HW03 – HW03 – forces: forces: 5P36, 5P46, 5P50, 5P52, 5P53, 5P57; 6P5, 6P11, 6P23, 6P49, 6P57. Chapter 5, problem 36: Holding on to a towrope moving moving parallel to a frictionless frictionless ski slope, slope, a 50 kg skier is pulled pulled up the slope, which is at an angle of 8.0° with the horizontal. What is the magnitude F rope rope of the force on the skier from the rope when when (a) the magnitude magnitude v of the skier’s skier’s velocity s i s constant at 2.0 2.0 m/s and (b) v = 2.0 m/s as v 2 increases at a rate of 0.10 m/s ? (a) This is a very easy FBD/schematic to draw; you should do this yourself. Newton’s 2nd law tells us
F 0 F
rope
solve for T Fg T m( g ) sin T mg sin (50kg)(9.81 sm ) sin 8.0 68N 2
(1.1)
Afterword: the sin ski -slope-plane. sin “picks out” the component of the force of gravity lying along the ski-slope-plane. (b) Repeating the calculation (1.1), and noting that we should anticipate T 68N as an error-check,
F ma T mg sin T mg sin ma (50kg)(9.81 solve for T
m s 2
) sin 8.0 (50kg )(0.10 sm2 ) 73N (1.2)
Chapter 5, problem 46: An elevator cab is pulled pulled upward by a cable. The cab cab and its single occupant occupant have a combined mass of 2000 kg. When that occupant drops a coin, its acceleration relative relative to the cab is 8.00 m/s2
downward. What is the tension in the cable? Let cable? Let the elevator be accelerating at aeg relative to the ground, the coin accelerating at acg g 9.81 m2 relative to the ground, and ace 8.00 m2 relative to the elevator. With this s s tricky statement out of the way, Newton’s 2 nd law applied to the elevator (accelerating (accelerating at aeg , recall) immediately immediately tells us the tension.
solve for T F maeg T Fg T m( g )
(1.3)
T m(aeg g ) m( g ace g ) mace (2000kg )(8.00 sm2 ) 1.60 104 N ;
Note that in the the 2nd line of (1.3), we used ace aeg acg . Also, since the coin’s acceleration’s magnitude is ls than g (e.g., ace g ), we expect that the tension should be less than mg , and indeed, we find that provides an error-check. mg (2000kg )(9.81 sm2 ) 1.6 104 N , which ballot Chapter 5, problem 50: In Fig. 5-46, three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The three masses are mA = 30.0 kg, mB = 40.0 kg, and mC = 10.0 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting Band C, and (b) how far does A move
in the first 0.250 s (assuming it does not reach the pulley)?
down as the +y-direction. Newton’s Newton’s 2nd law for each of the 3 blocks appears in both graphical1 Solution: (a) Chose down as and algebraic form as,
1
Equating illustrations to mathematical equations may be something you’re shy about, but it won Richard Feynman fame. Look up “Feynman diagram”, and “amplitudehedron”.
m A T AB
T m B m g T AB
B
BC
F
A
TAB m Aa;
FB mB g TBC TAB m Ba;
T mC FC m Cg T BC m C a ; m g BC
(1.4)
C
The system-acceleration a is therefore computed from adding up the three force-sums (1.4), solve for a a m A mB mC a TAB mB g TBC TAB mC g TBC (m B mC )g
Subsequently, we use (1.5) in
F
C
(m B mC ) g m A mB mC
6.125
m s2
; (1.5)
mC a of (1.4) to compute T BC as,
T BC mC ( g a ) mC 1
mC m A g 36.8N ; g m A mB mC mA mB mC m B mC
(b) We use Eq. 2-15 (choosing rightward as the + x direction): x = 0 +
(1.6)
1 2 at = 0.191 m. 2
Chapter 5, problem 52: An 85 kg man lowers himself to the ground from a height of 10.0 m by holding onto a rope that runs over a frictionless pulley to a 65 kg sandbag. With what speed does the man hit the ground if he started from rest? The system appears as follows (this is not obvious),
Pay attention to how up is positive for the sandbag, and down is positive for the man! Because the system accelerates in one direction only, we may as well make that direction positive. We compute the acceleration,
F
man
Ma Mg T ;
F
bag
ma
mg
Swit ched coordinates!
