Download Outline - Newton`s Laws

Outline - Newton's Laws
I.
II.
Newton's Laws
A.
First Law. An object at rest remains at rest and an object moving
with velocity v remains moving with velocity v if and only if no net
external force acts on the object.
B.
Second Law. The net force acting on an object is directly
proportional to and in the same direction as the acceleration. The
proportionality constant is the mass of the object. Fnet = ma
C.
Third Law. When two objects interact in whatever manner, the
force F1 2 exerted by the first object on the second is equal in
magnitude and opposite in direction to the force F2 1 of the second
object on the first. - F1 2 = F2 1
Application of Newton's Laws
A.
An object hanging motionlessly in Fig. 1 above has a weight of 800
N. Find the tensions T1 and T2 in the massless ropes. For the object
to be at rest, the sum of the forces acting on the object must be
zero. Since the tensions are neither in the X or Y-direction, take
components:
(Fnet)x = max = 0
(Fnet)y = may = 0
o
o
T2 cos 20 – T1 cos 20 T2 sin20o + T1 sin20o – 800N
=0
=0
o
o
T2 cos 20 = T1 cos 20 T2 sin20o + T1 sin20o = 800N
T2 = T1 = T
2T sin20o = 800 N
T = 1.17 x
3
10 N = T2 = T1
B.
C.
When you apply Newton’s second law, you must be very careful
about your choice of system.
1.
Imagine two masses m1 = 1.0 kg and m2 = 2.0 kg are
connected by a massless string and are pulled along a
frictionless surface by a force F = 9.0 N as shown in Fig. 2a
below. Find (a) the acceleration of the blocks and (b) the
tension in the string.
a.
In Fig. 2a above, I have isolated the entire system of
the two blocks. There are vertical forces acting on both
blocks: the normal force of the surface and the weight
of each block. Since there is no acceleration in the
vertical direction and there is no friction (we would
need to know the normal force in order to find the
frictional force on a block), we shall not consider forces
in the vertical direction. The only external horizontal
force acting on the system is F = 9.0 N. There are
internal horizontal forces acting on the blocks and the
string, but we are not interested in internal forces
because
Fnet external = msystem a
9.0 N = [(1.0 + 2.0)kg] a,
and
b.
a = 3.0 m/s2
Now we choose our system as shown in Fig. 2b and 2c
above. Again I only look at the horizontal forces. I
have isolated:
i.
the string experiencing F1 s, the force of block 1
on the string and F2 s, the force of block 2 on
the string
ii.
m1 experiencing Fs
block 1
iii.
m2 experiencing Fs 2, the force of the string on
block 2 and F = 9.0 N.
1
= the force of the string on
c.
For the string, Fnet external = mstringa
F2 s - F1 s = (0)3.0 m/s2 = 0
So F2 s = F1 s
By Newton's Third Law, Fs 2 = F2 s and Fs 1 = F1
Thus, Fs 2 = F2 s = F1 s = Fs 1.
We drop all this fancy notation and write
Fs 2 = F2 s = F1 s= Fs 1 = T as shown in Fig. 2c
above.
s
d.
For m1, Fnet external = m1a
T = 1.0 kg(3.0 m/s2) = 3.0 N
For m2, Fnet external = m2a
9.0 N - T = 2.0 kg(3.0 m/s2) = 6.0
N
9.0 N - 6.0 N = T = 3.0 N
e.
III.
D.
As you can see from the above example, Newton Third Law of
Motion forces always act on different objects. In other words,
they do NOT act on the same object.
E.
Practice problems in 104 Problem Set for Newton's Laws: 1 - 8.
Frictional Forces
A.
The frictional force f = µN, where the coefficient of friction µ is
labeled with a subscript s for static situations and k for objects in
motion. The coefficient of friction for static situations is always
greater than that for kinetic. The static frictional force can go from
0 to a value necessary to just about move the object.
B.
Finding the frictional force (Fnet)y = may = m(0) = 0
1.
2.
3.
C.
IV.
For Fig. 3a above, N – mg = 0. N = mg and f = µN = µmg
For Fig. 3b, N + F sin  – mg = 0. N = mg - F sin  and f =
µN
= µ(mg - F sin )
For Fig. 3c, N - mg cos = 0. N = mg cos  f = µN = µmg
cos 
Practice problem in 104 Problem Set for Newton's Laws: 10.
Free Body Diagrams
A.
Force Diagrams such as those shown in Fig. 4 below are called
Free-body diagrams. They are extremely important in solving
Newton Second Law of Motion problems.
B.
In each situation shown in Fig. 4a above, one or more forces act
upon an object. All drawings are in a vertical plane and friction is
negligible except in (b) and (d). Draw free body diagrams for the
figures, scale the forces as close as possible. Label all the forces
acting on the objects. If the object has an acceleration, show its
direction. If there is no acceleration, indicate that it is zero.
C.
In Fig. 4b above, (a) the only force that acts on the object is its
weight mg. Its Acceleration a is down. (b) For a constant velocity,
the net force acting on the object must be zero. The upward
frictional force equals the weight of the object. (c) There is no net
force perpendicular to the plane. The normal force is equal to the
component of the weight perpendicular to the plane. The
component of the weight parallel to the plane gives the
acceleration down the incline. (d) Now in addition to the forces
talked about in (c) there is a frictional force up the plane that is
equal in magnitude to the component of the weight parallel and
down the plane. There is no acceleration. (e) and (f) The only force
acting on an object in projectile motion (neglecting a frictional
force) is the weight of the object down.
V.
Gravitational force Fg = Gm1m2/r2, where m1 and m2 are
the masses of two point particles separated by a
distance r.
A.
Near the earth’s surface at height h very small compared to the
radius of the earth RE.
1.
You can treat a spherical mass ME as a point mass located at
the center of the earth. The force on an object of mass m at
height h above the earth by the earth is Fg = GmME/(RE +
h)2. Since RE is much greater than h, drop the h and Fg =
GmME/RE2.In general, Fnet = ma
2.
For a freely falling object, GmME/RE2 = ma
For free fall a = g
GmME/RE2 = mg = Weight of object
3.
(mg) is not a mass times an acceleration, it is a force. To
determine the direction of the force in a problem, decide on
a direction to call positive and then see if mg is in that
direction. After doing this, never substitute –9.8 m/s2 for g
in (mg).
4.
Practice Problem I. A classics student of mass 70.0 kg is 0.50
m from a physics student of mass 50.0 kg. Calculate the
approximate magnitude of the gravitational force that each
exerts on the other. Explain why your result is only
approximately equal to the magnitude of the actual
gravitational force.
Solution. The approximate magnitude of the gravitational
force that each exerts on the other is found from the
equation for point-like masses. F = Gmcmp/r2
= (6.67 x 10-11 N-m2/kg2)(70.0 kg)(50.0 kg)/(0.50 m)2
= 9.3 x 10-7 N.
The force is approximate because students are not pointmasses.
5.
Practice Problem II. An astronaut puts a bowling ball into a
circular orbit about the Earth at an altitude h of 350 km. Find
the ball's period of motion. The radius and mass of the earth
are M = 5.98 x 1024 kg and R = 6.37 x 106 m, respectively.
Solution. The radius of the orbit of the bowling ball r = the
radius R of the earth + the height h above the earth's
surface =
(6.37 + 0.35)106 m,
and the constant in Newton's law of gravitation G =
6.67 x 10-11 N-m2/kg2.
The gravitational force produces the centripetal acceleration:
GMm/r2 = mv2/r(1)
GMm/r2 = m(r/T)2/r
Since v = 2r/T, or T2 = 2 r3/GM, and
T = 2 (r3/GM)1/2
= 2{(6.72 x 106 m)3/6.67 x 10-11 N-m2/kg2 x 5.98 x
1024 kg}1/2
= 55 x 102 s
Review - Newton's Laws
1. The same constant net force acts on two different objects. A
plot of velocity as a function of time for the two objects is
given in Fig. 1 below. The graph labelled A is for object A; the
one labelled B is for object B. The mass of object A is 2.0 kg.
Find (a) the acceleration of object A, (b) the acceleration of
object B, (c) the mass of object B, and (d) the magnitude of
the net force.
2. An object of mass m, resting on a horizontal frictionless
surface is acted upon by an applied force F, the normal force
of the surface FN and the gravitational attraction of the earth
Fg, as shown in Fig. 2 below. Find Fg, FN and the acceleration
of the object when (a) F = 12 N and m = 2.0 kg, (b) F = 24 N
and m = 2.0 kg, (c) F = 12 N and m = 1.0 kg. Take g = 10
m/s2.
3. In Fig. 2 above, the surface now exerts a frictional force of 4.0
N on the object when it is in motion. If the mass of the object
m = 2 kg, find the acceleration of the object when (a) F = 12
N and (b) F = 24 N.
4. Two objects of mass mA = 2.0 kg and mB = 4.0 kg rest on a
frictionless surface (Fig. 3 below). An applied force F = 12 N
acts on object A, as shown in Fig. 3. Find (a) the acceleration
of the objects, (b) the force FAB of object B on A and (c) the
force FBA of object A on B.
