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Benton OSU Physics Dept.
Physics 2014: General Physics I Homework Assignment 3: Solutions
Due: Friday, 11 September 2009 at the beginning of your Recitation session.
- In a human cannon ball stunt at the circus, a circus performer, the âGreat DisAstro,â is fired out of an air cannon that is 20 m long. The mass of the âGreat DisAstroâ is 60 kg. If the average force on the âGreat DisAstroâ while he is being shot through the cannon barrel is 3500 N, determine his average acceleration inside the barrel and his velocity when he emerges from the end of the barrel.
Solution: We know the performerâs mass and we know the average force. Newtonâs second law says, F = ma. So,
3500N (^) 58 m s 2 60kg
F
a m
The second part of the problem is just a kinematic problem and we can use the equation
v f^2 = vi^2 + 2 a s ( â si )
We assume the performer starts from rest, v (^) i = 0, and that he starts at s (^) i = 0.
2 2 58 m s 2 20m 48 m s v (^) f = as = â
â
=.
- Can your car stop on a dime? Thatâs what the salesman said. If your car has a mass of 2000 kg and you are driving at a constant velocity of 120 km/hr when you slam on the breaks, what force must the breaks exert to bring the car to rest in a distance equal to 1.7 cm diameter of a dime? How many gâs is this and what effect is this force likely to have on you, the driver? Assume that the wheels donât lock and the car doesnât skid (admittedly not very realistic).
Solution: using a kinematic equation, we can find what value of acceleration will bring a car traveling at 120 km/hr to a stop in the 1.7 cm distance represented by the diameter of a dime. Once we have found the acceleration, we can calculate both the force and the number of gâs. First convert 100 km/hr to S.I. units.
0 km m 1 hour m (^120) hr 1000 km 33.3 s 3600 s
v = Ă Ă =
Also, we know that v (^) f = 0, s 0 = 0, and s (^) f = 1.7 cm = 0.017 m.
(^2 0 2) ( 0 )
2 0
f f
f
v v a s s
v as
Benton OSU Physics Dept.
2 2 (^0 )
7 2
33.3 m s (^) 32, 614 m 2 2 0.017 m s
2000 kg 32, 614 m 6.5 10 N s
f
v a s
F ma
= = â
â = â Ă
Since 1 g is equivalent to 9.8 m/s 2 ,
2
2
32, 614 m s (^) 3,328 g's 9.8 m s
This kind of force would turn your body into pulp. Even if the car lives up to the salesmanâs claim, it isnât something youâd likely want to driveâŚmaybe car salesmen should take physics.
- A 1000 kg mass is suspended from two chains as illustrated in the figure below. The chain on the left is at a 25° angle with respect to the ceiling, while the chain on the right is at a 45° angle with respect to the ceiling. Find the tension force in each of the two chains. How are the horizontal ( x ) components of force in the two chains related?
Solution: Since the mass and chains are all at rest, there is no net acceleration, meaning that this
is a case of static equilibrium and the â F = 0
G
. First we need to draw a free body diagram and
label all the forces. Since the tension forces in the two chains are at arbitrary angles, we need to find the x and y components of the tension forces.
1000 kg
25 °^45 °
Benton OSU Physics Dept.
- In an old warehouse, an 800 kg elevator is suspended from a worn steel cable that can withstand a maximum tension of 26,000 N. If you load the elevator with 1400 kg of cargo, what is the maximum value at which the elevator can accelerate upwards so that the cable doesnât break?
Solution: The total mass (800 kg + 1400 kg) will be subjected to both the force of gravity and the force accelerating the elevator. At maximum acceleration, the force accelerating the elevator will be equal to the 26,000 N tension in the cable minus the force due to gravity on the elevator.
F E = Ma = T â Mg
We can solve this for acceleration:
2 2
26, 000 N 2200kg 9.8 m s (^) 2.0 m 2200kg s
T Mg a M
â^ â^ â
- On July 30, 1958, Lieutenant Carter C. Collins, riding in a human centrifuge at Langley Research Center in Virginia, established to worldâs record for highest number of gâs without passing out. Lt. Collins (who coincidentally was the father of a good high school friend of mine) sustained an acceleration of 20.7 g for a period of 6 seconds. Given the fact that the human centrifuge was 30 ft (9.2 m) in diameter, what was the frequency with which the centrifuge rotated and what was Lt. Collins tangential velocity during the test?
Solution: The radius is half the diameter, 4.6 m. To get acceleration in m/s^2 , multiply the number of gâs by 9.8 m/s^2.
2 2 2
2
20.7g 9.8 m s 202.9 m s
202.9 m s 4.6m 30.6 m s
v a r v ar
Benton OSU Physics Dept.
- A medieval-style catapult has an arm that is 5 m long. When lobbing a 20 kg stone, the arms swings from the horizontal to the vertical (one quarter turn) in 0.15 s. a) What is the centripetal force exerted by the catapult arm on the stone? b) Neglecting air resistance and assuming the velocity of the stone is parallel to the level ground when the stone leaves the catapult, what is the maximum range of the stone?
Solution: The problem states that the arm of the catapult moves through a quarter turn in 0.15 s. This, together with the quarter arc arrow in the diagram, tells us that this is a circular motion problem. The other clue is the phrase âcentripetal force.â Again, if we must find the centripetal force, we must be dealing with circular motion.
a) We write down the equation for centripetal force (since thatâs what weâre asked to find). 2 2 c
mv F m r r
= = Ď
We know the mass of the rock and we know the radius of the circleâŚthatâs the length of the catapult arm. We donât know v or Ď. However, if the catapult arm moves through a quarter turn in 0.15 s, it would take 4Ă that time to move through a full circle and that we be equal to the period: T = 4 Ă 0.15 s = 0.6 s.
2 2
10.5 rad s 0.6s 20kg 10.5 rad s 5m 11, 025J
T
F m r
b) We can solve this problem kinematically. Since the initial velocity is all in the horizontal direction, the time it takes the rock to hit the ground will be the same as if we dropped the rock from rest with no horizontal velocity. The distance the rock will travel along the x-axis during this time will be the range. We assume the height of the arm is the height above the ground at which the rock is released, i.e. 5 m. We can find the initial velocity by:
Benton OSU Physics Dept.
6 6
2 6
6.37 10 m + 500 km 1000 m 6.87 10 m km 0.86 9.8 m s (^) 0.0011 rad 6.87 10 m s 2 5672 s = 1.58 hr 0.0011 rad s 24 hr day 15.2 times per day 1.58 hr
r
T
= Ă â
= Ă
Ă
