what does force have to do with motion
CHAPTER v
- FORCE AND Move - 1
- 5.i. Newtons Kickoff Constabulary
- 5.2. Force
- 5.iii. Newtons Third Police
- v.4. Mass and Weight
- 5.four.1. Measuring Mass
- 5.4.2. Measuring Weight
- 5.five. Applications
5. FORCE AND Motility - one
When an object all of a sudden changes its velocity and/or direction, nosotros can always observe an interaction between that object and its surround that is responsible for this change. We state that the environs exert a force on the object studied. Under the influence of a force, an object will accelerate. The forcefulness laws allow u.s. to summate the force acting on a body from the properties of the torso and its environment. The laws of motion are subsequently used to calculate the dispatch of the object nether influence of the forcefulness.
In this course we will exist discussing the laws of motion obtained by Newton. This is called Newtonian mechanics. It should be realized that Newtonian mechanics does non always provide correct answers. If the speed of the objects involved is an observable fraction of the speed of light, we must supercede Newtonian mechanics past Einstein'due south special theory of relativity. For problems on the scale of diminutive structure we must replace Newtonian mechanics by quantum mechanics.
5.1. Newtons Beginning Law
All around us we observe that all moving objects will come eventually to remainder, unless we apply a strength to them. We need to keep pedaling if we desire to continue a bicycle going with constant speed, nosotros need to have our engine running if we desire to keep driving with a speed of 55 miles/hr. In all these cases, friction will ultimately stop any moving object, unless the friction force is canceled past the force supplied past our legs, our engine, etc. If we reduce friction, the moving object volition take longer to slow down, and the strength needed to overcome the friction force will be less. In the limit of no friction, our object will keep moving with a constant velocity, and no force need to be applied. This conclusion is summarized in Newton's showtime law:
" Consider a body on which no internet force acts. If the body is at rest, information technology will remain at residual. If the trunk is moving with constant velocity, information technology will continue to exercise so. "
Newton'southward start law is really a statement about reference frames in that information technology defines the kinds of reference frames in which the laws of Newtonian mechanics concord. Reference frames in which Newton's first police force applies are called inertial reference frames.
I way to test whether a reference frame is an inertial reference frame, is to put a test body at rest and arrange things and then that no net strength acts in information technology. If the reference frame is an inertial frame, the body volition remain at rest; if the body does not remain at remainder, the reference frame is not an inertial frame. If you put a bowling ball at rest on a merry-go-round, no identifiable forces act on the ball, only information technology does not remain at residue. Rotating reference frames are non inertial reference frames. Strictly speaking, the earth is therefore also not an inertial frame, however, but if nosotros consider large scale motion such equally wind and ocean current do we need to take into account the non inertial character of the rotating earth.
5.2. Strength
If we exert the aforementioned force on several objects with dissimilar mass, we volition observe dissimilar accelerations. For example, one tin can throw a baseball game significantly further (and faster) than a ball of similar size fabricated of lead. The unit of measurement of forcefulness is the Newton (Due north), and a force of 1 N is defined every bit the force that when applied to an object with a mass of 1 kg, produces an acceleration of ane m/southward2. If we apply a force equal to 2 N, the corresponding acceleration is 2 k/sii.
Experiments have shown that the strength is a vector. This can be shown by demonstrating that the force has a magnitude and a direction. Suppose, we apply a strength of 3 N to our standard object (mass 1 kg). The force is applied such that the resulting acceleration of 3 m/sii is upward (positive y-management). In add-on, we utilize a force of 4 North in the horizontal direction (this force is practical such that the standard object will accelerate with an acceleration of 4 m/s2 in the direction of the positive x-centrality if this is the just strength applied). The situation is illustrated in Effigy 5.i. If both forces are acting on the standard mass simultaneously, the dispatch of the object is measured to be 5 thousand/due south2, and the direction of the acceleration coincides with the direction of the vector sum of the ii forces. The total strength is 5 N and is equal to the magnitude of the sum vector of the two forces (if nosotros assume that the direction of the strength is equal to the direction of the acceleration). We conclude that indeed the force is a vector and that both the force and the corresponding acceleration accept the aforementioned direction.
Figure five.1. The Acceleration of the standard trunk under the influence of two forces.
The dispatch produced by a certain force depends on the mass of the object. The acceleration of an object with twice the mass of the standard mass under the influence of a certain force is half that of the acceleration of the standard mass due to the same force. The following list summarizes what we have learned and so far about forces:
1. Force is a vector.
2. A force acting on an object produces an acceleration. The direction of the acceleration is the aforementioned as the direction of the forcefulness applied.
three. For a given forcefulness, the resulting acceleration of a body with a mass twice that of the standard mass, is half that of the acceleration of the standard mass nether the influence of the same strength.
The conclusions are summarized in Newton's second law:'
where [Sigma]F is the vector sum of all forces acting on an object with mass thousand, and 
Newton'southward second law includes a formal statement of Newton's first law: if there is no net force interim on an object ([Sigma]F = 0 North) the acceleration is zero (and the velocity of the object is abiding).
