11_20_2023---PHY 111 Lab #4_ Newtons First Law
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Jan 9, 2024
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Lab 4 Worksheet
Name: Elias Linn
Newton’s First Law
Section: #25904
Part 1: Confirmation of Newton’s First Law
Tape two straws to a level surface in such a way that the ends of the straws stick out past the edge of the surface. Connect the following number of paper clips to either end of a single string: Left side of string: 2
Right side of string: 3
Get a second string and tie 4 number of paperclips to one end of the string and then connect the other end of the string to the center of the first string. (If your instructor does not specify the number of paperclips then choose a combination of paperclips and fill in the numbers above.)
Place the string over your straw so that the center paperclips are between the two straws, and the others are on the outside ends of the straws. Something like this: Very gently tap the straws until the paperclips have reached equilibrium (i.e., the string no longer moves).
Use a protractor to measure the angles θ and β as indicated below: (It may be easier to take a picture of your system and then use the protractor on the picture.)
Complete the following data tables: Left side of system
Right side of system
Center of system
θ
β
Number of paperclips
2
3
4
Total mass (kg)
0.0015 kg
0.00365 kg
0.0047 kg Angle measured
Size of angle (degrees)
θ
42
β
73
Calculations
Show the calculations used to determine the weight acting on each string. (Do not simply record the answer. Make sure to show your work.)
String section
Weight of paperclips (N)
Left
W = m * g W= (0.0015 kg)*(9.8 N/kg) W = 0.0147 kg
Center
W = m * g W= (0.0047 kg)*(9.8 N/kg) W = 0.04606 kg Right
W = m * g W= (0.00365 kg)*(9.8 N/kg) W = 0.03577 kg Draw and label a quantitative force vector diagram for the knot that connects the two strings.
Show the calculations used to determine the horizontal and vertical components of the tension forces connected to the center string. (Do not simply record the answer. Make sure to show your work.)
Force component
Support rope with θ angle
Support rope with β angle
Horizontal force component
(N)
Cos(
θ) = F
x
/
F
1
Cos(42) = F
x
/(
0.0147)
F
x
= Cos(42)/(0.0147)
F
x
= 0.0109
Cos(
θ) = Fx
/
F
1
Cos(73) = F
x
/(
0.0147)
Fx
= Cos(73)/(
0.03577
)
F
x
= 0.0105
Vertical force component (N)
Sin(
θ) = F
y
/
F
Sin(42) = F
y
/(
0.0147)
Fy
= Sin(42)/(0.0147)
Fy
= 0.00984
Sin(
θ) = F
y
/
F
1
Sin(73) = F
y
/(
0.0147)
F
y
= Sin(73)/(
0.03577
)
Fy= 0.0342
Determine whether Newton’s first law holds true in the horizontal direction. Show your calculation and discuss your results. Be specific and describe how your calculation supports or does not support Newton’s first law.
Based on Newton’s First Law, in the sense of horizontal direction is where the two values are not equal to 0. Therefore, by looking at the horizontal directions values in terms of force, with that of the θ and β angles in which both equal to 0 in this case, for Newton's law
to be true. Therefore, based on Newton's First law isn’t satisfied, since the horizontal force of the left side was indeed equal to the overall magnitude of the horizontal force to the present right side.
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Determine whether Newton’s first law holds true in the vertical direction. Show your calculation and discuss your results. Be specific and describe how your calculation supports or does not support Newton’s first law.
Part 2: Unknown Mass
Remove the paperclips from the center string. Replace the paperclips with the object provided in your lab kit (it may be a cork or a small binder clip). Alternatively, if your instructor allows, you may use an item of your own choosing. The item should have a very small mass. Use trial and error to determine the number of paperclips to add to both the left and right side of the string so that when you place the string back over the straws, the system remains in equilibrium (i.e., does not shift when you gently tap the straws).
Once your system is in equilibrium, take a picture and include the picture in the space below. The picture should also include a card with your name, date, and section number. For newton's first law to be valid in the case of the vertical direction, W values must cancel out with that of the sum of the vertical elements on their respective side. In my case, Newton's first law wasn’t truly satisfied in this case as the overall force of the center’s inherent mass must be equal to the overall magnitude of the sum of the vertical left and right sides plus the value of the center. Therefore, the verticals right and left sides must be equal to 0 in that regard.
Complete the following data tables: Left side of system
Right side of system
Number of paperclips
3
4
Total mass (kg)
0.00365 kg
0.0047 kg Angle measured
Size of angle (degrees)
θ
35
β
50
Calculations
Show the calculations used to determine the horizontal and vertical components of the tension forces connected to the center string. (Do not simply record the answer. Make sure to show your work.)
Horizontal component (N)
Vertical component (N)
Support rope with θ angle Cos(
θ) = F
x
/
F
1
Cos(35) = F
x
/(
0.00365)
F
x
= Cos(35)(
0.00365
)
F
x
= 0.00298
Cos(
θ) = F
x
/
F
1
Sin(35) = F
x
/(
0.00365)
F
x
= Cos(35)(
0.00365
)
Fx
= 0.00209
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Support rope with β angle Cos(
θ) = Fy
/
F
1
Cos(50) = Fy
/(
0.0047)
Fy= Cos(50)(
0.0047
)
Fy
= 0.00302
Sin(
θ) = F
y
/
F
1
Sin(50) = F
y
/(
0.0047)
F
y
= Sin(50)(
0.0047
)
F
y
= 0.0036
Use the components and Newton’s first law to solve for the unknown mass of the object. Show your work below. Do not forget to include the units.
Discuss your final results. Is your answer reasonable answer? Explain. The horizontal portions of the orientation, in order to satisfy Newton's first order, must cancel out. The two F(x-values) don’t rule each other out in this instance. Regardless of this matter, we can state that the overall mass in which can be found is a result of the vertical direction of the object in reference to the unknown mass, in order to cancel out said values of the both sides of the vertical spectrum. Thus, we can solve for the center portion(W), then using the center portion to find said mass. Our weight of 0.0005806 kg is a little under valued due to the possibility of mis-calculations occurring in which could be related to the lack of results portrayed in the horizontal section,
where there wasn’t a clear “cancellation” of values because of an error on my part. The mistake in which I made is a possible a result of improper measurement as well as a lack of friction occurring.