Answer, Key { Homework 4 { Rubin H Landau 1 Which of the following best describes a free This print-out should have 18 questions. Check that it is complete before leaving the body diagram for the book? printer. Also, multiple-choice questions may 1. normal continue on the next column or page: nd all tension choices before making your selection. Note that only a few (usually 4) of the frictional problems will have their scores kept for a gravitational grade. You may make multiple tries to get a problem right, although it's worth less each 2. normal time. Worked solutions to a number of these tension problems (even some of the scored ones) may be found on the Ph211 home page. frictional
Falling Apple 02:05, trigonometry, numeric, > 1 min. 001
gravitational
An apple falls from a tree and hits the ground 10:5 m below. With what speed will it hit the ground? Correct answer: 14:3457 m=s.
correct 3.
tension frictional
Explanation:
When an object undergoes free fall from rest, its nal speed is given by p v = 2gh
Algorithm
6 1
h = 10:5 m 15 g = 9p :8 m=s2 v = p2:0 g h = 2:0 h9:8i h10:5i = 14 :3457 m=s q hm=si = hi hm=s2i hmi :
(1) (2) (3)
gravitational
4.
gravitational tension frictional normal
Explanation: normal tension
units
P303k conc forces1 05:01, noArithmetic, multiple choice, < 1 min. 002 A string is tied to a book and pulled at an angle as shown in the gure. The book remains in contact with the table and does not move. String
normal
frictional gravitational
The gravitational force points down. The frictional forces points to the left. The normal force points up. The tension in the string points in the direction the string is pulling.
003
Which forces on the book would change if the string were pulled twice as hard?
1. normal 2. gravitational
Book Table
Answer, Key { Homework 4 { Rubin H Landau 2 A 89:8 kg boxer has his rst match in the 3. tension in string Canal Zone with gravitational acceleration 9:782 m=s2 and his second match at the 4. friction North Pole with gravitational acceleration 9:832 m=s2. 5. normal and gravitational a) What is his mass in the Canal Zone? Correct answer: 89:8 kg. 6. normal, tension and friction correct
7. tension and friction 8. normal and friction 9. gravitational, tension and friction Explanation:
The force the table exerts on the book (the normal force) is less when the string is taut. The book will slide easier since the normal force has decreased. Thus the frictional force has also decreased. The string is taut indicating a tension that is not there when the string is not there.
Pulling Two Blocks 05:04, calculus, numeric, > 1 min. 004
Two blocks on a frictionless horizontal surface are connected by a light string. m2
m1
F
m1 = 14:7 kg and m2 = 20 kg. A force of 49:8 N is applied toward the right on the 20 kg block. Find the acceleration of the system. Correct answer: 1:43516 m=s2.
Explanation:
Let the tension in the string between the blocks be T , and apply Newton's second law to the each block: F1 = m1 a = T (1); F2 = m2 a = F , T (2); Adding these equations gives us (m1 + m2 )a = F F a= m + m2 1 net
Explanation: Basic Concept
An object's mass is constant, regardless of the gravitational acceleration.
Solution
An object's weight varies with gravitational position and is given by
W = mg
009
b) What is his weight in the Canal Zone? Correct answer: 878:424 N.
Explanation:
W = mg
010
c) What is his mass at the North Pole? Correct answer: 89:8 kg.
Explanation:
m = Wg
Weight of a Boxer 05:05, calculus, numeric, > 1 min. 011
d) What is his weight at the North Pole? Correct answer: 882:914 N.
