PY1052 Problem Set 2 – Autumn 2004
Solutions
(1) Suppose we have the two vectors ~a = (4.8 m)î + (2.2 m)ĵ and ~b =
(5.5 m)î − (1.7 m)ĵ. Find ~b − ~a in (a) unit-vector notation and (b) in terms
of their magnitudes and angles relative to the x axis.
In unit-vector notation, we have
~b − ~a = [(5.5 m)î − (1.7 m)ĵ] − [(4.8 m)î + (2.2 m)ĵ]
= (5.5 m − 4.8 m)î + (−1.7 m − 2.2 m)
= (0.7 m)î − (3.9 m)ĵ
The magnitudes and angles of these vectors relative to the x axis are
~b − ~a = (0.7 m)î − (3.9 m)ĵ:
magnitude =
tan θ =
−3.9 m
0.7 m
q
(0.7 m)2 + (−3.9 m)2 = 3.96 m
−→ θ = −79.8◦
20 deg
F = 35 N
1
F = 50 N
2
F = 15 N
3
45 deg
(2) Three forces act on a 0.5-kg object sitting on a frictionless surface as
shown. What is the object’s acceleration (a) in unit-vector notation and
(b) as a magnitude and direction relative to the x axis?
(a) In unit-vector notation, the three forces acting on the objectare
F~1 = (−35 N) cos(20◦ ) î + (35 N) sin(20◦ ) ĵ
F~2 = 50.0 N î
F~3 = (−15 N) cos(45◦ ) î + (−15 N) sin(45◦ ) ĵ
F~1 = −32.9 N î + 12.0 N ĵ
F~2 = 50.0 N î
F~3 = −10.6 N î − 10.6 N ĵ
The net (total) force on and acceleration of the object are
F~ = (−32.9 + 50.0 − 10.6) N î + (12.0 + 0 − 10.6) N ĵ = 6.5 N î + 1.4 N ĵ
F~
= 4.3 m/s2 î + 0.93 m/s2 ĵ
~a =
m
The magnitude and direction of the acceleration are given by
a=
q
(4.3 m/s2 )2 + (0.93 m/s2 )2 = 4.40 m/s2
tan θ =
0.93 m/s2
4.3 m/s2
−→ θ = 12.2◦
(3) An astronaut has a weight of 292.0 N on Mars, where g = 3.8 m/s 2 .
(a) What is his (a) mass on Mars? What are his (b) mass and (c) weight
on the Earth?
The astronaut’s weight is given by W = mg, where g is the corresponding gravitational acceleration.
292.0 N
(a) His mass on Mars will be, m = Wg = 3.8
= 76.8 kg
m/s2
(b) His mass on the Earth is the same, m = 76.8 kg
(c) His weight on the Earth is W = mg = (76.8 kg)(9.8 m/s2 ) = 752.8 N
(4) The “strong force” is the force that binds the protons and neutrons
in a nucleus together. If a free neutron collides with a nucleus and penetrates into it, the neutron will be captured if the strong force can bring
the neutron to a stop before it reaches the other side of the nucleus. (a)
If the fastest neutron that can be captured by a nucleus with a diameter
of d = 1.5 × 10−14 m has a velocity of 1.4 × 107 m/s, what is the magnitude of the strong force? Make the approximation that the strong force
is constant within the nucleus and zero outside it and that it always acts
opposite to the direction of the neutron’s velocity. The mass of a neutron
is 1.67 × 10−27 kg. (b) If a neutron with a velocity of 1.1 × 107 m/s makes a
direct hit with the nucleus, how long does the neutron move in the nucleus
before it is finally captured?
(a) The key idea here is that the neutron must be stopped by its acceleration due to
the strong force before it reaches the opposite side of the nucleus. In other words, the
neutron’s speed must become zero before it moves through a distance d. The faster
the neutron’s speed, the further the distance through which it can travel before being
stopped. Thus, we have for the fastest neutron that can be captured:
v 2 = v02 + 2ad
2
+ 2astrong d
0 = vmax
2
v
astrong = − max
2d
(1.4 × 107 m/s)2
= −
2(1.5 × 10−14 m)
Fstrong
Fstrong
= −6.53 × 1027 m/s2
= mn astrong = (1.67 × 10−27 kg)(−6.53 × 1027 m/s2 )
= −10.9 N
The negative sign indicates that the strong force acts opposite to the direction of
motion of the neutron.
