Download Physics 140 HOMEWORK Chapter 6B

Physics 140
HOMEWORK Chapter 6B
Q11. A person riding a Ferris wheel moves through positions at (1) the top, (2) the bottom, and
(3) midheight. If the wheel rotates at a constant rate, rank these three positions according to (a) the
magnitude of the person’s centripetal acceleration. (b) the magnitude of the net centripetal force on
the person, and (c) the magnitude of the normal force on the person, greatest first.
———
(a) acent,1 = acent,2 = acent,3 = v 2 /R, which is constant.
(b) Fcent,1 = Fcent,2 = Fcent,3 = mv 2 /R, which is constant. The centripetal force is F NET = macent .
Remember that “centripetal force” describes the EFFECT of the force; it’s the force that causes the
object to move in a circular path
(c) FN,2 > FN,3 > FN,1 . FBD at top: mg down; n up. ~a down. Therefore, n = mg − mv 2 /R < mg.
Similarly, n > mg at bottom. At midheight, ~a is horizontal, so n is at an angle (typically
slight on a
p
2
Ferris wheel), and its horizontal component supplies the centripetal force. |~n| = (mg) + (v 2 /R)2 ,
which puts it between m(g + v 2 /R) and m(g − v 2 /R).
P23. When the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of
0.500 m/s2 . 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?
———
Realize that the entire system must be analyzed. Let T1 be the tension in the left cord and T2 the
tension in the right cord. Since m3 = 2M > m1 = M , the acceleration is to the right. All blocks have
the same acceleration, called ax .
FBD for 1: T1 up; M g down. ~a up. +x upward.
FBD for 2: 2M g down (its mass is 2M ); n up; T2 to right; T1 to left; fk to the left. ~a to right. +x to
right.
FBD for 3: T2 up; 2M g down (its mass is 2M ). ~a down. +x down.
y-eq for 2: n − 2M g = 0 ⇒ n = 2M g ⇒ fk = µk n = µk (2M g).
x-eq for 1: T1 − M g = M ax
x-eq for 2: T2 − T1 − [2µk M g] = (2M )ax . Term in [] is fk .
x-eq for 3: 2M g − T2 = (2M )ax .
Add the three x-equations to (neatly!) eliminate T1 and T2 :
M g − 2µk M g = 5M ax . We are given ax and asked for µk :
µk = (g − 5ax )/2g = (9.8 − 2.5)/(2 · 9.8) = 0.372.
P29. In Fig. 6-34, blocks A and B have weights of 44 N and 22 N, respectively. (a) Determine the
minimum weight of block C to keep A from sliding if µs between A and the table is 0.20. (b) Block C
suddenly is lifted off A. What is the acceleration of block A if µk between A and the table is 0.15?
———
(a) We can consider blocks A and C as a unit.
FBD for AC: mA g + mC g down; n up; T to right; fs to left. ~a = 0. +x to right.
FBD for B: mB g down; T up. ~a = 0. +x downward (to be consistent with AC).
y-eq for AC: n − mA g − mc g = 0 ⇒ n = mA g + mC g ⇒ fs,max = µs (mA g + mC g).
x-eq for AC: T − [µs (mA g + mC g)] = 0. The term is [] is fs,max .
x-eq for B: mB g − T = 0. Add the two x-eq:
mB g − µs (mA g + mC g) = 0 ⇒
mC g = mB g/µs − mA g = (22 N)/0.20 − 44 N = 66 N.
(b) The FBD are similar: mC is now zero, and ~a is nonzero, equal for A and B, and in the direction of
each +x-axis.
y-eq for A: n − mA g = 0 ⇒ n = mA g ⇒ fk = µk mA g = (0.15)(44 N) = 6.6 N
x-eq for A: T − 6.6 N = mA ax .
x-eq for B: mB g − T = mB ax .
Add the x-equations to get:
mB g − 6.6 N = (mA + mB )ax ⇒
ax = (mB g − 6.6 N)/(mA + mB ) = (22 N − 6.6 N)/(4.49 kg + 2.25 kg) = 2.33 m/s2 .
P49. In Fig. 6-39, 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 70.0 kg. What is the magnitude of the normal force on the driver from the seat when
the car passes through the bottom of the valley?
———Recognize a circular motion problem on Newton’s 2nd Law.
~ N would usually be upward, but F
~ N = 0 as given; no other
FBD at the top of the hill: mg down; F
2
forces. ~a downward, and value known to be v /R. Make +x downward. Then Newton’s 2nd Law gives
mg = ma = mv 2 /R ⇒ v 2 /R = g.
~ N upward; no other forces. ~a upward, and known magnitude
FBD at the bottom of the hill: mg down; F
v 2 /R. Make +x upward. Then Newton’s 2nd Law gives
~ N = mv 2 /R ⇒ F
~ N = mg + mv 2 /R = 2mg = 1370 N, where the top-of-hill result has
−mg + F
been used.
P61. A block of mass mt = 4.0 kg is put on top of a block of mass mb = 5.0 kg. To cause the top
block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N
must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless
table (Fig. 6-47). Find the magnitudes of (a) the maximum horizontal force that can be applied to the
lower block so that the blocks will move together and (b) the resulting acceleration of the blocks.