Explanation:
W = mg
net
008
Algorithm
80 m = 89:8 kg 100 g = 9:782 m=s2 g = 9:832 m=s2 W =mg = h89:8i h9:782i C
N
C
C
(1) (2) (3) (4)
Answer, Key { Homework 4 { Rubin H Landau 3 that the spring scale reads 309:5 N. His true = 878:424 N 2 hNi = hkgi hm=s i units weight is 370 N, and the chair weighs 238 N. m =m (5) = 89:8 kg W =mg (6) = h89:8i h9:832i = 882:914 N hNi = hkgi hm=s2i units What is the magnitude of the acceleration of the system? 012 answer: 0:177303 m=s2. Find the weight in pounds of M = 694 grams Correct Explanation: of salami. T T Correct answer: 1:52891 lb. N
N
N
Explanation:
W = Mg 1 kg 9:8 m=s2 = 694 grams 1000 g = 6:8012 N = 6:8012 N 0:2248 lb=N = 1:52891 lb
Algorithm hkggi = 0:001 kg=g hlbNi = 0:2248 lb=N
WBrian + Wchair
Brian
(1) (2) (3) (4) (5)
g = 9:8 m=s2 400 M = 694 grams 1000 W = M hkggi g = h694i h0:001i h9:8i = 6:8012 N hNi = hgramsi hkg=gi hm=s2i units W = W hlbNi (6) = h6:8012i h0:2248i = 1:52891 lb hlbi = hNi hlb=Ni units N
lb
N
014
Consider the system of Brian and the chair. Note that two ropes support the system. Let T be the tension in the ropes. Hence applying Newton's law to the system gives X F = 2T , W , W = Ma where M represents the total mass of the system. +W = 62:0408 kg M=W g Hence solving for a, a = 2T , W M , W N , 370 N , 238 N) = (2 309:5 62 :0408 kg = 0:177303 m=s2
An inventive child named Brian wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley, he pulls on the loose end of the rope with such a force
Brian
chair
chair
Brian
chair
015
What is the force that Brian exerts on the chair? Correct answer: 67:1941 N.
Explanation:
T N WBrian
Answer, Key { Homework 4 { Rubin H Landau 4 N =m a+W ,T (9) Applying Newton's law to Brian only yields X = h37:7551i h0:177303i + h370i , h309:5i F =T +N ,W =m a = 67:1941 N where N is the force exerted on Brian by hNi = hkgi hm=s2i + hNi , hNi units the chair. Since m = W =g = 37:7551 kg, we obtain for N , Suspended in an Elevator N =m a+W ,T 05:05, calculus, numeric, > 1 min. 2 = 37:7551 kg (0:177303 m=s ) 016 ) + 370 N , 309:5 N The tension in a string from which a 7:9 kg = 67:1941 N object is suspended in an elevator is equal to N is the force the chair exerts on Brian, but by 46 N. Newton's third law, N is also the magnitude What is magnitude of the acceleration a of the elevator? of the force Brian exerts on the chair. Correct answer: ,3:97722 m=s2. Algorithm 2 g = 9:8 m=s (1) Explanation: Basic Concepts: W = 370 N 260 (2) 512 X 64 W = 238 N 256 (3) F = m a = F 10 = 11 20 (4) W = mg T = 0:5 (W + W + ) (5) = 0:5 (h370i + h238i + h11i) Solution: The weight W = mg of the object is less = 309:5 N than the tension in the string, so assume the hNi = hi (hNi + hNi + hi) units acceleration is upward. +W (6) M=W T g h238i a = h370ih+ 9:8i = 62:0408 kg Mg h N i + h N i hkgi = hm=s2i units =) a = T ,mmg a = 2:0 T , W M , W (7) Suspended in an Elevator 2 : 0 h 309 : 5 i , h 370 i , h 238 i 05:07, trigonometry, numeric, > 1 min. = h62:0408i 017 = 0:177303 m=s2 What is its direction? hi h N i , h N i , h N i hm=s2i = units 1. upward correct hkgi m =W g (8) 2. downward i 3. not moving = hh370 9:8i Explanation: = 37:7551 kg The acceleration, assumed upward, was h N i positive, units rect. so the upward assumption was corhkgi = hm=s2i Brian
Brian
Brian
Brian
Brian
Brian
Brian
Brian
chair
net
Brian
Brian
chair
chair
Brian