mg T ;
T Newton's 3rd Law reaction pair to T in
(1.7)
F
man
Using the above expression, we compute T from the 2nd (bag) equation and use that to eliminate T in the 1 st (man) equation, solve for a T m(a g ) Ma Mg T Mg m(a g ) a
M m M m
g;
(1.8)
which yields a = 1.3 m/s2. Since the system starts from rest, Eq. 2- 16 determines the speed (after traveling 2 2 v v0 2ad 0 2
M m M m
2 gd 0 2
85 65 85 65 no need to indicate units: mass divided by mass is unitless!
y =
10 m) as follows,
m
9.81 10.0m 5.1 s m s 2
(1.9)
Chapter 5, problem 53: Three blocks are connected on a horizontal frictionless table and pulled to the right with a force T 3 = 65.0 N? Assume that m 1 = 12.0 kg, m2 = 24.0 kg, and m3 = 31.0 kg. (a) Calculate the
acceleration of the system. (b) and (c) Calculate the tensions T1 and T2 in the interconnecting cords.
(a) The system acceleration comes from the external tension T3 accelerating the entire system’s mass; accordingly, newton’s 2nd law is for m1 m2 m3 as,
F Ma (m m 1
2
solve for a m3 )a T3 a
T 3 m1 m2 m3
65.0N (12.0 24.0 31.0)kg
0.970
m s2
; (1.10)
Thus, for only being interested in the acceleration of the whole system, we can pretend the system is just one big block of mass m1 m2 m3 . Note this is exactly what we do without thinking if we compute the acceleration of a ball undergoing a force: the ball is obviously made of bunches and bunches of atoms, and we only concern ourselves with the total mass of all those atoms making up the ball. (b) We must now write Newton’s 2nd Law for the individual blocks. However, the system acceleration remains the same (1.10) we calculated in part-(a), the T1 is external tension times a mass-fraction
F m a T 1
1
1
(12.0kg )(0.970 sm ) 11.6 N m1a 2
m1 m1 m2 m3
T 3
(1.11)
(c) Now in part c, treat blocks 1 and 2 as a system (as we did in (1.10) for the three blocks) and compute,
F
solve for T m2 a T2 T1 T2 m2 a T1 (24.0kg)(9.70 sm ) 11.6 N 34.9 N 2
2
2
Some fun you could have: put stuff in symbol-form: T2 m2 a T1
m3T3 m1 m2 m3
m1 m1 m2 m3
Chapter 5, problem 57: A block of mass m1 = 3.70
kg on a frictionless plane inclined at angle 30.0 is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.30 kg (Fig. 552). What are (a) the magnitude of the acceleration of
T3
m1 m3 m1 m2 m3
T3 34.9 N
(1.12)
each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?
2 Solution: The free-body diagram for each block is shown below as,
2
Warning: Again, for block 1, we take the + x direction to be up the incline and the + y direction to be in the direction of the normal
force F N that the plane exerts on the block. This is in contrast to block 2, where we take the + y direction to be down! However, this benefits us in that you have a common system acceleration a (“conservation of string”), which is unambiguously positive ().
The free body diagram (1.13) subsequently implies the force balances,
F
1, x
m1a T m1g sin ;
F
1, y
(1.13)
m1a 0 FN m1g cos ;
F
2
m2a m2g T ;
(1.14)
respectively. The first and third of these equations provide a simultaneous set for obtaining values of a and T . The second equation is not needed in this problem, since the normal force is neither asked for nor is it needed as part of some further computation (such as can occur in formulas for friction). (a) The question is really asking for the system-acceleration. Consider the force-balance in the direction of the string, and notice that you have a Newton’s 3 rd Law pair T which goes away if we sum the equations,
F
1, x
solve for a F2 (m1 m2 )a T m1g sin m2 g T Newton's 3rd Law pairs in red
a
m2 m1 sin m1 m2
g
[2.30 kg (3.70 kg)sin30.0] 9.80 m/s2 3.70 kg 2.30 kg
(1.15)
0.735m/s2 ;
Sanity Check: Notice what happens if we put 90 in the symbolic expression a
m2 m1 sin m1 m2
of (1.15): we
reproduce Eq. (1.8) from Chapter 5, problem 52 (remember doing that?! Never forget the physics problems you have worked!), as we should , since this value of corresponds to the incline of (1.13) “tilting” up into a pulleysystem, which makes us super certain that we didn’t make a mistake! . (b) The result for a is positive, indicating that the acceleration of block 1 is indeed up the incline and that the acceleration of block 2 is vertically down. (c) Having the system-acceleration makes finding the tension a bit of a victory lap. Looking at the 1st equation in (1.14), we write,
T m1a m1g sin 3.70 kg 0.735 m/s 2 3.70 kg 9.80 m/s 2 sin 30.0 20.8N. Chapter 6, problem 5: A baseball player with mass m = 79 kg, sliding into second base, is retarded by a frictional force of magnitude 470 N. What is the
coefficient of kinetic friction k between the player and the ground? Solution: The free-body diagram for the player is shown to the right. F N is the normal
force of the ground on the player, mg is the force of
gravity, and f is the force of friction.
