Download Physics 101 Homework 7 Due February 2 1 1. The coefficient of

Physics 101
Homework 7
Due February 2
1. The coefficient of static friction between hard rubber and normal street pavement is about 0.8.
On how steep a hill (maximum angle) can you leave a car parked?
Solution:
From lecture we know that
s  tan max  0.8  max  tan 1 0.8  39o  40o
More detailed solution (optional):
See the included free-body diagram. To find the maximum angle, assume that the car is just
ready to slide, so that the force of static friction is a maximum. Write Newton’s 2nd law for both
directions. Note that for both directions, the net force must be zero since the car is not
accelerating.
y
F
F
y
 FN  mg cos   0  FN  mg cos 
x
 mg sin   Ffr  0  mg sin   Ffr   s FN   s mg cos 
 s 
mg sin 
mg cos 
 tan   0.8    tan 1 0.8  39 o  40 o
x
FN
Ffr


mg
2. A mass m is placed on an inclined plane (m > 0) and slides down the plane with constant
speed. How would move a similar block (same m) of mass 2m if it placed on the same
incline?
Solution:
The component of gravity acting down the plane is double for 2m. However, the normal force
(and hence the friction force) is also double (the same factor!). This means the two forces still
cancel to give a net force of zero. So, the block would slide down at constant speed. Please look
lecture 5, slide 6.
3. A box sits on a flat board. You lift one end of the board, making an angle with the floor. As
you increase the angle, the box will eventually begin to slide down. Why? (Note: there are
two reasons!)
Solution:
1) The component of the gravity force parallel to the plane increased.
2) The normal force exerted by the board decreased. Since friction depends on Normal force, we
see that the friction force gets smaller.
1
Physics 101
Homework 7
Due February 2
4. Suppose a car moves at constant speed along a hilly road. Where does the car exert the
greatest and least forces on the road: (a) at the top of a hill, (b) at a dip between two hills, (c)
on a horizontal road?
Solution:
The force that the car exerts on the road is the Newton’s 3rd law reaction to the
normal force of the road on the car, and so we can answer this question in terms
of the normal force. The car exerts the greatest force on the road at the dip
between two hills. There the normal force from the road has to both support the
weight AND provide a centripetal upward force to make the car move in an
upward curved path. The car exerts the least force on the road at the top of a hill.
We have all felt the “floating upward” sensation as we have driven over the crest
of a hill. In that case, there must be a net downward centripetal force to cause the
circular motion, and so the normal force from the road does not completely
support the weight.
FN
mg
FN
mg
5. Suppose the space shuttle is in orbit 400 km from the Earth’s surface, and circles the Earth
about once every 90 minutes. Find the centripetal acceleration of the space shuttle in its orbit.
Express your answer in terms of g, the gravitational acceleration at the Earth’s surface.
Solution: The orbit radius will be the sum of the Earth’s radius plus the 400 km orbit height.
The orbital period is about 90 minutes. Find the centripetal acceleration from these data.
 60 sec 
r  6380 km  400 km  6780 km  6.78  10 6 m
T  90 min 
  5400 sec
 1 min 
2
4 2  6.78  106 m 
 2 
r

 9.18 m s 2

2
 T 
 5400 sec 
arad   2 r  
 1g

g  9.80 m s 2  arad   9.18 m s 2  
 0.9 g's
2 
 9.80 m s 
Notice how close this is to g, because the shuttle is not very far above the surface of the Earth,
relative to the radius of the Earth.
6. Calculate the force of Earth’s gravity on a spacecraft 2 Earth radii above the Earth’s surface
(3 radii from the center of the Earth) if the spacecraft’s mass is 1800 kg.
Solution: We can use equation: g  G
M
r2
The spacecraft is three times as far from the Earth’s center as when at the surface of the Earth. Therefore,
since the acceleration due to gravity decreases as the square of the distance, the force of gravity on the
spacecraft will be one-ninth of its weight at the Earth’s surface.
1800 kg  9.80 m s 2
1
FG  9 mg Earth's 
 1.96  103 N
9
surface

Also, we can use equation: Fg 

Gm1 m2
and mass of Earth.
r2
2
Physics 101
Homework 7
Due February 2
7. Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, that
Mars’ radius is 3400 km, that Earth’ radius is 6380 km , and that Earth’ mass is 5.97 1024 kg ,
determine the mass of Mars.
Solution:
The expression for the acceleration due to gravity is g  G
G
M Mars
2
Mars
R
 0.38 G
M Earth
2
REarth
R2
. For Mars, g Mars  0.38 g Earth . Thus

2
M Mars
M
R 
 3400 km 
23
 0.38M Earth  Mars   0.38  5.97  10 24 kg  
  6.4  10 kg
R
6380
km


 Earth 
2
8. At the surface of a certain planet, the gravitational acceleration g has a magnitude of
12.0 m s 2 . A 21.0-kg brass ball is transported to this planet. What is (a) the mass of the brass
ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the
planet?
Solution:
(a) Mass is independent of location and so the mass of the ball is 21.0 kg on both the Earth and
the planet.
(b) The weight is found by W  mg .


  21.0 kg  12.0 m s   252 N .
WEarth  mg Earth   21.0 kg  9.80 m s 2  206 N
WPlanet  mg Planet
2
3