Force Lab- Physics

VPL Lab ah-Force Table 1 Rev 7/17/14

Name School ____________________________________ Date

The Force Table – Vector Addition and Resolution

“Vectors? I don’t have any vectors, I’m just a kid.” – From Flight of the Navigator

PURPOSE

 To observe how forces acting on an object add to produces a net, resultant force

 To see the relationship between the equilibrant force relates to the resultant force

 To develop skills with graphical addition of vectors and vector components

 To develop skills with analytical addition of vectors and vector components

EXPLORE THE FORCE TABLE APPARATUS; THEORY

See Video Overview at: http://virtuallabs.ket.org/physics/apparatus/04_1stequil/

 

Figure 1 – The Force Table Apparatus

Important Note: For each direction answer enter an angle between 0° and 360° with respect to the force table.

Working with vectors graphically is somewhat imprecise. You should expect to see lower precision in this lab. But working

with vectors graphically will give you a better understanding of vector addition and vector components. You can improve

your precision by using the Zoom features of the apparatus.

 

VPL Lab ah-Force Table 2 Rev 7/17/14

PROCEDURE

I. Addition of Two Dimensional Vectors

In this part you’ll add two forces to find their resultant using experimental addition, graphical addition, and analytical

addition. The two forces you’ll work with are:

F2 = .300 g N at θ2 = 320°

F4 = .180 g N at θ4 = 60°

There are two things that we should clarify about these two statements. We’ll refer to Figure 1 in the following.

 The force subscripts on forces and angles (F2, θ2) refer to the hanger numbers. For example, in Figure 1,

θ1 = 20°

 The notation for forces in this lab is a bit unusual. From the “Masses on Hangers” table we know that hanger #1 has

a 50 g, and a 200 g mass on it. The hanger’s mass is 50 g, so the total, 300 g, is shown in the box beside the hanger.

Be we want to record forces in newtons. From W = mg, we can easily calculate the weight of a 300-g mass.

W = .300 kg × 9.8 N/kg = 2.94 N

But there’s a much more convenient way of handling this. We can abbreviate this calculation to the simpler

W = .300 g N

A. Experimental Addition; The Equilibrant

By experimental addition we mean that we will actually apply the two forces to an object and measure their total, resultant

effect. We’ll do that by first finding the equilibrant, E, the force that exactly balances them. When this equilibrium is

achieved the ring will be centered. The resultant, R, is the single force that exactly balances the equilibrant. That is, either F2

and F4, or their resultant R can be used to balance the equilibrant.

Note: When adjusting the angle of a pulley to a prescribed angle, pay no attention to the string. Drag the pulley until the

color-coded pointer is at the desired setting. Another way that works well is to disable all the hangers while adjusting the

angles. When you do this, the ring is automatically centered so the strings can be used to help adjust the angle.

1. Enable pulleys 2 and 4 and disable pulleys 1 and 3 using their check boxes. A disabled pulley is greyed out. You can

drag it anywhere you like since any masses on it are inactive.

2. Drag pulleys 2 and 4 to their assigned angles θ2 and θ4. Use the purple and blue pointers of the pulley systems, not the

strings, to set the angles. You should zoom in to set the angles more precisely.

3. Add masses to hangers 2 and 4 to produce forces F2 and F4.

4. Enable pulley 3.

5. Using pulley 3, experimentally determine the equilibrant, E to just balance (cancel) F2 and F4. Do this by

adjusting the mass on the hanger and the angle until the yellow ring is approximately centered on the central

pin. You can add and remove masses to home in on it. You add masses by dragging and dropping them on a

hanger. You remove them using the Mass Total/Mass Removal tool. This tool displays the total mass on the

hanger, including the hanger’s mass. Clicking on the tool removes the top mass on the hanger.

 

MT/MR

Tool

Experimental Equilibrant, E: g N, °

6. From your value of E, what should be the resultant of F2 and F4? (Same force, opposite direction.)

Predicted Resultant, R: g N, °

7. Disable pulleys 2 and 4 and enable pulley 1. Experimentally determine the resultant by adjusting the mass on pulley 1

and its angle until it balances the equilibrant. That is, the ring should not move substantially when you change between

enabling F2 and F4, and enabling just F1.

