exerted on the
machine by the ground, Ffriction is the
frictional force exerted on the machine by the ground, and
Fplayers is the force exerted on the machine by
the players as they push it across the field.
We have several unknowns at this point. We can start by
determining Wmachine. This is found by recalling
Now we need to find Nplayers. To do this, we
need to draw a free body diagram of the players:
where W players is the weight of the players
and N machine is the normal force exerted on
the players by the scrum machine. Since the players are not accelerating
in the y-direction, we can sum the forces in this direction, and by
Newton's Second Law, that sum must be zero. This implies
We can find the weight of the players using the same definition as used
in Equation (1). This would give us
Now that we know this, we know that N machine = 6272 N.
However, we need to know Nplayers. This might be
a problem, except we know that these two forces are an
action-reaction pair, and therefore are equal but opposite
according to Newton's Third Law. This means
Now we can find the normal force exerted on the machine by the
ground, Nground. We do this by summing the
forces on the machine in the y-direction, and noting that the machine
is not accelerating, so that sum must equal zero. Mathematically, this
looks like
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Nground + Nplayers + Wmachine |
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- [ Nplayers + Wmachine ] |
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Nground = - [ (-6272 N) + (-686 N) ] |
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Recalling that we can relate this force to the frictional force using
the equation
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F k = F friction = mk Nground , |
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we can calculate the value of F friction . Since we know
mk = 0.78, we get
Finally, we can find the force applied by the players using
Newton's Second Law in the x-direction. We are told the sled is not
accelerating, so again the net force in this direction must be zero.
This tells us that the players must exert a force of
Follow-up problems
5-26. A 50-kg parachute jumper descends at a steady
40 km/hr. What is the force of the air on the parachute?
Solution:
We have two objects we must consider, both of which are moving
at a constant speed of 40 km/h: the jumper and the parachute. So we
begin by drawing a free-body diagram for each, shown here:
where FEarth on jump represents the weight of
the jumper, Fpar on jump represents the cords
of the harness pulling up on the jumper,
Fjump on par represents the jumper pulling down on the parachute, and
Fair on par represents the air pushing up on
the parachute. Since we know the mass of the jumper, we can find their
weight using:
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W
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= |
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Earth on jump = m |
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g
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Plugging in our known values, we see the jumper weighs
W = (50 kg)(9.8 m/s2) = 490 N.
Now we know that an object moving at a constant velocity is
not accelerating. Then by Newton's Second Law,
we know that the net force on that object must be zero. From this, we
can quickly see
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par on jump + |
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Earth on jump |
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par on jump - 490 N |
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We know from Newton's Third Law that
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jump on par = - |
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par on jump |
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So we see
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jump on par = - 490 N |
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Finally, we know that the parachutist, like the jumper, is descending at
a steady 40 km/h, meaning that it is undergoing no net acceleration, and
therefore
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air on par + |
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jump on par |
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5-33. At liftoff, a space shuttle with
2.0 ×106 kg total mass undergoes an upward acceleration of 0.60g.
(a) What is the total thrust force developed by its engines?
(b) What force does the seat back exert on a 60-kg astronaut
during liftoff?
Solution:
We begin by drawing a free-body diagram for the space shuttle, shown here:
where FEarth on s represents the weight of
the space shuttle and Fengine on s
represents the thrust provided by the engines of the space shuttle.
(a) Since we know the mass of the space shuttle, we can
find its weight using:
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W
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= |
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Earth on s = ms |
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g
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Plugging in our known values, we see the space shuttle weighs
W = (2.0 ×106 kg)(9.8 m/s2) = 1.96 ×107 N.
Now we can use Newton's Second Law to find the net force
acting on the shuttle, recalling that its acceleration of
0.6g = 5.88 m/s2, and is in the positive direction. Then
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(2.0 ×106 kg)(5.88 m/s2 |
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(b) To find the force of the seat back on the
astronaut, we must draw a free body diagram for the astronaut:
Again we can find the weight of the astronaut using
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Earth on a = ma |
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g
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Plugging in our known values, we see the astronaut weighs
W = (60 kg)(9.8 m/s2) = 588 N. Finally, we know also that the astronaut
must be accelerating at the same rate as the space shuttle, as he is
remaining in the shuttle. This means we can use Newton's Second Law to
find the net force acting on the astronaut.
