Download Physics 140 HOMEWORK Chapter 05A Q3. In Fig. 5

Physics 140
HOMEWORK Chapter 05A
~ 1 and F
~ 2 are applied to a lunchbox as it slides at constant velocity over
Q3. In Fig. 5-21, forces F
a frictionless floor. We are to decrease angle θ without changing the magnitude of F1 . For constant
velocity, should we increase, decrease, or maintain the magnitude of F2
———
Increase. As θ decreases, the horizontal component of F1 increases. Since the horizontal forces were
balanced before, F2 must be increased if they are to remain in balance. (“velocity constant”).
Q7. July 17, 1981, Kansas City: The newly opened Hyatt Regency is packed with people listening and
dancing to a band playing favorites from the 1940s. Many of the people are crowded onto the walkways
that hang like bridges across the wide atrium. Suddenly two of the walkways collapse, falling onto the
merrymakers on the main floor. The walkways were suspended one above another on vertical rods and
held in place by nuts threaded onto the rods. In the original design, only two long rods were to be
used, each extending through all three walkways (Fig. 5-24a). If each walkway and the merrymakers
on it have a combined mass of M , what is the total mass supported by the threads and two nuts on
(a) the lowest walkway and (b) the highest walkway? Threading nuts on a rod is impossible except at
the ends, so the design was changed: Instead, six rods were used, each connecting two walkways (Fig.
5-24b). What now is the total mass supported by the threads and two nuts on (c) the lowest walkway,
(d) the upper side of the highest walkway, and (e) the lower side of the highest walkway? It was this
design that failed.
———
Note: The answers in the back of the book are correct for the way the question is phrased, in terms of
“mass” and “total of the two rods on each side.” My answers are in terms of the tension – a force – on
each individual rod, so there is a factor of 2/g to convert my answers to the way the question is actually
phrased. I like my way better because the tension in each individual member (rod or nut/thread) is the
way an engineer is more likely to view it.
(a) In the original design, each nut on the lowest walkway supports M g/2, since there are two of them.
(b) In the original design, each nut on the highest walkway supports M g/2, since there are two of them.
The tension in each of the rods above the top walkway is 3M g/2, but the tensile strength of the rods
is greater than the strength of the nut/threads.
(c) M g/2. Draw the FBD: T3 up (at left end), T3 up (at right end); M g down. ~a = 0 so F NET =
0 ⇒ 2T3 = M g ⇒ T3 = M g/2.
(d) Let T3 be the tension in the lowest rod, T2 in the middle, and T1 the tension in the top rod. Tensions
are assumed equal on the left and right rods, although they differ as you change level. The tension in
the rod is equal to the force on the nut at the end of the rod.
FBD for middle walkway: T3 down twice, T2 up twice, and M g down, representing the gravitational
force on the MIDDLE walkway. Note that the grav force for the lowest walkway does not appear
directly on this FBD.
2T2 − 2T3 − M g = 0 ⇒ T2 = M g, knowing that T3 = M g/2.
(e) FBD for top walkway: 2T1 up, 2T2 down, M g down. This M g is the weight of the top walkway.
2T1 − 2T2 − M g = 0 ⇒ T1 = 3M g/2.
Anyone going into mechanical engineering or civil engineering can see why I included this problem.
Over 100 people were killed. The original design was actually inadequate as well as impractical to
build; it had a safety factor of about 2, instead of the required value of about 3. The revised design
obviously had no margin at all.
P3. If the 1 kg standard body has an acceleration of 2.00 m/s2 at 20.0◦ to the positive direction of
an x-axis, what are (a) the x-component and (b) the y-component of the net force acting on the body,
and (c) what is the net force in unit-vector notation?
———
~ NET = m~a, so F
~ NET = (1 kg)(2 m/s2 ) = 2 N. This force is in the same direction
Newton’s Second Law: F
as the acceleration.
(a) Fx = (2 N)(cos(20◦ ) = +1.88 N.
(b) Fy = (2 N)(sin(20◦ ) = +0.684 N. The vector is in quadrant I.
~ NET = (1.89ı̂ + 0.684̂) N.
(c) F
P17. In Fig. 5-36, let the mass of the block be 8.5 kg and the angle be 30◦ . Find (a) the tension in
the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the
resulting acceleration of the block.
———
~ NET = 0. Let the +x-axis point up-plane and +y
FBD: T up-plane; n away from plane; mg down. F
perpendicular away from plane.
Newton’s 2nd law gives:
x-eq: T − mg sin θ = 0 ⇒
(a) T = mg sin θ = (8.5 kg)(9.8 m/s2 )(0.5) = 41.7 N
(b) y-eq: n − mg cos θ = 0 ⇒
n = mg cos θ = (8.5 kg)(9.8 m/s2 )(0.866) = 72.1 N.
(c) If the cord is cut, the net force on the block is 41.7 N down-plane, so the acceleration is
a = (41.7 N)/(8.5 kg) = 4.90 m/s2 . Note units.
P20. A car traveling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a
distance of 65 cm (with respect to the road) while being brought to rest by an inflated air bag. What
magnitude of force (assumed constant) acts on the passenger’s upper torso, which has a mass of 41 kg?
———The force will be ma, but we need the acceleration to use this. Convert v0 = 53 km/hr = 14.7 m/s.
v 2 − v02 = 2a∆x ⇒ a = (v 2 − v02 )/(2∆x) = (0 − 14.72 )/(2 · (0.65 m)) = −167 m/s2 .
Now: F = ma = (41 kg)((−)167 m/s2 ) = 6830 N (drop the negative sign for “magnitude”).
P31. A block is projected up a frictionless inclined plane with initial speed v0 = 3.50 m/s. The angle
of incline is 32◦ . (a) How far up the plane does the block go? (b) How long does it take to get there?
(c) What is its speed when it gets back to the bottom?
———
(a) FBD for block: mg down; n away from plane. ~a is down-plane. Make +x down-plane (so that a
will be positive, noting that v0 = −3.5 m/s in this coordinate system.
The y-eq doesn’t bear on acceleration in frictionless problems.
x-eq: mg sin θ = ma ⇒ a = g sin θ = (9.8)(0.530) = +5.19 m/s2 .
At its maximum height, v = 0, so ∆x = (v 2 − v02 )/2a = (0 − 3.52 )/(2 · 5.19) = −1.18 m. Negative
means up-plane (!).
(b) t = ∆x/vavg = (−1.18 m)/(−1.75 m/s) = 0.674 s. (remember that vavg = (v0 + vf )/2)
(c) v 2 = v02 + 2a∆x = v02 , since ∆x = 0. v = ±v0 = −v0 = +3.5 m/s, where we realize that the block
is now moving down-plane (in the positive direction).