Experimental Resultant, R: g N, °

 

 

VPL Lab ah-Force Table 3 Rev 7/17/14

B. Graphical Addition

With graphical addition you will create a scaled vector arrow to represent each force. We add vector arrows by connecting

them together tail to head in any order. Their sum is found by drawing a vector from the tail of the first to the head of the last.

Using the default 2×102 g vector scale, create vector arrows (± 3 grams) for F2, F4. For example, drag the purple vector by its

body and drop it when its tail (the square end) is near the central pin. It should snap in place.

You now want to adjust its direction and length to correspond to direction and magnitude of F2. It’s hard to do both of these

at the same time. That’s where the Resize Only and Rotate Only tools come into play. They let you first adjust the direction

and then adjust the length. Let’s do F2.

Zoom in. Drag the head (tip) of the vector arrow in the assigned direction force F2. That is, θ2. Zoom out. Click the Resize

Only check box. It’s direction is now fixed until you uncheck that box. You can now drag the head of the arrow and adjust

just its magnitude without changing its direction. The current magnitude of F2 is displayed at the top of the screen next to the

home location of that purple vector. You need to adjust its length to 310 ± 3 g. Repeat for F4 using the blue arrow.

The resultant, R, is the vector sum of F2 and F4. Add F2 and F4 graphically by dragging the body of F4 until its tail is over

the tip of F2 and releasing it. Form the resultant, R, by creating an orange vector, F1, from the tail of the first vector, F2, to

the head of the last vector, F4. (You can actually connect F2 and F4 in either order.) You can read the magnitude and

direction of R directly from the length and direction of F1. You’ll likely find some disagreement between these results and

your results from part A. This is because both methods are somewhat imprecise.

1. Graphical Resultant, R: g N, °

The equilibrant, E should be the same magnitude as R,

but in the opposite direction. Produce E with the green

vector arrow. It should be attached to the central pin and

in the direction of the force F3.

2. Graphical Equilibrant, E: g N, °

3. Draw each of the four vector arrows on Figure 2.

Vectors F2 and F4 should be shown added in their tail to

head arrangement. The E and R vectors should be shown

radiating out from the central pin.

Label each vector with its force value in g N units.

4. Take a Screenshot of the force table showing your four

force vectors. Make it large enough to include the

MT/MR text boxes displaying the mass values. Disable

pulley 1 so as to show the system in equilibrium. Save

the figure locally as “FTable_2.png”, print it out, and

paste it in the space provided.

Figure 2: Graphical Addition; Equilibrant

5. Describe what we mean by the terms resultant and equilibrant in relation to the forces acting in this experiment?

 

VPL Lab ah-Force Table 4 Rev 7/17/14

6. From your figures, which two forces, when added would

equal zero?

Circle two F2 F4 E R

7. Show this graphical addition in the space to the right. You’ll

want to offset them a bit since they would otherwise be on

top of one another.

8. Create the same arrangement somewhere on the screen on

your apparatus. Take a Screenshot, save it locally as

“FTable_3.png”, print it out, and paste it in the space

provided.

9. What three vectors, when added would equal zero?

Circle three F2 F4 E R

10. Show this graphical addition in the space to the right.

11. Create the same arrangement somewhere on the screen on

your apparatus. Take a Screenshot, save it locally as

“FTable_4.png”, print it out, and paste it in the space

provided.

 

Figure 3: Two Forces Adding to Zero

 

Figure 4: Three Forces Adding to Zero

 

12. Vector E should appear in both Figure 3 and Figure 4. Why?

 

C. Analytical addition with components

You’ve found experimentally and graphically how to add the forces F2 and F4 to produce the resultant, R. Hopefully you

found that these two methods make it very clear to you what is meant by the addition of vectors. But you also found that both

methods are pretty imprecise – sort of the stone tools version of vector addition. We can get much better results using

trigonometry.

