Applikasi Newton

APPLICATION OF  NEWTON’S SECOND LAW

TABLE 5-2

VARIATIONS OF g WITH LATITUDE AND ELEVATION

Station	North latitude	Elevation, m	g, cm / sec2	g, ft / sec2
Canal Zone	9o	0	978.243	32.0944
Jamaica	18o	0	978.591	32.1059
Bermuda	32o	0	979.806	32.1548
Denver	40o	1638	979.609	32.1393
Cambridge	42o	0	980.398	32.1652
Standard station			980.665	32.1740
Greenland	70o	0	982.534	32.2353

or  less than the average  density  of  the earth.  The greatest variation,  however,  arises from the rotation of the  earth, and will be discussed in Chapter 6.

                From a survey   of variations in the   value of g, conclusions can be drawn   as to the presence of   deposits of ore or oil beneath the earth’s surface.  Hence the precise measurement of g is one of the methods   of   geophysical   prospecting. Some representative values of g are given   in Table 5-2.

                5-8   Applications   of Newton’s  second law.  We now give a number of  examples of the application of Newton’s second law to specific problems. In all these examples, and in the problems at the end of the of the chapter, it wil be assumed that the acceleration of  gravity is 32.0 ft/sec² or 9.8m/sec², unless otherwise specified.

                EXAMPLE 1.  What is the resultant   force  on a body weighing 48 lb when its acceleration  is 6 ft/sec² ?

                We  must first find the mass of the  body. From the relation m =w/g,

                    .

                Hence

                  

               

EXAMPLE 2. What is the resultant force on a body of mass 48 kgm when its acceleration is 6 m/sec²  ?

                The mass of the body  is given, and hence

                 

                EXAMJPLE 3. A block whose mass is 10 Kgm rests  on a horizontal  surface. What constant horizontal force T is  required to give it a velocity  of 4 m / sec in 2 sec ,starting from rest, if the friction force between the block and the surface is constant and is equal to 5 Newton’s ?  Assume that  all forces act at the center of the block. ( See Fig. 5-4.)

                 The mass  of the block is given. Its y-  acceleration is zero. Its x- acceleration can be found from the data on the velocity acquired in a given time. Since the forces are constant,  the x- acceleration is constant  and from the equations  of motion with  constant acceleration,

                                .

            Hence from  Newton’s second law in  component  form,

                                                 

           

                       

                                               

            EXAMPLE  4. An elevator and its load weigh a total of 1600 lb. Find the tension  T in the supporting cable when  the elevator, originally moving  downward  at  20 ft/sec, is brought to rest  with constant acceleration in a distance of 50 ft.  ( see Fig. 5-5.)

                The mass of the elevator  is

                                               

            From the equations of motion with constant acceleration,

                                       

                The initial velocity v0 is – 20 ft/sec ; the velocity v  is zero. If we take the origin at  the point  where  the deceleration  begins, then y  = 

                Hence

                                 .

                Fig. 5-5. The resultant force has                                 Fig.5-6. The resultant force has the magnitude T – w .                                                     the magnitude P –w .

                The acceleration is therefore positive ( upward ).

                From the free-body diagram ( Fig. 5-5 ) the resultant force is

                                 

                Hence

                                               

                                   

                                                T  = 1800 lb.

                EXAMPLE 5. With what force will the feet of a passenger press  downward on the elevator floor when the elevator  has the acceleration above, if the passenger weighs 160 lb ?

                This example illustrates a problem that is frequently encountered, in which it is necessary to find a desired force by first computing the force that  is the reaction to the one desired, and then using Newton’s third law. That is, we first calculate the force with which the elevator floor pushes upward on the passenger ; the force desired is the reaction to this.

Figure  5-6 shows the force acting on the passenger. The resultant force is P—𝛚, and since the mass of the passenger  is 5 slugs and his acceleration is the same as that of the elevator,

∑F = ma

P – 160 lb = s slugs x 4   = 20 lb

P= 180 lb.

The passenger exerts an equal  and opposite  force downward on the elevator floor.

Example 6

What is the acceleration of a block on a frictionless plane inclined at an angle ɵ with the horizontal ?

The only forces acting on the block are its weight w  and the normal force N exerted by the plane (fig. 5-7). Since neither the weight nor the mass of the block is given, a letter must be  used to represent  one or the other. Let us call the mass  m; the weight is then mg. take exes parallel and perpendicular to the surface of  the plane and resolve the weight into x- and y- components. Then

∑F_y  =N- mg cos q

∑F_z  =N- mg sin q

But we know that a_y = 0, so from the equation ∑F_y  = ma_y we find that N= mg cos q

. From the equation ∑F_x = ma_x , we have

mg sin q  = ma_x,

a_x  = g sin q

The mass does not appear in the final result, which mean that any block, regardless of its mass, will slide on a frictionless inclined plane with an acceleration down the plane of g sin q . (note that the velocity is not necessarily down the plane).

