EXPERIMENT 1 (101): Force Table

Objective: To show the resolution of force vectors into X and Y components for the caseof static equilibrium.Apparatus: Force table, pulley clamps (3), weight holders,weight set, graph paper.Theory: When several forces act at one point and the net force is zero, then the point

is in equilibrium: its acceleration is zero. In this experiment the forces are due to a mass onthe end of a string. The strings pass over the pulleys and are connected to a point (knot)in the center of the table. If the net force due to all the forces is zero, the point will be atrest in the center of the table. If the point is pulled away and released, it will come again torest at the table center. The magnitude of each force is Fi =Mig where Fi has the units ofnewtons when Mi is in Kg, and g = 9:8 m=sec2:The angular position of each string �i can be measured by a scale along the rim of the

table so each force is recorded as a pair of numbers ( Fi, �i) .Each force can also be written in a rectangular (X; Y )coordinate system. TheX�component

is Fi cos �i and the Y�component is Fi sin �i . If there are n�forces acting at a point, thatpoint will be in equilibrium if:a)PFX =

Pi=ni=1 Fi cos �i = 0;

b)PFY =

Pi=ni=1 Fi sin �i = 0:

Writing forces as ( Fi, �i) is more useful when forces are to be entered on a graph.Procedure:1) The instructor will give each table a card on which is written the forces to use in their

experiment.2) First set the two pulley clamps to 90� angle . On the two weight hangers put the

masses M1 and M2 as given on your card.3) Experimentally �nd the third force which brings the central point to rest on the table.

Record all three angles and masses. Remember to include the mass of the weight hangeritself.Analysis:1) Calculate each force in Newtons by F1 = M1g; F2 = M2g; and F3 = M3g in which

g = 9:8 m=s2:

2) The forces�!F1;�!F2 and

�!F3 will be arranged as shown in Fig.2 .

�!F1 and

�!F2 are given while�!

F3 and �3 will be determined so that�!F1;�!F2 and

�!F3 will be in balance. From the diagram we

havejF1j = jF3j cos �3 (1)

jF2j = jF3j sin �3 (2)

From here we obtain upon division

tan �3 =

����F2F1���� (3)

and we �nd �3 theoretically. Compare this �3 that you found from the formula (3) with the�3 that you found experimentally. Also check from 2a (or 2b) that the magnitude of

�!F3 ,

i.e.����!F3��� is correct.

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3) The three forces will be added graphically on a graph paper as follows

(Fig. 1.1)

The result must satisfy �!F1 +

�!F2 +

�!F3 = 0 (4)

vectorially, which is equivalent to

F1X + F2X + F3X = 0 (5)

andF1Y + F2Y + F3Y = 0: (6)

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EXPERIMENT 3 (101): Static and Kinetic Friction Co-e¢ cients

Objective: To test several of the laws of friction by sliding objects on an inclined planeand to measure the static and kinetic friction coe¢ cients.

(Fig. 3.1)

Apparatus: Inclined plane with variable angle, sliding objects (carts) electronic timerto �nd speed, ruler.Theory: The laws of friction are empirical - that is they are based on experimental stud-

ies. What really happens when two surfaces that are in contact slide is not well understood.There are some general rules which seem to be true and will be tested in this experiment.These rules are only approximate in that they hold only over a limited range of applications.The rules:1) The forces of static and kinetic friction are proportional to the normal force.2) The force of friction does not depend on contact area.3) The force of friction does not depend on the relative velocity of the two surfaces.An incline will be used which can be set to any angle � . On this incline are put heavy

carts with bottom surfaces covered with di¤erent materials. The force due to gravity on thecart is mg where m is the mass of the cart. This force is directed vertically down as shownin �g 1. This force can be resolved into its components along the incline and normal to theincline. Rule 1 says there will be a frictional force proportional to the normal force. Thedirection of this force is against the direction of motion so it is directed up the incline. Theproportionality constant � between the normal force and the force of friction is known asthe coe¢ cient of friction. In mechanics there are two coe¢ cient of friction: one for the casewhen the object is moving (�K) and one for when the object is at rest (�S).These are known as the kinetic and the static coe¢ cients of friction. Suppose a cart is

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placed on the incline as in �g.1

(1)

and the incline angle � is slowly increased from � = 0 . At some critical angle the cart willstart to slide. At that angle , the force of static friction must be just equal to the force downthe incline mg sin �

mg sin � = �Smg cos � (2)

which implies�S = tan � (3)

For di¤erent carts you will read di¤erent angles and di¤erent �S: To �nd the coe¢ cient ofkinetic friction �K our procedure will be as follows ( we shall do only for one of the carts):Set each cart into motion with zero initial speed from di¤erent heights measured by �y. Theformulas

v2 = 2a (�y) (4)

a = g (sin � � �k cos �)

can be used for our purpose. The speed can be found from Eq. (3). If we plot v2 versus �y,the slope will be 2a = 2g (sin � � �k cos �) . The angle � can also be found easily, and fromthe slope the only unknown will be �K , which can be determined.