You’ve just verified that the resultant effect of two (or more) vectors can be found by attaching them together tail to head and

drawing a new vector from the tail of the first to the head of the last. That single new vector is equivalent to the two original

vectors acting together. So the one achieves the same result as the original two. The reverse is also true. We can simplify the

process of addition of vectors if we replace each vector with a pair of vectors whose sum equals the original vector. At least if

we do it strategically.

1. Put all four vectors back at the center of the force table with their tails snapped to the central pin. Click the check box

beside the purple arrow under “Show Components.” Zoom in for a better look. The two new purple vectors are the x and

y-components of the purple vector, F2. We would call them F2x and F2y.

2. Drag the x-component, F2x and add it to the F2y. That is, connect its tail to the head of F2y. Clearly they add up to F2. But

just to be sure, drag F2x back to the center and add F2y to it. Same result.

So F2 = F2x + F2y. A

Note that this is all vector addition. We’re using bold text for our vector names to emphasize that this is not scalar

addition, which doesn’t take direction into account. F2 equals the vector sum of F2x and F2y because when we connect

the components together tail to head, the vector from the tail of the first to the head of the last is F2.

So, we can use F2 and F2x + F2y interchangeably. That’s the strategic part.

3. Turn on the blue components.

 

 

VPL Lab ah-Force Table 5 Rev 7/17/14

4. We’ll dim the main vectors. Drag the large blue dragger on the Vector Brightness tool almost all the way to the left. Only

the components will remain bright. Leaving one of them, say F2x where it is, add the other three components to it in any

order you like. This might be a good time to Hide the Table. There’s a button at the top. Feel free to hide and show the

force table as needed.

5. Notice where the head of the last component ends up. Notice how the four components add up to R! Try another order of

addition. The order doesn’t matter. So we can say,

R = F1 = F2x + F2y + F4x + F4y in any order.

Remember, this is vector addition, not scalar addition. So we’re still having to connect them together graphically to find

the resultant. We’re still using stone tools.

6. Turn the components for F1. You now have three sets of components.

7. Drag all the y-components out of the way.

8. Add F2x and F4x together.

9. How do these two vectors relate to F1x that is, Rx?

 

10. Repeat with the y-components. What similar statement can you make about the y-components?

 

11. Finally, how is R related to Rx and Rx?

 

To summarize,

Rx = F2x + F4x and Ry = F2y + F4y and R = Rx + Ry

 

Figure 5: Vector Components

We can now leave our stone tools behind and take advantage of this new formulation by using trigonometric functions and

the Pythagorean Theorem.

Here’s our task. You were previously asked to find the resultant, R of F2 and F4 graphically. You now want to find the same

result without the using the imprecise graphical methods.

Knowing F2 and θ2 we can calculate F2x and F2y analytically as follows.

F2x = F2 cos(θ2) (1)

F2y = F2 sin(θ2) (2)

We can do the same for F4. We can then find Rx and Ry using

Rx = F2x + F4x (3)

Ry = F2y + F4y (4)

R = Rx + Ry (5)

 

 

VPL Lab ah-Force Table 6 Rev 7/17/14

There’s one slight problem with Equation 5. Like Equations 3, and 4, it’s a vector equation, but in Equations 3, and 4, the

vectors are collinear, so they can be added with signs indicating direction. But since vectors Rx and Ry are perpendicular, we

have to use the Pythagorean theorem instead. So we can find the magnitude and direction of the resultant, R with the

magnitudes of R’s components.

R2 = Rx2 + Ry2 (6)

tan(θ) = Ry/Rx (7)

12. You’ve found experimentally and graphically how to add the forces F2 and F4 to produce the resultant, R. Now let’s try

analytical addition. Using the table provided, find the components of F2 and F4 and add them to find the components of

R. Use these components of R to determine the magnitude and direction of R. Note that one of the four components will

have a negative sign. The components now displayed on the force table should make it clear why.

13. Show all your calculations leading to your value for R below. Remember, a vector has both a magnitude and a direction.