Sample Trouble 5-ane
A student pushes a loaded sled whose mass is 240 kg for a distance of 2.3 m over the frictionless surface of a frozen lake. He exerts a horizontal force equal to 130 N. If the sled starts from rest, what is its final velocity ?
Effigy 5.2. Coordinate System Sample Trouble 5-ane.
This is a one-dimensional problem. The coordinate system is defined such that the origin coincides with the position of the sled at fourth dimension t = 0 s, and the force is applied in the positive direction (see Figure 5.ii). Since the force is abiding, the resulting dispatch a is besides abiding, and can be calculated by applying Newton's second law:
The constant acceleration is only practical over a distance d (= two.3 1000). In the coordinate system chosen, the equation of motion tin be written as follows:
From this equation, the time at which the sled has covered a distance d tin be calculated:
and the velocity of the sled at that fourth dimension is equal to
Sample Problem 5-two
In a two dimensional tug war, Alex, Betty and Charles pull on ropes that are tied to an automobile tire. The ropes make angles every bit shown in Figure v.three, which is a view from above. Alex pulls with a force FA = 220 N and Charles with a force FC = 170 N. With what force must Betty pull in order to go along the tire stationary ?
Since the tire is stationary, the net strength on the tire must be zip. This also means that the internet force forth the 10 and y management must exist nil:
Substituting the known values for FA, FC and [theta] in the first equation, we tin can calculate [phi]:
Substituting this value for [phi] into the second equation we can calculate FB:
Figure 5.three. Forcefulness Diagram Sample Trouble 5-ii.
v.3. Newtons Tertiary Law
If a hammer exerts a force on a blast, the nail exerts an equal but oppositely directed force on the hammer. This is true in general, and is described by Newton's 3rd law:'
" Suppose a body A exerts a force (FBA) on torso B. Experiments testify that in that example trunk B exerts a forcefulness (FAB) on body A. These two forces are equal in magnitude and oppositely directed:
"
Note: Since the two members of an action-reaction pair e'er act on different bodies, they can not cancel each other.
5.4. Mass and Weight
The mass of a body and the weight of a trunk are totally different properties. The mass m of a trunk is a scalar; its SI unit the kilogram. The mass of a body tin can be adamant by comparison information technology to the standard kilogram. Mass is an intrinsic belongings of a body; it is the same on the earth's surface, in an orbiting satellite, on Mars, or in interstellar space. The weight 
where g is the free-fall acceleration at the location of the body. Since the complimentary-fall acceleration varies from point to betoken, the weight of an object depends on its location, and is therefore not an intrinsic property of a body.
5.4.ane. Measuring Mass
The mass of a trunk can be determined via comparison with the standard mass. The equal-arm rest is designed for this purpose (run across Effigy five.4). The equal-arm residual is balanced if the force on the left equals the force on the right:
Effigy v.4. Equal-Arm Balance.
These 2 forces are the gravitational forces acting on m1 and mii and can be calculated easily:
If the costless-fall dispatch is constant at the location of the balance, we tin conclude that if the artillery are balanced:
yard1 = one thousandtwo
Therefore, the equal-arm residue determines the relative mass of two objects by comparison their weight.
5.4.2. Measuring Weight
The measurement of the weight of an object can exist carried out using a spring scale (see Figure 5.5). The spring scale uses a jump to measure the weight of the object. There is a one-to-i relation between the stretch of the bound and the applied strength (responsible for the stretch). In general, bound scales are calibrated and show the mass of the object. However, information technology should exist stressed that the mass of the object is obtained from the measured weight, and in this process information technology is assumed that the gratis-fall acceleration equals nine.8 m/s2. Therefore, a spring scale will merely point the correct mass if it is used at a location at which the free-autumn acceleration is equal to that at the calibration site (note: a spring calibration will incorrectly make up one's mind the mass of an object if it is used on the moon, or in an accelerating elevator).
Effigy 5.5. Spring Calibration.
v.5. Applications
Sample Problem 5-vii
Effigy five.6 shows a block of mass m = fifteen kg hanging from iii cords. What are the tensions in these cords ?
The mass m experiences a gravitational force equal to mg. Since the mass is at balance, cord C must provide an opposing forcefulness equal to mg. Applying Newton'south third law, we conclude that cord C exerts a force on the knot whose magnitude is equal to mg (and pointed in the direction shown in Effigy five.vi). Since the system is at residuum, the cyberspace force on the knot must be equal to zero:
This vector equation can be rewritten in terms of its components forth the 10-axis and y-axis, using the following information:
Figure v.6. Sample Problem five-seven.
Using these expressions we can write downwards the equations for the x and y-components of the net forcefulness:
The offset expression can be used to express TA in terms of TB:
Substituting this expression into the equation for [Sigma]Fy we obtain:
from which we tin calculate TB:
Knowing TB, nosotros can now calculate TA:
In the case of sample problem 5-half dozen, the tensions in the cords are:
TA = 100 N
TB = 140 N
TC = 150 N
Problem
Figure 5.7 shows a block with mass m on a frictionless airplane, tilted by an angle [theta]. What is the acceleration of the block ?
Figure 5.7. Mass m on an inclined plane.