Example 7

Refer to fig 2-3 (chapter 2). Let the mass of the block be 4 kgm and that of the rope be 0.5 kgm. If the force F₁ is 9 Newton’s what are the forces  F₁^´ , F₂ dan F₂^´ ? the surface on which the block moves is level and frictionless.

We know from Newton’s third law that F₁= F₁^´ and that F₂= F₂^´.

Hence F₂^´= 9 Newton’s.  The force F₂ could be computed by applying Newton’s second law to the block, if its acceleration were known, or the force F₂^´could be computed. By applying this law to the rope if its acceleration were known.  The acceleration in not given, but it can be found by considering the block and rope together as a single system.  The vertical forces on this system need not be considered. Since there is not friction, the resultant external force acting on  the system is the force F₁. (The forces F₂ and F₂^´.are  internal  forces when  we consider block and rope as a single  system, and the force  F₁^´  does not act on the system, but on the man) then, from Newton’s second law,

∑F =ma

9 n=  ( 4 kgm  + 0.5 kgm ) x a

 a = 2 .

We can  now apply Newton’s second law to the block.

∑F =ma

F₂ = 4 kgm x 2 .= 8 newtons.

Considering the rope alone, the resultant force on it is..

∑F =  F₁- F₂^´  = 9 n - F₂^´.

And from the second law

9 n - F₂^´ = 0.5 kgm x 2  = 1 n

F₂^´ = 8 Newton

In agreement with Newton’s Third law, which was  tacitly used when the forces F₂ and F₂^´ were omitted in considering the system as a whole, we find  that F₂  and F₂^´ are equal in magnitude. Notice, however, that the forces F₁ and F₂ are not equal and opposite (the rope is not in equilibrium) and that these forces are not an action- reaction pair.

Example 8

In fig.5-8, a block  of weight w₁ = 16 lb moves on a level  frictionless surface, connected by a light flexible cord passing over a small frictionless pulley to a second hanging block of weight  w₂ =  8 lb. what is the acceleration  of the system, and what is the tension in the cord connecting the two blocks ?

The diagram shows the forces acting on each block. The forces exerted on the blocks by the cord can be considered an action-reaction pair, so we have used the block on the surface,

∑F_x  =T  = 0.5 slugs x  a

∑F_y  =N-w₁ =  N – 16 lb  =0

For the hanging block ,

∑F_y  = w₁- T = 8 lb  - T

                      =0.25 slugs X a

Since a is the same for both blocks, on solving the first  and third equations simultaneously, we obtain

a = 10.7  ,      T = 5.3 lb

Note carefully that although the earth pulls on the hanging block with a  force of 8 lb, the force exerted on the 16 lb block is only 5.3 lb. it is  the connecting cord which pulls on  the 16 lb  block, and the tension in this cord must  be less  than  8 lb or the hanging block would  not accelerate  downward.

Example 9

Up to this point, applications of Newton’s  second law  have been limited to cases where  the resultant  force acting on a body was constant, thereby imparting to the body a constant acceleration. Such cases are very important, but make only  light demands  on one’s mathematical knowledge.  When the resultant  force is variable, however, the acceleration in not apply. We conclude this section  with three examples of motion under the action of a variable force.

An automobile  of mass 50 slugs is traveling at 30 ft/sec. the driver applies  the brakes in such a  way that velocity  decreases  to zero according  to the relation.

v = v₀ –kt²

where v₀  = 30 ft/ sec , k = 0.30 ft/sec³, and t is the time  in seconds after the  brakes are applied. Find the resultant force decelerating the automobile, 5 sec after the brakes are applied

the mass of automobile is given. To find the resultant force on it from Newton’s second law we must first compute its acceleration. This is given by

a =   = (v₀ – kt²)  = -2 kt

hence when  t = 5 sec,

a = -2  X 0.30    X  5 sec = -3

therefore at this instant

∑F  = ma  =  50 slugs X (-3 ) = -150 lb.

(what is the significance of  the negative sign)

Example 10

An automobile of mass 50 slugs accelerates from rest. During the first 10 sec, the resultant force acting on it is given by  ∑F   = F₀ – kt, where F₀ =200 lb, k = 10 lb/sec, and t  is the time in seconds after the start. Find the velocity at the end of 10 sec, and the distance covered in this time.