(Fig. 3.2)

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Procedure: Put each cart ( C1, C2 and C3) on the incline and raise the angle until eachone starts to slide. Record each angle as (�1; �2 and �3) and make the following table.

Cart No Critical angle � �s = tan �C1 = �1 = tan �1 =C2 = �2 = tan �2 =C3 = �3 = tan �3 =

(5)

In this way we shall obtain the static friction coe¢ cient for each of the carts.To �nd the kinetic friction coe¢ cient �K we shall make use of Eq. (3) , to plot the graph

of v2 versus �y and to �nd the slope of the resulting straight line. Note that �K for onlyone of the carts is enough.A �ag of length l = 10 cm is �xed on the cart. As this �ag passes through the electronic

timer we shall read the time as t , so that the speed of the cart will be

v =l

t(6)

By changing �y each time we shall record a di¤erent velocity and as a result we shall plotthe graph of v2 versus �y , ( take at least �ve di¤erent data). Note that �y can easily befound by using a ruler. The curve will that you plot must be straight line.

From the slope of the line we shall �nd �K , since � is known and g = 9:8 m=s2 .

Question: If you add extra masses to the carts will the result for �S and �K change?Does friction force depend on the area in contact? The race cars have large tires, what

must be the reason?

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EXPERIMENT 4 (101): Finding the value of "g " byusing an Air Track

Objective: To �nd the acceleration due to gravity "g " by studying the acceleration of anobject on an air track.Apparatus: Air track, glider, �ags, photo sensor, timer, riser, meter stick.

Theory: An air track can be used to study motion with nearly zero friction. Metalgliders move along the air track on a layer of air. In this experiment the velocity of the glideris measured by passing it through a photocell gate which is connected to an electronic clock.On the top of the glider is mounted a �ag of length L. As this �ag �rst breaks the light beamin the gate, the clock begins to measure time. When the �ag passes the photosensor and thelight beam is again unbroken, the clock stops so the length of time that the �ag broke thelight beam is accurately measured. In this experiment the air track is �rst carefully leveled.Then one end- the end with a single support- is lifted by a riser of height H. The air trackwill be inclined at a small angle �.

sin � =H

L0(1)

where L0 = the distance between the air track suspension points. The glider -with its �agmounted- is released at the high end of the air track and accelerates down to the otherend. As it passes through the gate the �ag breaks the light beam for a time t. So for eachmeasurement we record t, L, L0, H and D. D is the distance between the front edge of the�ag and the photosensor. This is the distance the glider travels before starting the clock.What can be calculated from this data? Let us see!! For motion in one dimension withconstant acceleration we have the equations:

x =1

2at2 + v0t+ x0 (2)

v = at+ v0 (3)

v2 � v20 = 2a (x� x0) (4)

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Q1: What is the velocity of the glider when its �ag �rst breaks the light beam?A1: In this case x = D; a = g sin �; x0 = v0 = 0: When these are put into Eq. (4) we

�nd

v2 = 2gD sin � (5)

or

v2 =2gD

L0H (6)

The glider then moves through the gate in a time t.Q2: What is the velocity of the glider as it passes the gate?A2: The clock reads t, the length of the �ag is L so v = L

t. By Eq. (3) if v2 is graphed

againstH, a straight line should result with a slope of 2gDL0. SinceDand L0are known constant

s, by setting the experimentally found slope equal to 2gDL0we can �nd the acceleration due to

gravity g. It is possible that the straight line drawn through the experimental points doesnot go through the exact origin on the graph. That means to get v2 zero seems to requiresome small non-zero value of H. Therefore in this case the track was not initially level.Procedure1) Carefully level the air track. Make this adjustment with the glider on the track with

the air ON. Adjust the support screw until the glider sits at rest in the center of the track.2) The wires from the photoelectric gate plug into the back of the timer in the START

position.3) Set the timer decimal point to 0.001 second. When the �ag on the glider passes

through the light beam the timer should operate.4) Set the gate at the lower end of the track.5) Fix the �ag on the glider so that it will not slide out of position when the glider is

stopped after the gate. Use the 10cm �ag. Hold the glider in its position . Record theposition of the edge of the glider, using the scale on the air track. Move the glider to theposition where the counter just begins to count. Record the position of the same edge ofthe glider, again using the scale �xed on the air track. The di¤erence of these two readingsshould be the distance D.6) Measure the height H of each riser.7) measure the spacing of the track supports L0:8) Put one of the riser under the single air track support screw. Zero the timer. Release

the glider from rest and record the timer t through the gate. Repeat this four (4) times.9) Repeat this procedure for a total of six di¤erent values of riser height H.Analysis:1) In each case, from the average time t calculate the velocity of the glider by v = L

t.

2) Make a graph of v2 (vertical axis) against H (horizontal axis) . Draw the best straightline through your data points.3) Find the slope of this straight line, including units.4) By theory this slope must be equal to 2gD

L0. Set this equal to your experimental slope

and �nd the acceleration due to gravity g.5) Does the straight line go through the origin of the graph? If not, decide what this

means.