Looking at our free-body diagram, we know that the net force is the sum
of all forces shown in the diagram. When we do this, we see we can
solve for the force of the seat back on the astronaut:
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seat on a + (3.53 ×104 N |
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seat on a - (588 N |
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(353 N |
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5-45. A 2200-kg airplaine is pulling two gliders, the
first of mass 310 kg and the second of mass 260 kg, down the runway with
an acceleration of 1.9 m/s2 (Fig. 5-34 on page 115). Neglecting the
mass of the two ropes and any frictional forces, determine (a)
the horizontal thrust of the plane's propellor; (b) the tension
force in the first rope; (c) the tension force in the second
rope; and (d) the net force on the first glider.
Solution:
(a) The thrust of the airplane is responsible for the
acceleration of the airplane and both gliders. In order to determine the
magnitude of this thrust, we must consider the plane and both gliders as
a single object. The free body diagram of this composite object looks
like this:
where Wall represents the weight of the plane and both
gliders, Nall represents the normal force of the ground
pushing up the plane and both gliders, and Fp represents the
thrust of the propellor.
We know that we can relate the net force on all three objects
to their acceleration using Newton's Second Law,
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(net on all) = mall |
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where mall = (2200+310+260) = 2770 kg and aall = 1.9 m/s2.
If we note that the only force acting in the x-direction is the
thrust, we see that it must be the net force in that direction. This
allows us to relate the thrust to the acceleration using Newton's Second
Law, or
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(net on all) = |
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p = mall |
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all |
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Plugging in our known values, we see
(b) Now we must find the tension in the first rope. In
order to do this, we need to consider the plane and gliders as separate
objects. We can see that the plane and the first glider are affected by
the force of the tension in the first rope, so we choose to look at the
plane, as we know the thrust produced by the plane's engine. The free
body diagram for the plane looks like:
where Wplane represents the weight of the plane,
Nplane represents the normal force of the ground
pushing up the plane, Fp represents the thrust of the
propellor, and T1 represents the tension in the first rope.
We know that we can relate the net force on the plane to its
acceleration using Newton's Second Law,
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(net on plane) = mplane |
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where mplane = 2200 kg and aplane = 1.9 m/s2.
We see we must again sum the forces in the x-direction to
determine the net force that causes the acceleration. Thus, we can solve
our problem using Newton's Second Law, which becomes
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[ (2200 kg) (1.9 m/s2) - (5263 N) ] |
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(c) Now we must find the tension in the second rope.
We again must treat the plane and the gliders as separate objects. In
this case, the first glider and the second glider will be affected by
the tension force in the second rope. Since we know the force applied
to the first glider by the first rope, let us look at the free body
diagram for this glider:
where Wg1 represents the weight of the first
glider, Ng1 represents the normal force of the
ground pushing up the glider, T1 represents the tension in
the first rope, and T2 represents the tension in the second
rope.
We know that we can relate the net force on the first glider to its
acceleration using Newton's Second Law,
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(net on g1) = mg1 |
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g1 |
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where mg1 = 310 kg and ag1 = 1.9 m/s2.
We see we must again sum the forces in the x-direction to
determine the net force that causes the acceleration. Thus, we can solve
our problem with Newton's Second Law, which becomes
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[ (310 kg) (1.9 m/s2) - (1083 N) ] |
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(d) Now we need to find the net force acting on the
first glider. Since we have solved for all of the forces shown in its
free body diagram, this task is relatively easy. We see quickly that
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(1083 N |
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which was what we wanted.
5-74. An F-14 jet fighter has a mass of
1.6 ×104 kg and an engine thrust of 2.7 ×105 N. A 747 jumbo jet
has a mass of 3.6 ×105 kg and a total engine thrust of
7.7×105 N. Is it possible for either plane to climb vertically,
with no lift from its wings? If so, what vertical acceleration could it
achieve?
Solution:
We begin by drawing a free-body diagram for either plane flying straight up,
shown here:
where W represents the weight of either plane and T
represents the thrust provided by the engines of that plane.
First we will look at the F-15. With a mass of
mF14 = 1.6 ×104 kg, we see that the weight is given by
Summing the forces for the F-14 in vertical flight shows us
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(2.7 ×105 N) - (1.57 ×105 N) |
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But we know that the net force is related to the acceleration of object
by Newton's Second Law
This is a positive acceleration, indicating that the F-14 can
climb vertically, with no lift from its wings.
Now let's look at the 747. With a mass of
m747 = 3.6 ×105 kg, we see that the weight is given by
Summing the forces for the 747 in vertical flight shows us
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(7.7 ×105 N) - (3.53 ×106 N) |
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This is a negative acceleration, indicating that the 747 cannot
climb vertically, with no lift from its wings.
Solutions translated from TEX by TTH, version 1.57.