The mass of the automobile and the resultant force acting on it are both given. The acceleration can be found from Newton’s second law and the  principles of kinematics can then be used to find the velocity and the position. From the second law,

a =   = .

therefore

  dv = .

 ∫  dv = .

 

v  =                       (5-7)

Where  C₁ is a constant of integration.  Since v = 0 when t  =  0,   C₁ is zero also.

Hence when t  = 10 sec.

 v  =  

  =  

To find the position. Return to  Eq (5-7)

v =

∫  dx =

x =

if the origin is at the starting point, x = 0 when   t  = 0 and  hence C₂  =0. When t  = 10 sec,

x =

= 200 ft – 33 ft  = 167 ft

Example 11

Discuss the motion of a freely falling body (or a body projected vertically upward), taking into account the variation of the gravitational  force on the body with its distance from the earth’s center. Neglect  air resistance.

The gravitational force on the body at a distance r  from the earth’s  center is Gmm_E / r²  , and from Newton’s second law its acceleration is 

Where the positive  direction is upward (or better, radially outward).

It was shown in Eq (4-8) that the acceleration can be expressed as

Then

Where v₁ and v₂  are the velocities at the radial distances r₁ and r₂. It follows that

v₂²  - v₁²  = 2 Gm_E (1/r²  1/r²)                  (5-8)

                                                                                                              

as an illustration, let us find the initial velocity v₁ required to project a body vertically upward so that  it rises to a height above the earth’s  surface equal to the  earth’s radius R .then  v₂  = 0, r₁  = R,   r₂= 2 R, and

v₁²  =  Gm_E/ R                           (5-9)

let g₀  represent the acceleration of gravity  at the  earth’s  surface, where  r = R. then from Eq. (5-6),

Gm_E   = g₀ R²

and  Eq. (5-9) can be written

v₁²  =  g₀ R.                         (5-10)

how does this compare with the velocity that would be required if the acceleration had the constant value g₀ ?                          

5-9 The equal-arm analytical balance. As with many other physical quantities, the mass of a body can be measured in several different ways. One is to use the relation by which the quantity is defined, which in this case is the ratio of the force on the body to its acceleration. A measured force is applied to the body, its acceleration is measured, and the unknown mass is obtained by dividing the force by the acceleration. This method is used exclusively to measure masses of anatomic particles.

The second method consists of finding by trial some other body whose mass (a) is equal to that of the given body, and (b) is already known. Consider first a method of determining when two masses are equal. It will be recalled that at the same point on the earth’s surface all bodies fall freely with the same acceleration g. since the weight 𝛚 of a body equals the product of its mass m and the acceleration g, it follows that if, at the same point, the weights of two bodies are equal, their masses are equal also. The equal-arm balance is an instrument by means of which one can determine very precisely when the weight of two bodies are equal, and hence when their masses are equal.

The essential feature of the balance, shown schematically in Fig. 5-9, is a light, rigid beam on which are firmly mounted tree equally spaced agate knife-edges, parallel to one another and perpendicular to the length of the beam. The central knife-edge at O rests on a polished plane agate plate supported from the floor of the balance case. The scale pans are hung from two similar plates at A and B. A pointer fastened to the beam swings in front of the scale S. the knife-edges and plates act as practically

Frictionless pivots, since the scale pans can swing freely about their supporting knife-edges, the center of gravity of the pans and of any bodies placed on them will always be directly below the knife-edges. The weight of the beam is w, and its center of gravity is directly below the central knife-edge when the beam is horizontal. The weights of the scale pans are equal

In using the balance, the body whose mass m₁ is desired is placed in the left scale pan, and a combination of known masses m₂ is placed in the right pan. If the combined weights of  bodies and scale pans, w₁ and w₂, are exactly equal, the beam remains in stable equilibrium in a horizontal position as in Fig 5-8, under the action of the four parallel forces w₁, w₂,w, and the upward force P exerted at the central knife-edge. Then the weight of the unknown body equals that of the bodies in the right scale pan, and hence their masses m₁ and m₂ are equal also.

There remains the question as to how a set of bodies of known mass (usually referred to as a set “weights”) is obtained in the first place. This is also done with the equal-arm balance. Let us start with a standard kilogram and make two bodies (a) whose masses are equal, as determined by the balance, and (b) which when combined will just balance the standard kilogram. Each of these then has a mass of one-half kilogram, or 500 grams. Other submultiples, or multiples, of the standard can be prepared by a similar procedure.