(b) If the initial speed of the neutron is 1.1 × 107 m/s and it is subject to the constant
force Fstrong , the time it will travel before coming to rest is given by
v = v0 + at
v = v0 + astrong t
0 = v0 + astrong t
v0
t = −
astrong
1.1 × 107 m/s
= −
−6.53 × 1027 m/s2
= 1.68 × 10−21 s
(5) A 58-kg skier skies down a frictionless ski slope inclined at an angle
of 18◦ to the horizontal while a strong wind blows horizontally against her.
Determine the (a) magnitude of the force of the wind and (b) the normal
force due to the slope if her speed down the slope is increasing at a rate
N
18 deg
of 1.4 m/s2 .
Fwind
18 deg
mg
(a) There are three forces acting on the skier – her weight, the normal force from the
ski slope, and the force of the wind. In the diagram, each of these has been broken
up into components along the slope and perpendicular to the slope. The sum of the
forces perpendicular to the slope must be zero (no acceleration in that direction), and
the sum of the forces along the slope must be equal to the skier’s acceleration down
the slope. The sum of the forces along the slope is
mg sin(18) − Fw cos(18) = ma
m(g sin(18) − a)
Fw =
cos(18)
(58 kg)[(9.8 m/s2 ) sin(18) − 1.4 m/s2 ]
Fw =
cos(18)
Fw = 98.4 N
The sum of the forces perpendicular to the slope is
N − mg cos(18) − Fw sin(18) = 0
N = mg cos(18) + Fw sin(18)
= (58 kg)(9.8 m/s2 ) cos(18) + 98.4 N sin(18)
N = 571 N
We can see that the normal force is more than it would be without the wind, because
the wind has a component that pushes the skier down into the slope, and the normal
force must balance this as well as the component of the skier’s weight in that same
direction.
(6) Four penguins are being pulled along very slippery (frictionless) ice
by an antarctic explorer. The masses of three of the penguins and two of
the tensions are known. (a) Find the mass of the remaining penguin. (b)
Find the remaining two tensions T1 and T2
Tension=111 N
M
12 kg
15 kg
Tension = 222 N
20 kg
(a) The key to solving this problem is to understand that all the penguins experience
the same acceleration, since they are all tied together. We can most easily find the
penguin’s mass by considering (i) the force F1 acting on the last two penguins in the
row and (ii) the total force FT acting on all the penguins:
F1 = 111 N = (12 kg + M )a
FT = 222 N = (47 kg + M )a
You can now eliminate a and solve for M any way you like; for example, dividing the
second by the first equation yields:
47 kg + M
12 kg + M
24 kg + 2M = 47 kg + M
M = 23 kg
2 =
(b) Now that we have the missing penguin mass, we can find the acceleration of the
penguins. Newton’s 2nd law tells us that
F~ = m~a
We can consider either the last two penguins or all four penguins – either way we
obtain the same acceleration:
111 N = (12 kg + 23 kg)a = (35 kg)a −→ a = 3.17 m/s2
222 N = (47 kg + 23 kg)a = (70 kg)a −→ a = 3.17 m/s2
We can now find the tensions T1 and T2 by applying Newton’s 2nd law to the last
penguin and the last three penguins:
T1 = (12 + 23 + 15) kg ∗ (3.17 m/s2 ) = 158.5 N
T2 = (12 kg)(3.17 m/s2 ) = 38.0 N
(7) A block of mass m1 = 1.20 kg on a frictionless ramp inclined at an
angle of 35◦ is connected by a cord over a massless, frictionless pulley to
a second block of mass m2 = 3.00 kg sitting on a horizontal frictionless
surface. A force F2 = 2.3 N is exerted horizontally on m2 as shown. (a)
What is the acceleration of the blocks? Does m1 move up or down the
ramp? (b) What is the tension in the cord?