A summary of the steps is provided to get you started.

x-Components y-Components

F2x = g N F2y = g N

F4x = g N F4y = g N

Rx = g N Ry = g N

R = g N, ° (State your angle between 0 and 360°.)

F2x = F2 cos(θ2) = .300 g N cos(θ2) F2y = F2 sin(θ4) = .300 g N sin(θ4)

Rx = F2x + F4x Fy = F2y + F4y R2 = Rx2 + Ry2 tan(θ) = Ry/Rx

 

II. SIMULATION OF A SLACKWIRE PROBLEM.

Let’s model a realistic system similar to what you might find in your homework. Figure 6 shows a crude figure of slackwire

walker, Elvira, making her way across the wire. She weighs 450 N. (About 100 lbs.) At a certain instant the two sides of the

rope are at the angles shown. Only friction allows her to stay in place. The gravitational force is acting to pull her down the

steeper ‘hill.’ If she was on a unicycle she’d tend to roll to the center. It’s complicated.

When friction is holding her in place the single rope acts like two separate sections of rope in this situation with different

tension forces on either side of her. To understand this it helps to imagine her at the extreme left where the left section of rope

is almost vertical and the right one is much less steep. In this case T3 is providing almost all of the vertical support, while T1

pulls her a little to the right. It helps to just try it. Attach a string between two objects in the room. Leave a little slack in it.

Now pull down at various points. You’ll feel the big frictional tug on the more vertical side and less from the more horizontal

side.

We want to explore this by letting her move along the rope.

 

 

VPL Lab ah-Force Table 7 Rev 7/17/14

Figure 6: Elvira Off-center on the Slackwire

 

A. Experimental Determination of T1 and T3

1. Preliminary prediction: Which tension do you think is greater, T1 or T3? Circle one.

First send each of your vectors “home” by clicking on each of their little houses. Then remove all the masses from your

four pulleys by activating all of them using their click boxes and then clicking in the total mass boxes beside each hanger

until they read 50.

Elvira weighs N.

We’ll let one gram represent 1 N on our force table and set our vector scale to 4×102 N. (Note the label to the left of each

vector.) We’ll picture our force table as if it were in a vertical plane with 270° downward and 90° upward. We’ll use

pulleys 1, 3, and 2 to provide our two tensions, T1 and T3, and the weight of Elvira respectively.

Disable all pulleys and set all the angles to match Figure 6. Use Zoom.

Turn pulleys 1, 3, and 2 back on.

2. Prediction: Since θ3 = 2 × θ1, do you think T3 will be about twice T1?

3. Set Elvira’s weight to the value you were assigned. If you’re assigned a weight of 450 N, use 450 g including the mass of

the hanger. So you’d add 400 g.

4. Adjust the masses on each pulley until you achieve equilibrium. It’s best to alternate adding one mass to each side in turn

until you get close to equilibrium.

5. T1 = N at 10°

6. T3 = N at 160°

How’d that prediction go (#2)? If trig functions were linear we wouldn’t need them.

7. Write a statement about the relationship between the steepness of the rope and the tension in that rope for this vertical

arrangement.

 

VPL Lab ah-Force Table 8 Rev 7/17/14

B. Graphical Check of our Results

1. Using the vector scale of 4×102 N, create

vector arrows for each force, T3, T1, and W.

Don’t forget to use the Resize Only tool

after you get the angles set.

2. Take a Screenshot, save it locally as

“FTable_7.png”, print it out, and paste it in

the space provided.

3. Draw your three vectors on Figure 7.

4. How can we graphically check to see if our

values for T1 and T3 are reasonably correct?

How are T1, T3, and W related? What would

happen if any one of them suddenly went

away? Elvira would no longer be in

.Figure 7: Elvira – Asymmetrical Vectors

 

Thus the three vectors are in equilibrium and must add to equal zero! What would that look like? The sum, resultant, of

three vectors is a vector from the tail of the first to the head of the last. If they add up to zero, then the resultant’s

magnitude would be zero which means that the tip of the final vector would lie at the tail of the first vector.