Brian
Brian
chair
Brian
Answer, Key { Homework 4 { Rubin H Landau 5 For the mass m1 , T1 acts up and the weight Algorithm 2 g = 9:8 m=s (1) m1 g acts down, with the acceleration a di4 m = 7:9 kg 8 (2) rected upward: T = 46 N 20 (3) F 1 = m1 a = T1 , m1 g (1) 50 (4) For the mass on the table, a is directed to a = T ,mm g the right, T2 acts to the right, T1 acts to the h 46 i , h 7 : 9 i h 9 : 8 i left, and the motion is to the right so that the = h7:9i frictional force m2 g acts to the left: = ,3:97722 m=s2 F 2 = m2 a = T2 , T1 , m2 g (2) 2i h N i , h kg i h m = s 2 units For the mass m , T acts up and the weight hm=s i = hkgi 3 2 m3 g acts down, with the acceleration a directed downward: Acceleration with Friction 05:08, calculus, numeric, > 1 min. F 3 = m3 a = m3 g , T2 (3) 018 a Adding these equations yields m (m1 + m2 + m3 )a = m3 g , m2 g , m1 g µ 2 , m1 )g a = (m(m3 ,+m m2 + m3 ) 1 m m net
net
net
2
1
3
The suspended 3:95 kg mass on the left is accelerating up, the 2:4 kg mass slides to the right on the table, and the suspended mass 7:3 kg on the right is accelerating down. If the coecient of friction is 0:109, what will be the acceleration of the system? Correct answer: 2:21731 m=s2.
Algorithm
g = 9:8 m=s2 (1) 2 m1 = 3:95 kg 4 (2) 1 2 m2 = 2:4 kg 3 5 (3) 6 m3 = 7:3 kg 9 (4) 0 1 = 0:109 0 2 (5) (m3 , m2 , m1 ) g Explanation: (6) a = m1 + m2 + m3 Basic Concepts: 109i h2:4i , h3:95i) h9:8i = (h7:3i , hh30::95 F = ma 6= 0 i + h2:4i + h7:3i 2 = 2:21731 m=s The acceleration a of each mass is the same, but the tensions in the two strings will be hm=s2i = (hkgi , hi hkgi , hkgi) hm=s2i units di erent. hkgi + hkgi + hkgi : :
: :
net
Solution:
Let T1 be the tension in the left string and T2 be the tension in the right string. Consider the free body diagrams for each mass: T1
T2
a
a T1
a T2
µm2g m1g
m3g
021
A rider in a \barrel of fun" nds herself stuck with her back to the wall.
Answer, Key { Homework 4 { Rubin H Landau Which diagram correctly shows the forces m v2 sin 2. F = acting on her? r 2 1. 3. F = mrv tan 2 4. F = rmcosv 2. 2 5. F = rmsinv 2 6. F = rmtanv 3. 7. F = m g tan
6
8. F = m g cos correct 4.
5.
9. F = mrv cos correct 2
r
10. F = m g2 + vr2 Explanation: Basic Concepts: To keep an object mov4
ing in a circle requires a force directed toward the center of the circle; the magnitude of the force is 2 F = m a = mrv Also remember: c
6. None of the others Explanation:
The normal force of the wall on the rider provides the centripetal acceleration necessary to keep her going around in a circle. The downward force of gravity is equal and opposite to the upward frictional force on her.
Car on a Banked Curve 022
c
X F~ = F~
i
i
Solution: Solution in an Inertial Frame:
Watching from the Point of View of Someone Standing on the Ground
05:08, trigonometry, numeric, > 1 min. A curve of radius r is banked at angle so that a car traveling with uniform speed v can round the curve without relying on friction to keep it from slipping to its left or right. Find the component X of the net force parallel F~ k . to the incline
1. F = mg cot
N
mg θ
The car is performing circular motion with a constant speed, thus its acceleration is just the centripetal acceleration, a = v2=r. The net force on the car is: c
2 v F = ma = m r net
c
Answer, Key { Homework 4 { Rubin H Landau 05:09, trigonometry, numeric, > 1 min.
