University Physics: Mechanics Ch5. Newton’s Law of Motion Lecture 7 Dr.-Ing. Erwin Sitompul .
University Physics: Mechanics
description
Transcript of University Physics: Mechanics
University Physics: Mechanics
Ch5. Newton’s Laws of MotionLecture 8
Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com
2013
8/2Erwin Sitompul University Physics: Mechanics
Homework 6: 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?
8/3Erwin Sitompul University Physics: Mechanics
a
Solution of Homework 6
TT
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 gm m
(3.7sin 30 2.3) 9.8
(3.7 2.3)
20.735 m s • What is the meaning of negative sign?
8/4Erwin Sitompul University Physics: Mechanics
Solution of Homework 6(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
Down Assumption : 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
8/5Erwin 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
8/6Erwin 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
8/7Erwin 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
8/8Erwin 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
a = –g
8/9Erwin 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
8/10Erwin 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 gm m
1 2
1 2
2mmT gm m
Applying Newton’s Law: Problem 4
8/11Erwin 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 gm m
(2.8) (1.3) (9.8)(2.8) (1.3)
23.59 m s
1 2
1 2
2mmT gm m
2(2.8)(1.3) (9.8)(2.8) (1.3)
17.40 N
• What happen if m2 > m1?
8/12Erwin Sitompul University Physics: Mechanics
Virtual Experiment: Atwood Machine
Two masses of 550 g and 560 g are connected through a pulley to construct an Atwood machine. If the masses are released from rest, how much time is required for the masses to move a distance of 1 m?
8/13Erwin 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 maxdxvdt
24 2 6ct t
xxdvadt
2 12c t
2 12(3)xa c (at = 3 s)t2 36c
net, (2)(2 36)xF c (at = 3 s)t 36
2 36 18c 2 18c
29 m sc
8/14Erwin Sitompul University Physics: Mechanics
Hanging ChairA man sits in a chair that dangles from a massless rope, which runs over a massless, frictionless pulley and back down to the man’s hand. The combined mass of man and chair is 95 kg.With what force magnitude must the man pull on the rope if he is to rise (a) with a constant velocity and (b) with an upward acceleration of 1.3 m/s2?
(a) 0a
(b)
2 0A gT F 12AT mg
21.3 m sa 2 B gT F ma 12 ( )BT m g a
TA
Fg = mg
TA
a
TB
Fg = mg
TB
12 (95)(9.8) 465.5 N
12 (95)(9.8 1.3) 527.25 N
8/15Erwin Sitompul University Physics: Mechanics
Hanging ChairContinuing the same problem, if the rope on the right extends to the ground and is pulled by a co-worker, with what force magnitude must the co-worker pull for the main to rise (c) with a constant velocity and (d) with an upward acceleration of 1.3 m/s2?
(c) 0a
(d)
0C gT F
CT mg
21.3 m sa D gT F ma ( )DT m g a
Fg = mg
TC
a
Fg = mg
TB(95)(9.8) 931 N
(95)(9.8 1.3) 1026 N
• What is the magnitude of the force on the ceiling from the pulley system in each of 4 cases?
8/16Erwin 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 7: The Traffic Light
8/17Erwin Sitompul University Physics: Mechanics
Homework 7A1. Two forces are applied to a ring that is
lying on a frictionless surface. It is shown on the figure from the bird’s-eye-view. If the mass of the ring is 6 kg, determine the position of the ring 2 seconds after the two forces are applied.
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?
8/18Erwin Sitompul University Physics: Mechanics
θ
Homework 7B1. As your jet plane speeds down the runway on
takeoff, you decide to determine its acceleration, so you take out your yo-yo and note that when you suspend it, the string makes an angle of 22° with the vertical (see figure). (a) What is the acceleration of the plane? (b) If the mass of the yo-yo is 40 g, what is the tension in the string?Hint: Decompose the tension T.
2. In the next figure, let the mass of the block be 8.5 kg and the angle θ be 30°. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the elevation decrease of the mass after moving for 0.8 s (assuming it does not reach the floor).
a