In gild to determine the acceleration of the cake we have to determine the full strength interim on the block along the inclined aeroplane. Two forces human activity on the block: the gravitational forcefulness exerted past the globe on the block, and a forcefulness, called the normal force exerted by the plane on the block (see Effigy 5.8). This force must exist nowadays since in its absence mass m will experience free fall (instead of sliding motility). Since the normal force is normal to the inclined plane it does not have a component along it. The component of the gravitational force forth the inclined plane is given by
The acceleration produced by this force tin can be determined from Newton'south second law
Figure 5.viii. Forces interim on mass 1000.
Sample Problem 5-8
Effigy 5.9 shows a block with mass m held by a cord on a frictionless plane, tilted by an angle [theta]. What is the tension in the cord ? What force does the plane exert on the block ?
Figure five.nine. Sample Trouble 5-8.
This problem tin can be solved easily if the coordinate system is called carefully. The best choice of coordinate organisation is shown in Effigy 5.10. Since the block is at residual, the net force on it must exist zip:
Due to the choice of the coordinate arrangement, both North and T only take components forth the y-axis and x-axis, respectively:
The mass will stay at balance if all components of the net force are nothing:
From these equations nosotros can obtain N and T:
Figure 5.10. Coordinate System # 1 used in Sample Problem v-8.
Figure 5.eleven. Coordinate System # two used in Sample Trouble v-viii.
The standard pick of coordinate system with the x-axis coinciding with the horizontal direction and the y-centrality coinciding with the vertical management (encounter Effigy 5.eleven) would accept made the problem significantly more difficult. In this coordinate system, N and T take component along both the x and y direction:
In this case, N and T tin can be obtained by solving the post-obit equations:
Of course, the solutions for N and T are identical to those derived previously, but the derivation is harder.
Sample Problem 5-10
Two blocks are connected by a string that passes over a (weightless) caster (see Figure v.12). Observe the tension in the cord and the (common) dispatch.
The blocks are moving with a constant acceleration. Since the cord is assumed to rigid, the acceleration of mass 1000 has to be equal to the acceleration of mass M. Notwithstanding, since the pulley reverses the direction of motion, the direction of the acceleration of mass m is opposed to the direction of the acceleration of mass M. For each of the masses nosotros can write downward the following force equations:
The first equation tin exist used to limited T in terms of a:
Substituting this expression for T into the 2nd equation, we obtain:
Figure five.12. Setup Sample Problem five-10.
The acceleration a can now be calculated:
Note that a is positive when M > m, and a is negative when Grand < m. The acceleration is zero if m = M. This of grade agrees with what our expectations. The tension in the cord can at present be calculated:
Problem
A block of mass thou1 on a shine inclined aeroplane of angle [theta] is connected by a cord over a modest frictionless pulley to a 2d block of mass thousandii hanging vertically (meet Effigy v.13). The mass of the string and the pulley can be neglected.
a) What is the acceleration of each cake ?
b) What is the tension in the string ?
Effigy 5.13. Inclined plane and caster.
In order to make up one's mind the acceleration and the tension, nosotros have to identify all forces acting on both masses. The following forces human action on thousandone (run into Figure 5.xiv):
* The gravitational strength W1 = m1 m. This force is pointing downwards in the vertical direction.
* The normal force N. This force is exerted by the surface of the inclined airplane on the mass and is pointing in a management perpendicular to the inclined information technology.
* Tension T. The cord exerts this strength on the mass. Its direction is parallel to the inclined plane.
In general, the net force acting on yardi volition be non-nil and m1 volition have a not-zero acceleration. The acceleration will be along the 10-centrality (run across Figure 5.14) and is defined to be positive if the acceleration is in the aforementioned management as the tension T. The components of the net force interim on m1 are given by
Effigy v.14. Forces interim on m1.
Figure 5.15. Forces acting on m2.
The following forces human activity on m2 (see Figure v.fifteen):
* The gravitational force Westwardii = mtwo g. This force is pointing downwards forth the vertical.
* The tension T. The cord exerts this force on the mass. This force is pointing upwardly along the vertical. The tension in the string is the aforementioned at each point, and the magnitude of this force is therefore equal to the one interim on 10001 although it points in a different direction.
The net force on m2 will exist non-nothing and the mass will accelerate. Since thousand1 and m2 are connected via a cord, they will accept the aforementioned acceleration. If the direction of the acceleration of m1 is along T, the direction of the acceleration on mii will be forth Westwardtwo (encounter Figure 5.15). None of the forces acting on mtwo has a component along the 10-axis and we will therefore just consider the net forcefulness forth the y-axis:
Equations (1) and (iii) are two equations with 2 unknown (T and a), and can be solved. Equation (three) tin can be rewritten as
Substituting equation (4) for T in equation (1), we tin determine a:
Substituting equation (5) into equation (4) nosotros obtain the tension T:
Ship comments, questions and/or suggestions via electronic mail to wolfs@pas.rochester.edu and/or visit the dwelling page of Frank Wolfs.
Source: http://teacher.pas.rochester.edu/phy121/lecturenotes/Chapter05/Chapter5.html
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