Try it in with your apparatus. Leave T1 where it is, then drag T3’s tail to T1’s head. Then drag W’s tail to the head of T3.

5. What about your new figure says (approximately) that the three forces are in equilibrium?

 

6. Draw your new figure with

the three vectors added

together on Figure 8.

7. Take a Screenshot, save it

locally as “FTable_8.png”,

print it out, and paste it in the

space provided.

 

Figure 8: Graphical Addition – Asymmetrical

 

VPL Lab ah-Force Table 9 Rev 7/17/14

C. Analytical Check of Your Results

If our three vectors add up to zero, then what about their components?

1. When you add up the x-components of all three vectors the sum should be

2. When you add up the y-components of all three vectors the sum should be

3. Test your predictions using the following table. Don’t forget the direction signs!

x-Components y-Components

F1x = N F1y = N

F3x = N F3y = N

Wx = N Wy = N

ΣFx = N ΣFy = N

4. Show your calculations of the eight values in the table in the space provided below.

 

Before we continue… Why all the tension?

Why does the tension on each side have to be so much larger than the weight actually being supported? All the extra tension

is being supplied by the x-components. Then why not just get rid of it? You’d have to move the supports right next to each

other to make both ropes vertical. That would be pretty boring and frankly nobody would pay to see such an act, even if they

tried calling it “Xtreme Urban Slackwire.”

Similarly, wih traffic lights, a huge amount of tension is required to support a fairly light traffic light. The simpler solution of

a pair of poles in the middle of the intersection wouldn’t go over much better than Xtreme Urban Slackwire.

Next time you see poles or towers supporting large electrical wires look for places where the wire has to change direction.

(South to East for example.) The support structures in straight stretches (in-line) don’t have to be really sturdy since they

have horizontal forces pulling equally in opposite directions. Thus they just have to support the weight of the wire. When the

wires have to change directions the poles at the corners have to provide these horizontal forces. Thus they’re much sturdier

and larger to give them a wider base. They often have guy wires to the ground to help provide these forces. To make things

more difficult the guy wires will be usually be very steep so they have to have a lot of extra tension in them to supply the

horizontal force components.

 

Figure 9: Power Lines – In Line and at Corners

 

 

VPL Lab ah-Force Table 10 Rev 7/17/14

III. SIMULATION OF A SYMMETRICAL SLACKWIRE PROBLEM.

In Figure 10, Elvira has reached the center of the wire. This is where she would be if she rode a unicycle and just let it take

her to the “bottom.” The angle values are just guesses. They’d be between the 10° and 20° we had before, but the actual value

would depend on the length of the wire and its elastic properties.

 

Figure 10: Elvira at the Center of the Slackwire

Without changing the masses used in part II, adjust both angles to 15°.

1. What does the symmetry of the figure suggest about how the tensions T1, and T3 should compare?

 

2. Similarly what can you say about comparative values of the x-components T1x and T3x? (Ex. Maybe T1x = 2× T3x?)

 

3. What can you say about comparative values of the y-components T1y and T3y?

 

4. Knowing that the weight being supported is 450 N, what can you say about actual values of the y-components T1y and

T3y?

 

5. T1y = T3y = N

You can now easily calculate T1 and hence T3.

6. T1 = T3 = N at 15°

Show the calculation of T1 below.

 

 

7. Change each T to as close as you can get to this amount. Does this produce equilibrium?

8. With your apparatus, create the two vector arrows to match this amount – one for T1 and one for T3. Add T1 + T3 + W

graphically. Draw or paste a copy of this graphical addition below.

 

 

VPL Lab ah-Force Table 11 Rev 7/17/14

9. Create the two vector arrows to match

this amount – one for T1 and one for

T3. Add T1 + T3 + W graphically. Draw

or paste a copy of this graphical

addition in Figure 11.

10. Take a Screenshot, save it locally as

“FTable_11.png”, print it out, and

paste it in the space provided.