———We recognize that in (a) and (b), the force which accelerates mt is the static friction force.
~ bNon t upward; 11.999 N to
Take first the scenario when the mb is held fixed. FBD for mt : mt g down; F
right; fs to left. ~a = 0 because it isn’t slipping yet. fs = fs,max = 12 N.
~ N upward. ~a to right. We know from the initial
(a) FBD for mt as drawn: fs to right; mt g down; F
scenario that fs = fs,max = 12 N, so the x-eq is:
12 N = mt a = (4 kg)a ⇒ a = 3.0 m/s2 .
This is the acceleration of both blocks. Since they move as a unit, and since their total mass is 9 kg,
~ 0 = (9 kg)(3 m/s2 ) = 27 N.
F
(b) 3.0 m/s2 , as shown above.
P70. Figure 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the
cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.)
The bob has a mass of 0.040 kg, the string has length L = 0.90 m and negligible mass, and the bob
follows a circular path of circumference 0.94 m. What are (a) the tension in the string and (b) the
period of the motion?
———Sketch the bob looking edge-on to its circular path, with the bob at its leftmost position.
FBD for bob: T up/right at angle θ to vertical; mg down. ~a to right. +x to right.
We are given the circumference C = 0.94 m. Therefore, r = C/2π = 0.1496 m and θ = 9.569◦ .
(a) y-eq: T cos θ = mg ⇒ T = mg/ cos θ = (0.040 kg)(9.8 m/s2 )/0.9861 = 0.3975 N = 0.40 N.
(b) x-eq: T sin θ = mv 2 /r ⇒ v 2 = rT sin θ/m = (0.1496 m)(0.3975 N)(0.1662)/(0.040 kg) =
0.2471 p
m2 /s2 .
v = 0.2471 m2 /s2 = 0.4971 m/s.
Use P for the period (since T is the tension).
P = C/v = (0.94 m)/(0.4971 m/s) = 1.9 s.
P34. (Opt) In Fig. 6-37, a slab of mass m1 = 40 kg rests on a frictionless floor, and a block of mass
m2 = 10 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and
the coefficient of kinetic friction is 0.40. A horizontal force of magnitude 100 N begins to pull directly
on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block
and (b) the slab?
———It’s easy to work this problem and get the right answer while missing a key point. See the part (c) I
add after solving.
The first question is: do the blocks stick together? The right way to answer this is to assume they do,
then calculate fs , then compare it to fs,max .
Assuming they stick, the only interesting force is F0 = 100 N and the total mass is mtotal = 50 kg.
a = F0 /mtot = (100 N)/(50 kg) = 2.00 m/s2 .
Continuing the stick-toghether assumption, focus on the the slab. The only force of interest is fs which
accelerates it.
fs = ms a = (40 kg)(2 m/s2 ) = 80 N.
Now, considering the block, fs,max = µs mb g = (0.6)(10 kg)(9.8 m/s2 ) = 58.8 N.
So the required fs exceeds fs,max , and the assumption is invalid.
Therefore, work the problem with sliding friction, and the blocks have different accelerations.
I assume you can draw the FBD’s and get that fk = 39.2 N.
(a) x-eq for block: F0 − fk = mb ab ⇒ ab = (100 − 39.2)/10 = 6.08 m/s2 .
(b) x-eq for slab: fk = ms as ⇒ as = 39.2/40 = 0.98 m/s2 .
OK, here’s the source of my opening remark: F0 = 100 N is clearly greater than fs,max = 58.8 N, so
why couldn’t we immediately work the problem assuming they slip?
(c) Let F0 = 70 N. This clearly exceeds fs,max = 59.8 N. Yet, the block does not slip – both blocks
accelerate as a unit. Prove it.
P59. (Opt) In Fig. 6-45, a 1.34 kg ball is connected by means of two massless strings, each of length
L = 1.70 m, to a vertical, rotating rod. The strings are tied to the rod with separation d = 1.70 m
and are taut. The tension in the upper string is 35 N. What are the (a) tension in the lower string,
(b) magnitude of the net force on the ball, and (c) speed of the ball? (d) What is the direction of Fnet ?
———The strings and rod form a nice equilateral triangle. But the forces aren’t symmetric - the trig is just
a little simpler.
FBD for ball: upper string T1 angled up/left; T2 angled down/left; mg down. ~a to left; make +x to
left. Let θ be the angle from horizontal of each string, noting that θ = 30◦ .
(a) y-eq: T1 sin θ − T2 sin θ − mg = 0. Solve for the only unknown T2 :
T2 = T1 − mg/ sin θ = (35 N) − (1.34 kg)(9.8 m/s2 )/(0.500) = 8.736 N = 8.7 N.
(b) x-eq: T1 cos θ + T2 cos θ = mv 2 /R.
Since the y forces are balanced, the net force is the the LHS of this last eq:
Fnet = (T1 + T2 ) cos θ = (43.74)(0.8660) = 37.88 N = 38 N.
(c) From
p geometry, R = L cos θ = 1.472 m. Solve the y-eq above for v:
v = RFnet /m = 6.45 m/s.
(d) Toward the center of the circle; horizontally toward the axle (rod).