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University Physics: Mechanics. Ch 5 . Newton’s Law of Motion. Lecture 7. Dr.-Ing. Erwin Sitompul. http://zitompul.wordpress.com. Homework 5: Two Boxes and A Pulley. - PowerPoint PPT Presentation
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University Physics: Mechanics
Ch5. Newton’s Law of Motion
Lecture 7
Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com
7/2Erwin Sitompul University Physics: Mechanics
Homework 5: Two Boxes and A PulleyA 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?
7/3Erwin Sitompul University Physics: Mechanics
a
Solution of Homework 5
T
T
m1gm2g
a
m1gcosθ
m1gsinθ
FN
Forces on m1 along the x axis:
Forces on m2 along the y axis:
net, 1x xF m a
net, 2y yF m a
1 1sinT m g m a
2 2T m g m a
1 2
1 2
( sin )
( )
m ma g
m m
(3.7sin 30 2.3)
9.8(3.7 2.3)
20.735 m s
• What is the meaning of negative sign?
7/4Erwin Sitompul University Physics: Mechanics
Solution of Homework 5
(a) The magnitude of the acceleration of each block
(b) The direction of the acceleration of the hanging block
(c) The tension in the cord
2 m s0.735a 20.735 m s
DownAssumption : The acceleration points upwardResult : Negative valueConclusion : The true acceleration points downward
1 1sinT m g m a
2 2T m g m a
(3.7)(9.8)(sin 30 ) (3.7)( 0.735)
(2.3)(9.8) (2.3)( 0.735)
20.85 N
20.85 N
7/5Erwin Sitompul University Physics: Mechanics
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?
Applying Newton’s Law: Problem 3
7/6Erwin Sitompul University Physics: Mechanics
(a) Find a general solution for the scale reading, whatever the vertical motion of the cab.
N gF F ma
N gF F ma
N ( )F m g a
• The scale reading is equal to FN, which is the force exerted by the surface of the scale towards the passenger
→
(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
0a N (71.43)(9.8) 700 NF mg
Applying Newton’s Law: Problem 3
7/7Erwin Sitompul University Physics: Mechanics
(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
• It the cab accelerates downward, the magnitude of acceleration is negative
23.2 m sa N ( )F m g a
23.2 m sa N ( )F m g a
(71.43)(9.8 3.2) 928.59 N
(71.43)(9.8 3.2) 471.44 N
Applying Newton’s Law: Problem 3
7/8Erwin Sitompul University Physics: Mechanics
What does the scale read if, in case accident happens, the cab falls vertically downward?
N ( )F m g a 29.8 m sa
Applying Newton’s Law: Problem 3
7/9Erwin Sitompul University Physics: Mechanics
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.
Applying Newton’s Law: Problem 4
• Atwood Machine
7/10Erwin Sitompul University Physics: Mechanics
T
T
m1gm2g
a
a
net, 1y yF m a
1 1 yT m g m a net, 2y yF m a
2 2 yT m g m a
Mass m1 Mass m2
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).
1 1T m g m a 2 2T m g m a
1( )T m g a 2 ( )T m g a
1 2
1 2
m ma g
m m
1 2
1 2
2mmT g
m m
Applying Newton’s Law: Problem 4
7/11Erwin Sitompul University Physics: Mechanics
Applying Newton’s Law: Problem 4
(a) The magnitude of the blocks’ acceleration
(b) The tension in the cord
1 2
1 2
m ma g
m m
(2.8) (1.3)
(9.8)(2.8) (1.3)
23.59 m s
1 2
1 2
2mmT g
m m
2(2.8)(1.3)
(9.8)(2.8) (1.3)
17.40 N
• What happen if m2 > m1?
7/12Erwin Sitompul University Physics: Mechanics
Example: Particle MovementA 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? 2 33 4 2x t ct t
net,x xF max
dxv
dt 24 2 6ct t
xx
dva
dt 2 12c t
2 12(3)xa c (at = 3 s)t
2 36c
net, (2)(2 36)xF c (at = 3 s)t 36
2 36 18c 2 18c
29 m sc
7/13Erwin Sitompul University Physics: Mechanics
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?
Homework 6: The Traffic Light
7/14Erwin Sitompul University Physics: Mechanics
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?