N2
F2
N1
T
T
m2g
35 deg
m g
(a) The key here is that the accelerations of the two1 blocks are the same; also, the
tension pulling on m2 is the same as that pulling on m1 . For each block, there is no
acceleration perpendicular to the surface on which it sits, so that the sum of forces
perpendicular to the surface must be zero. The acceleration along the surface is equal
to some value a, which is the same for the two blocks. We have for the forces on m 2
and m1 :
N2 − m 2 g = 0
−F2 + T = m2 a
Perpendicular to surface
Along surface
N1 − m1 g cos(35) = 0
m1 g sin(35) − T = m1 a
Perpendicular to surface
Along surface
Above, we have defined the +x direction to be to the right (i.e. down the ramp for
m1 ). We can find the acceleration of the blocks by adding the second and fourth
equations:
−F2 + T = m2 a
m1 g sin(35) − T = m1 a
m1 g sin(35) − F2 = (m1 + m2 )a
m1 g sin(35) − F2
a =
m1 + m2
(1.20 kg)(9.8 m/s2 ) sin(35) − 2.3 N
=
4.20 kg
2
a = 1.06 m/s
Because the acceleration is positive, it is in the +x direction – m 1 moves down the
ramp.
(b) We can now find the tension in the cord by plugging the acceleration from (a)
back into either the second or the fourth equation above:
−F2 + T = m2 a
T = m 2 a + F2
= (3.00 kg)(1.06 m/s2 ) + 2.3 N
T = 5.48 N
(8) A horse is used to pull a barge along a canal. The horse pulls on the
rope with a force of 7500 N at an angle of 20◦ to the direction of the canal.
The mass of the barge is 9200 kg, and it moves straight along the canal
with an acceleration of 0.10 m/s2 . What are the magnitude and direction
of the force on the barge by the water?
Here, we have two forces acting on the barge: the force due to the horse F H and
the force due to the water FW . What we know is that the sum of these forces must
point in the direction of motion of the barge, straight along the canal. We also know
that the magnitude of the sum of these forces must be the barge’s mass times its
acceleration. Look at the forces in the x and y directions, where the +x direction is
in the direction of the motion of the barge and the +y direction is toward the bank
where the horse is walking:
ΣFx = FHx + FW x = FH cos(20◦ ) + FW x = mB aB
ΣFy = FHy + FW y = FH sin(20◦ ) + FW y = 0
Now we can find the x and y components of the force on the barge due to the
water:
FW x = m B a B − F H
cos(20◦ )
= (9200 kg)(0.10 m/s2 ) − (7500 N)(0.940)
FW x = −6130 N
FW y = −FH sin(20◦ )
= −(7500 N)(0.342)
FW y = −2565 N
The magnitude and direction of FW are given by
q
2
FW
x
+
FW =
FW = 6645 N
2
FW
y
=
q
(−6130 N)2 + (−2565 N)2
θW = arctan(−2565 N)(−6130 N) = 203◦
The direction of F~W is 203◦ = −157◦ from the +x direction.
(1b) A spaceship approaches the surface of Jupiter’s moon Callisto.
When the ship’s engine provides an upward thrust of 3260 N, the ship
descends at constant velocity, and when the engine provides an upward
thrust of 2200 N, the ship accelerates downward at 0.39 m/s 2 . (a) What is
the weight of the spaceship near Callisto’s surface? (b) What is the mass
of the ship? (c) What is the free-fall acceleration of Callisto?