7
024
N
Fnet F|| mg θ
The component of this force parallel to the incline is X F~ k = m g sin = F cos 2 = mrv cos In this reference frame, the car is at rest, which means that the net force on the car (taking in consideration the centrifugal force) is zero. Thus the component of the net \real" force parallel to the incline is equal to the component of the centrifugal force along that direction. Now, the magnitude of the cen2 trifugal force is equal to F = m vr , so
With what frictional force must the road push on a 1200 kg car if the driver exceeds the speed for which the curve was designed by v = 11 km=hr? Correct answer: 1657:11 N.
Explanation:
The speed of the car is greater, so its centripetal acceleration is greater. N
net
F||
The free body diagram showing the forces acting on the car is N Fnet F|| θ
c
Fk = F
net
2 cos = F cos = mrv cos c
023
Explanation:
Fk is the component of the weight of the car parallel to the incline. Thus 2 m v m g sin = Fk = r cos 2 tan = gv r km=hr )2 = (9:8(150 m=s2 )(71:8 m )
m 2 hr 2 1000 km 3600 s = 2:46733 = arctan(2:46733) = 67:9375
Car on a Banked Curve
mg
Thus, the net force parallel to the incline is Fk = m g sin + F where F is the friction force. On the other hand, the component of the acceleration parallel to the incline is still 2 ak = (v +rv) cos Then, 2 m g sin + F = m ak = m (v +rv) cos or 2 ( v + v ) F =m cos , g sin r = 1657:11 N : Alternative Solution: Solution in a Noninertial Frame: Watching from the Driver's Point of View. Note: The centrifugal \force" is a \ ctitious force" that comes from the accelerating car being a non-inertial frame. fr
fr
If r = 71:8 m and v = 150 km=hr, what is ? Correct answer: 67:9375 .
mg
θ
fr
fr
Answer, Key { Homework 4 { Rubin H Landau 8 When the velocity is increased to v + v, hm=si = hi hkm=hri units the "no sliding" implies that hi m = 1200:0 kg (11) 2 m ( v + v ) 2 0 2 0 (mg)k + F = cos = (v + v ) , v (12) r 20 = (h41:6667i + h3:05556i) , h41:6667i2 0 2 = 263:966 m2=s2 Using: (m g)k = mrv cos , hm2=s2 i = (hm=si + hm=si)2 0 , hm=si2 0 units 2 ( ) ( v + v ) f = m cos (13) F =m cos , g sin : r r i cos (h1:18573i) = h1200i h263:966 Algorithm h71:8i (1) hdeg rad i = 57:2958 deg=rad = 1657:11 N hkgi hm2=s2 i cos (hradi) g = 9:8 m=s2 (2) units h N i = 50 hmi r = 71:8 m 100 (3) 90 v = 150 km=hr 160 (4) :0 v (5) v = 1000 3600:0 :0 h150i = 1000 3600:0 = 41:6667 m=s =hri hm=si = hi hkm units hi 20 (6) p = vr g :6667i2 0 = h41 h71:8i h9:8i = 2:46733 20 units hi = hhmmi=hsmi =s2i = arctan (p) (7) = arctan (h2:46733i) = 1:18573 rad hradi = arctan (hi) units = hdeg (8) radi = h1:18573i h57:2958i = 67:9375 h i = hradi hdeg=rad i units v = 11 km=hr 10 (9) 20 :0 v (10) v = 1000 3600:0 :0 h11i = 1000 3600:0 = 3:05556 m=s fr
:
v
u
u
:
u
:
:
v
fr
u
: u
:
:
r
r
u
r
:
:
