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Lecture 7
Ch5. Newton’s Law of Motion
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Homework 5: Two Boxes and A Pulley
A block of mass m1 = 3.7 kg on a frictionless plane inclined at
angle θ = 30° is connected by a cord over a massless,
frictionless pulley to a second block of mass m2 = 2.3 kg.
What are:
(a) the magnitude of the acceleration of each block,
(b) the direction of the acceleration of the hanging block, and
(c) the tension in the cord?
Erwin Sitompul
University Physics: Mechanics
7/2
Solution of Homework 5
FN
T
T
a
a
m1gsinθ
m1gcosθ
m2g
m1g
Forces on m1 along the x axis:
Fnet,x  m1ax
T  m1 g sin    m1a
Forces on m2 along the y axis:
Fnet,y  m2 a y
T  m2 g  m2 a
Erwin Sitompul
(m1 sin   m2 )
g
(m1  m2 )
(3.7 sin 30  2.3)

9.8
(3.7  2.3)
 0.735 m s2
a
• What is the
meaning of
negative sign?
University Physics: Mechanics
7/3
Solution of Homework 5
(a) The magnitude of the acceleration of each block
2
a  0.735 m s
 0.735 m s2
(b) The direction of the acceleration of the hanging block
Down
Assumption : The acceleration points upward
Result
: Negative value
Conclusion : The true acceleration points downward
(c) The tension in the cord
T  m1 g sin   m1a
 (3.7)(9.8)(sin 30)  (3.7)(0.735)
 20.85 N
T  m2 g  m2 a
 (2.3)(9.8)  (2.3)(0.735)
 20.85 N
Erwin Sitompul
University Physics: Mechanics
7/4
Applying Newton’s Law: Problem 3
A passenger of mass 71.43 kg stands on a platform scale in an
elevator cab. We are concerned with the scale reading when
the cab is stationary and when it is moving up or down.
(a) Find a general solution for the
scale reading, whatever the
vertical motion of the cab.
(b) What does the scale read if the
cab is stationary or moving
upward at a constant 0.5 m/s?
(c) What does the scale read if the
cab acceleration upward 3.2 m/s2
and downward at 3.2 m/s2?
Erwin Sitompul
University Physics: Mechanics
7/5
Applying Newton’s Law: Problem 3
(a) Find a general solution for the scale reading, whatever the
vertical motion of the cab.
→
• The scale reading is equal to FN, which is the
force exerted by the surface of the scale
towards the passenger
FN  Fg  ma
FN  Fg  ma
FN  m( g  a)
(b) What does the scale read if the cab is stationary or moving
upward at a constant 0.5 m/s?
• In stationary condition or when moving upward with a
constant velocity, the acceleration of passenger is zero
a  0  FN  mg  (71.43)(9.8)  700 N
Erwin Sitompul
University Physics: Mechanics
7/6
Applying Newton’s Law: Problem 3
(c) What does the scale read if the cab acceleration upward
3.2 m/s2 and downward at 3.2 m/s2?
• If the cab accelerates upward, the
magnitude of acceleration is positive
a  3.2 m s 2  FN  m( g  a)  (71.43)(9.8  3.2)  928.59 N
• It the cab accelerates downward, the
magnitude of acceleration is negative
a  3.2 m s2  FN  m( g  a)  (71.43)(9.8  3.2)  471.44 N
Erwin Sitompul
University Physics: Mechanics
7/7
Applying Newton’s Law: Problem 3
What does the scale read if, in case accident happens, the
cab falls vertically downward?
FN  m( g  a)
a  9.8 m s2
Erwin Sitompul
University Physics: Mechanics
7/8
Applying Newton’s Law: Problem 4
The figure below shows two blocks connected by a cord (of
negligible mass) that passes over a frictionless pulley (also of
negligible mass). One block has mass m1 = 2.8 kg; the other
has mass m2 = 1.3 kg. Determine:
(a) the magnitude of the blocks’ acceleration.
(b) the tension in the cord.
• Atwood Machine
Erwin Sitompul
University Physics: Mechanics
7/9
Applying Newton’s Law: Problem 4
Mass m1
Mass m2
Fnet,y  m2 a y
T  m2 g  m2 a y
Fnet,y  m1a y
T  m1 g  m1a y
 The acceleration of m1 and m2 have the
same magnitude a, oppose in direction.
 We take the acceleration of m1 as
negative (downward) and of m2 as
positive (upward).
T
a
T
a
m1g
m2g
Erwin Sitompul
T  m1 g  m1a
T  m1 ( g  a)
T  m2 g  m2 a
T  m2 ( g  a)
m1  m2
a
g
m1  m2
2m1m2
T
g
m1  m2
University Physics: Mechanics
7/10
Applying Newton’s Law: Problem 4
(a) The magnitude of the blocks’ acceleration
a
m1  m2
(2.8)  (1.3)
g 
(9.8)  3.59 m s2
m1  m2
(2.8)  (1.3)
• What happen if m2 > m1?
(b) The tension in the cord
T
2m1m2
2(2.8)(1.3)
g 
(9.8)  17.40 N
m1  m2
(2.8)  (1.3)
Erwin Sitompul
University Physics: Mechanics
7/11
Example: Particle Movement
A 2 kg particle moves along an x axis, being propelled by a
variable force directed along that axis. Its position is given by
x = 3 m + (4 m/s)t + ct2 – (2 m/s3)t3, with x in meters and t in
seconds. The factor c is a constant. At t = 3 s, the force on the
particle has a magnitude of 36 N and is in the negative direction
of the axis.
What is c?
x  3  4t  ct 2  2t 3
Fnet,x  max
Fnet,x (at t = 3 s)  (2)(2c  36)  36
2c  36  18
2c  18
c  9 m s2
Erwin Sitompul
dx
 4  2ct  6t 2
dt
dvx
 2c 12t
ax 
dt
vx 
ax (at t = 3 s)  2c  12(3)
 2c  36
University Physics: Mechanics
7/12
Homework 6: The Traffic Light
A traffic light weighing 122 N hangs from a cable tied to two
other cables fastened to a support, as in the figure below. The
upper cables make angles of 37° and 53° with the horizontal.
These upper cables are not as strong as the vertical cable and
will break if the tension in them exceeds 100N.
Will the traffic light remain hanging in this situation, or will one
of the cables break?
Erwin Sitompul
University Physics: Mechanics
7/13
Homework 6
New
1. What is the net force acting on the
ring in the next figure?
2. Joe’s Advertising wishes to hang
a sign weighing 750.0 N so that
cable A, attached to the store
makes a 30.0° angle, as shown in
the figure. Cable B is horizontal
and attached to and adjoining
building.
What is the tension in cable B?
Erwin Sitompul
University Physics: Mechanics
7/14