In each case, there are two forces acting on the spaceship: the upward thrust and the
downward force due to Callisto’s gravity. Thus, taking upward to be + and downward
to be –:
3260 N − M g = 0
2200 N − M g = −M a
a = 0.39 m/s2
(a) From the first of these equations, the weight of the spaceship near Callisto’s
surface is
3260 N = M g
(b) We can solve for M by substituting this value for M g into the second equation:
2200 N − (3260 N) = −M a
3260 N − 2200 N
M =
a
3260 N − 2200 N
= 2.72 × 103 kg
=
0.39 m/s2
(c) Now plug M into either of the original two equations to solve for g:
3260 N = M g
3260 N
3260 N
=
= 1.20 m/s2
g =
M
2.72 × 103 kg
(2b) A man is sitting in a bosun’s chair dangling from a massless rope
that runs over a massless, frictionless pulley and back down to the man’s
hand. The combined mass of the man and chair is 105 kg. With what
force must the man pull the rope if he is to rise (a) with constant velocity
and (b) with an upward acceleration of 1.40 m/s2 ? Suppose instead that
the rope extends to the ground, where a co-worker pulls on it. With
what force must the co-worker pull for the man to rise (c) with a constant
velocity and (d) with an upward acceleration of 1.40 m/s 2 ? What is the
force on the ceiling from the pulley in parts (a) and (c)?
T
T
F
(a) First consider the case when the man rises with constant velocity – this means
the acceleration is zero. There are three forces acting on the chair+man – the upward
tension in the rope Tl acting at the top of the chair, the upward tension in the rope
Tr exerted where the man holds on to the rope, and the downward force of gravity
on the chair+man M g. Because the tension is the same all along the rope, these two
tensions are equal, Tl = Tr = T . Thus, since there is no acceleration:
T + T − Mg = 0
2T − M g = 0
1
M g = (52.5 kg)(9.8 m/s2 ) = 514 N
T =
2
Looking at the rope where the man holds on, there are two forces there – the
upward tension T and the downward force exerted by the man F ; since these must
add to zero, the force exerted by the man is equal to the tension, or half the weight
of the man and chair.
(b) If the chair’s acceleration is upward and has the value a, we have instead (remembering we have defined up to be + and down to be –, so that the net acceleration is
positive):
T + T − Mg = Ma
1
M (g + a)
T =
2
= 52.5 kg(9.8 m/s2 + 1.40 m/s2 ) = 588 N
(c) If instead a co-worker pulls on the rope and the man lets go, the only upward
force acting on the man and chair is the tension acting at the top of the chair; if the
acceleration is constant:
T − Mg = 0
T = M g = (105.0 kg)(9.8 m/s2 ) = 1029 N
Again, the downward force exerted by the co-worker must be equal to the upward
force due to the tension, so that Fco = T = 1029 N – twice as much as when the
worker pulls himself up.
(d) If the upward acceleration is again a = 1.40 m/s2 , provided by the co-worker on
the ground, we have
T − Mg = Ma
T = M (g + a)
= 105.0 kg(9.8 m/s2 + 1.30 m/s2 ) = 1176 N
Again, the downward force exerted by the co-worker must be equal to the upward
force due to the tension, so that Fco = T = 1176 N.
In both cases (a) and (c), the force on the ceiling from the pulley will be equal to 2T ,
since the rope on each side exerts a downward force equal to T . Thus, in part (a),
the force on the ceiling is 2Ta = 1029 N, while in part (c) the force on the ceiling is
2Tc = 2058 N.
(3b) In the cable-car system shown, the maximum permissible mass of
each car loaded with occupants is 2800 kg. The cars, riding on a support
cable, are pulled by a second cable attached to each support tower. Assuming the cables are straight, what is the difference in tension between
adjacent sections of pull cable if the cars have the maximum permissible
mass and are being accelerated up the 35◦ incline at 0.80 m/s2 ?
Tup
T
down
mg
35 deg
35 deg
mg sin 35
The direction of the acceleration is up, parallel to the direction of the cable. Therefore, the sum of the forces acting along this direction must add to produce the net
acceleration of the cable cars. Looking at any arbitrary car, we have from Newton’s
2nd law
Tup − Tdown − M g sin(35) = M a
Tup − Tdown = M (g sin(35) + a)
= 2800 kg(9.8 m/s2 sin(35) + 0.80 m/s2 ) = 1.800 × 104 N
Thus, although we cannot find Tup or Tdown individually, and these can be different
for different cars, the difference of the two will be given by the result above.