5. A rope is attached to a hook on a wall. You pull on the rope,
but it does not move. Discuss the forces acting on the hook,
the rope, and you. Which of these forces are Newton third law
of motion forces?
6. The two objects in Fig. 3 above are now connected by a
massless string between A and B, string 1, and pulled to the
right by exerting a force of 12 N to the right through another
massless string, string 2, attached to block B. Find (a) the
acceleration of the blocks and (b) the tension in string 1.
7. (a) An object of mass m = 3.0 kg is set into motion vertically
and then raised with a constant velocity of 3.0 m/s. What
applied force F is needed? (b) The object is now given an
upward acceleration of 3.0 m/s2. Now what applied force is
needed? Take g = 10 m/s2.
8. A system consisting of a rope of mass 0.10 kg between two
blocks each of mass 0.10 kg is lifted by an applied force F =
9.0 N (Fig. 4 below). (a) Find the acceleration of the system.
Find the tension at (b) the top of the rope, and (c) the bottom
of one-fifth of the rope. Take g = 10 m/s2.
9. A block is pulled along a rough horizontal surface. Figure 5
below is a plot of the applied force F as a function of the
acceleration of the block Find (a) the force that must be
applied to move the block with constant velocity, (b) the mass
of the block, (c) the force that must be applied to give the
block an acceleration of 1.0 m/s2. (d) Suppose the experiment
is repeated with a block of a greater mass. If the new block is
pulled along the same surface, how will this change the graph?
(e) Suppose the experiment is repeated with the original
block, changing only the surface on which the block is pulled.
How will this change the graph?
10. A block of mass m = 2.0 kg is moved to the right on a
horizontal surface by a force of 26 N. The coefficient of kinetic
friction µk between the surface and the object is 1/5. (a) Draw
the object and show all of the forces acting on it. Find (b) the
frictional force on the block and (c) the acceleration of the
block.
Take g = 10 m/s2.
11. Repeat problem 10 when the applied force makes an angle of
22.6o above the horizontal.
12. An object of mass m = 3.0 kg accelerates down a frictionless
inclined plane that makes an angle of 370 with the horizontal.
(a) Show all the forces acting on the object. Draw an X-axis
parallel and down the plane and a Y-axis perpendicular to the
plane and upward. (b) Find and draw the components of the
forces on the X and Y-axes. (c) Find the acceleration of the
block. Take g = 10 m/s2.
13. Repeat Problem 12 for a coefficient of kinetic friction µk
between the surface and the object of 0.154.
14. The coefficient of kinetic friction between the block of mass m
= 2.5 kg and the plane in Fig. 6 below is 1/6. Find (a) the
acceleration of the block and (b) the tension in the string.
Take g = 10 m/s2.
15. An object of mass m = 2 kg moves in a circle on a table with
uniform circular motion. The speed of the object is 2 m/s and
the radius of the circle is 0.5 m. Find the (a) frictional force
acting on the object and (b) total force of the table on the
object.
16. A man of mass 60 kg pushes on a sled of mass 10 kg and the
man and the sled accelerate forward with an acceleration of
2.0 m/s2. Neglect a frictional force on the sled and find (a)
the force of the man on the sled, (b) the force of the sled on
the man and (c) the frictional force of the snow on the man.
17. A wooden rod of negligible mass is connected to the shaft of a
motor. An object with mass m = 2 kg is attached to the other
end of the rod. As the shaft rotates, the object moves in a
vertical circle of radius 0.5 m with a constant speed of
magnitude v = 3 m/s. (a) Find the magnitude of the
centripetal acceleration for this object. The vertical circle path
is shown in Fig. 7 below for an object that moves counterclockwise. Redraw the figure and show on it, the direction of
the velocity at (b) B, the bottom of the circle, (c) T, the top of
the circle and (d) S, the side of the circle. Now draw the
direction of the acceleration for (e) point B, (f) point T and (g)
point S Find the force of the rod on the object For for (h) point
B, (i) point T and (j) point S.
18. A 10-kg object rests on a scale in an elevator. Find the
reading on the scale when the elevator moves with (a) a
constant velocity of 5 m/s and (b) a constant acceleration of
5 m/s2 upward. Take g = 10 m/s2.
19. A person of mass m stands on a bathroom scale in an
elevator. Figure 8 below is a rough plot of scale reading as a
function of time. Describe the motion of the elevator.
21. A block with a mass m = 2.0 kg initially at rest is pushed up the incline of 37o by a horizontal force F
= 60 N (Fig. 14 below). The coefficient of kinetic friction between the block and the incline µ k =1/2.
Find the acceleration of the block.