You’ll notice that this figure doesn’t differ

much from the previous one. The graphical

tool we’re using is not very precise. The

same goes for “the real world.” Measuring

the tension in a heavy electrical cable or

bridge support cable is very difficult. One

good method involves whacking it with a

hammer and listening for the note it plays!

Figure 11: Graphical Addition – Symmetrical

IV. VECTOR SUBTRACTION

Here’s the scenario. We have a three-person kinder, gentler tug-o-war. The goal is to reach consensus, stalemate. Two of our

contestants are already at work.

Darryl1 = 800 N at 0°

Darryl2 = 650 N at 240°

The question is – how hard, and in what direction, must Larry3 pull to achieve equilibrium?

1. As before, empty all the hangers, and then set up pulleys 1 and 2 to represent these forces using 1 gram/1 newton.

2. Send all the vector arrows home to get a clean slate. Then create orange and purple vectors to match using the 4×102 N

for your vector scale. Attach them to the central pin.

One method of solving the problem is to add the Daryl forces to find the resultant. Larry’s force would be the equilibrant in

the other direction. Another way is to add the two Daryl forces and then draw a third force, Larry, to complete the triangle

which would leave a resultant of zero.

The following vector equation describes that method.

ΣF = Darryl1 + Darryl2 + Larry3 = 0

This is a vector equation. It means that you’ll get a resultant of zero if you connect the three vectors together tail to head. To

find the Larry vector we need to subtract the two Darryl vectors from both sides. We know how to add vectors but how do

we subtract them?

With scalar math it would look like this:

ΣWorth = Darryl1$ + Darryl2$ + Larry3$ = 0 (Yes, that’s net worth)

Larry3$ = –Darryl1$ – Darryl2$

If Darryl1$ = $800 and Darryl2 = $650, they we’d get

Larry3$ = –$800 –$650 (Note that we’re adding negatives here.)

Larry3$ = –$1450

 

 

VPL Lab ah-Force Table 12 Rev 7/17/14

Negative dollars don’t exist but we would interpret this as something like a debt. That is to make the three brothers have zero

net worth, Larry3 needs to be $1450 in debt.

With vectors it works the same way but we interpret the negatives signs as indications of direction. A pull of -50 N to the left

means a pull of 50 N to the right.

ΣF = Darryl1 + Darryl2 + Larry3 = 0

Larry3 = –Darryl1 – Darryl2

Larry3 = (–Darryl1) + (–Darryl2)

So to find Larry3 we need to create the two negative Darryl vectors and add them.

This means to draw the vector (–Darryl1) and (–Darryl2) which are the opposites of Darryl1 and Darryl2 and add them. That

is, we can subtract vectors by adding their negatives.

So if

Darryl1 = 800 N at 0°

Darryl2 = 650 N at 240°

then the negatives of these are vectors of the same lengths but in the opposite directions. Thus,

–Darryl1 = 800 N at 180°

–Darryl2 = 650 N at 60°

3. Ruining the hard work you just did, create

these two –Darryl vector arrows. (Don’t

change the mass hangers. Just the vectors.)

You’ll do this by just reversing the directions

of both the vector arrows you initially created.

Leaving the –Darryl1 vector pointing at 180°,

add the –Darryl2 vector in the usual tail to

head fashion.

Larry3 is the sum of these two. Create the

green Larry3 vector from the tail of -Darryl1

to the head of -Darryl2.

Draw these three vectors on Figure 12.

4. Take a Screenshot, save it locally as

“FTable_12.png”, print it out, and paste it in

the space provided.

Figure 12: Larry and 2 Darryls

 

5. Larry3 (graphical) = N ° (from the length and direction of your green, Larry3 vector)

Move pulley 3 to the position indicated by the direction of Larry3. You’ll want to temporarily turn off the hangers to set

the angle correctly.

Place the necessary mass on hanger three to produce the Larry3 force.

6. Larry3 (experimental) = N ° (from the mass on hanger 3 and the location of the pulley)

Ta Da! I hope this has made vectors as little bit less mystifying.