University Physics: Mechanics Ch5. Newton’s Law of Motion Lecture 7 Dr.-Ing. Erwin Sitompul .

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?


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?


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


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


(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



(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



What does the scale read if, in case accident happens, the cab falls vertically downward?

N ( )F m g a 29.8 m sa



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


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




(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?


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


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


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?

University Physics: Mechanics Ch5. Newton’s Law of Motion Lecture 7 Dr.-Ing. Erwin Sitompul .

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