The force of friction is related to the normal force by f = k F N . We use Newton’s second law applied to the vertical axis to find the normal force. The vertical component of the acceleration is zero, so we obtain F N – mg = 0; thus, F N = mg . Consequently,
k
f
F N
470N
79 kg 9.8 m/s
2
0.61.
(1.16)
crate? (a) The free-body diagram for the crate is shown,
• chapter 6, problem 11 (copied from solution manual – no commentary from me): A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.50, what minimum force magnitude is required from
.
the rope to start the crate moving? (b) If k 0.35 , what is the magnitude of the initial acceleration of the
T is the tension force of the rope on the crate, F N is the normal force of the floor on the crate, mg is the force of
gravity, and f is the force of friction. We take the + x direction to be horizontal to the right and the + y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are: T cos – f = 0
T sin F N mg 0 where = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos and the second gives F N = mg – T sin . If the crate is to remain at rest, f must be less than s F N , or T cos < s (mg – T sin ). When the tension force is sufficient to just start the crate moving, we must have T cos = s (mg – T sin ). We solve for the tension:
T
s mg cos s sin
0.50 68 kg 9.8 m/s2 cos 15 0.50 sin 15
304 N 3.0102 N.
(b) The second law equations for the moving crate are T cos – f = ma F N + T sin – mg = 0. Now f = k F N , and the second equation gives F N = mg – T sin , which yields f k (mg T sin ) . This expression is substituted for f in the first equation to obtain T cos – k (mg – T sin ) = ma, so the acceleration is
a Numerically, it is given by
a
T cos k sin
g
m
g b0.35gc9.8 m / s h 13. m / s .
304 N cos15 0.35 sin 15 68 kg
k g .
2
2
•• chapter 6, problem 23 (copied from solution manual – no commentary from me): When the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of 0.500 m/s 2. Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coefficient of kinetic friction between block 2 and the table?
Let the tensions on the strings connecting m2 and m3 be T 23, and that connecting m2 and m1 be T 12, respectively. Applying Newton’s second law (and Eq. 6-2, with F N = m2 g in this case) to the system we have
m3 g T23 m3 a T23 k m2 g T12 m2 a T12 m1 g m1a Adding up the three equations and using m1 M , m2 m3 2M , we obtain 2 Mg – 2 k Mg – Mg = 5 Ma . 2
With a = 0.500 m/s this yields k = 0.372. Thus, the coefficient of kinetic friction is roughly k = 0.37. •• Chapter 6, Problem 49: In the figure, a car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver's mass is 52.1 kg. What is the minimum magnitude of the normal force on the driver from the seat when the car passes through the bottom Solution: Intuitively, the correct answer is F N
of the valley?
2mg 2(52.1kg)(9.81 sm ) 1.02 103 N . The justification for this, 2
top of hill F ma F N mar FN m
v
2
r
m(g ); FN 0 m
v
2
r
mg ;
2 v bottom of hill F N m( g ) m m( g ) mg 2mg ; r
(1.17)
•• Chapter 6, problem 57: A puck of mass m = 1.50 kg slides in a circle with radius r = 20.0 cm on a frictionless table while attached to a hanging cylinder of mass M = 2.50 kg by a cord through a hole in the table. What speed keeps the cylinder at rest? Solution: newton’s 2nd law says the tension due to gravity equals the tension due to centripetal acceleration; solving for “v” in this equation,
T mar m
v
2
r
Mg v
Mgr m
(2.50 kg)(9.81 sm2 )(0.200 m) 1.50 kg
1.81
m s
;
(1.18)
The +/- ambiguity in (1.18) means the mass-m could be circling clockwise or counterclockwise, and still exerting the same tension.