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EXPERIMENT 5 (101): Conservation of Linear Mo-mentum and Kinetic Energy

Objective: In a collision process on an air track to check that linear momentum and totalkinetic energy are conserved.Apparatus: Air track, gliders with �ags, electronic timers monted to gates.( photo

sensors)

(Fig. 4.1)

Theory : We are already familiar with an air track from Experiment 4. On an airtrack, in principle, the friction is reduced to zero. For this reason it becomes very suitable tomake collision experiments on an air track. For this purpose we use two gliders with variablemasses on the air track and two photo sensors to record electronically the time intervals.Each glider has a �ag on it with a length of l = 10 cm. (see �g.1). Prior to the collisionwe denote the speeds of objects ( the gliders) by v1 (with mass M1) and v2 (with massM2). After the collision these become v01 and v

02, respectively. Accordingly , the conservation

principle of linear momentum is expressed by

M1v1 +M2v2 =M1v01 +M2v

02: (1)

Since the air track is a one-dimensional object the process takes place in one-dimension only,say the X-axis. Depending on the directions of velocities , the velocities can at most be � insign. Similarly , the conservation principle for the total kinetic energy in an Elastic Collisioncan be expressed by

1

2M1v

21 +

1

2M2v

22 =

1

2M1v

021 +

1

2M2v

022 (2)

Let us remind that each kinetic energy term is automatically positive. In order to make surethat no potential energy due to gravity is involved the air track must be perfectly horizontal.This must be checked in the beginning before the experiment is started. The masses of eachglider will be provided by the Assistant in the Laboratory. What we aim in this experimentessentially is to check the validity of eq.(1) and eq.(2).The most important part of the experiment is to �nd the velocities correctly. For this

purpose we use photo sensors that electronically record time passage for each glider . If thetime is recorded by t1and the length of the �ag on the glider is l1 then from the relation

v1 =l1t1

(3)

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we can �nd the speed v1 and similarly other speeds. In order to make the experimentsimpler, so that we can conduct better measurements we assume throughout the experimentthat v2 = 0 , that is one of the objects is at rest initially. This casts our equations of linearmomentum and kinetic energy conservations into the forms

M1v1 =M1v01 +M2v

02 (4)

1

2M1v

21 =

1

2M1v

021 +

1

2M2v

022 (5)

so that we have to �nd only three velocities instead of four.Procedure: First you have to check that the air track is perfectly horizontal with zero

inclination angle. Then , provide the masses to be located on each glider, the Assistant willhelp you in this regard. Make the arrangement as in �g.(1). One of the gliders (M2) inat rest is between the two photo sensors while the other one (M1) lies outside the photosensors. Before we move M1 with initial speed v1 we turn on the pump of the air track sothat friction becomes zero. As M1 passes through one of the photo sensors its time will berecorded. The two gliders will collide and they will pass through di¤erent photo sensors. Inthis way their times will be recorded and velocities found from eq.(3). We shall be knowingnow, v1; v01; and v

02 to be substituted into eq.(s) (4) and (5). If eq(4) is satis�ed this will

mean that in the process of collision linear momentum is conserved. Similarly, eq(5) will tellus that the total kinetic energy is conserved in the process. In such an experiment we mustadmit that a few percent error between the left and right hand sides of eq.(s) (4) and (5)are tolerable. The students must be able to explain about the possible numerical di¤erencesin those equations. For example, is the collision exactly elastic? Is there some energy losstaking place in the collision process? How?Caution: If the di¤erence between the left and right hand sides of eq.(s) (4) and (5) are

big that means the data taken is not recorded correctly. So, please repeat the experimentand reduce your errors as much as possible. For example, ifM1 comes to a step after collisionyou should increase its initial speed so that it will gain enough speed to pass through thetime sensors after the collision.Questions: What are the di¤erences between, Elastic, Inelastic and completely Inelastic

Collision?Can you give an example for each di¤erent case?

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EXPERIMENT B (101): Finding the value of � exper-imentally

It is known from mathematics that � is the ratio of circumference (C) to diameter (D) in acircle. By simple measurement in the Physics laboratory we shall determine the value of �.The procedure goes as follows:1) By using a meter stick measure the circumference C of �ve di¤erent round objects.

Record these in a table as shown below.2) Measure also the diameter D of the same �ve objects and add these to your table.3) From the relation � = C

Dcalculate �ve values of � from your data. ( Note that you

need a calculator for these calculations). Show these also in your table. In this calculationbe careful your result has the correct number of signi�cant �gures.

C D C=DC1 = D1 = �1 = C1=D1 =C2 = D2 = �2 = C2=D2 =C3 = D3 = �3 = C3=D3 =C4 = D4 = �4 = C4=D4 =C5 = D5 = �5 = C5=D5 =

(Table 1)

4) From the �ve results for � �nd the average value and compare it with the true valueof � (= 3:14).From these results �nd the average value for � ; which is

�av =�1 + �2 + �3 + �4 + �5

5: (1)

Using the value of � which is known to be 3:14 from the formula

3:14� �av3:14

� 100 =? (2)

�nd the percentage error in your experiment for the value of �:

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