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E1 Mechanics and Forces - Laboratory Report Experiment report Student name Student no Grade Preliminary Exercise Complete the following questions on this sheet, either typed or handwritten. In these...

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E1 Mechanics and Forces - Laboratory Report
Experiment report
    Student name
    Student no
    Grade
    
    
    
Preliminary Exercise
Complete the following questions on this sheet, either typed or handwritten.
In these exercises use g = 9.8 m/s2
1:    In the theory document, you will find the experimental set-up for the statics lab (three suspended masses connected by a string). Ignoring friction and taking the special case where m1 = m2 = 0.1 kg and m3 = 0.15 kg.
Find 1 and 2 at the equili
ium position.
2:    In Part B: Attwood’s Machine, ignoring friction and the rotational inertia of     the pulley (ie. Assume the pulley is mass-less) calculate the acceleration of m2     (downwards is positive) for the case where m2 = 120 g and m1 = 100 g.
3:    In an experiment using m2 = 120 g and m1 = 100 g, the measured acceleration     was 0.85 m/s2. In this case assume that there is friction at the pulley but that     the mass of the pulley is negligible. Determine the additional mass in     grams that must be added to m2 in order to overcome this friction.
Online Experiment - Part 1 (see Static Video for demonstration)
1. The pulley system shown in Figure 1 is set up in the “Part 1 - static experiment” video with m1 = 200 g, m2 = 250 g and m3 = 300 g. A sheet of paper was placed behind the strings so that the string positions could be marked on it. This is shown below with the angles of the strings (relative to vertical) and masses (in grams) included.
2.    Draw the free body diagram (as in Figure 2 of the Theory section) for this apparatus with forces, coordinate system and x/y force components clearly marked. Calculate horizontal and vertical components of forces based on equili
ium and insert these into the table at the bottom of this page. Note that due to friction the horizontal and vertical components will generally not add to zero. Use your results and equations (3) and (4) to estimate the two (x and y) components of the net frictional force.
FREE BODY DIAGRAM HERE:
Results for Part A: Addition of Forces    
(Record your results for the force components below)
    1
    54
    deg
    2
    43
    deg
    F1 = m1g
    
    N
    F2 = m2g
    
    N
    F3 = m3g
    
    N
    F1sin 1
    
    N
    F1cos 1
    
    N
    F2sin 2
    
    N
    F2cos 2
    
    N
    fnet x
    
    N
    fnet y
    
    N
Atwood’s Machine - Part 2 (see Atwood’s Machine Video for demonstration)
Note that to save time the mass of the washers used in the experiment is given:
mwasher = 1.155 g
    Enter all results in the resource spreadsheet available for download on Canvas and paste the tables/graphs into report where indicated. Use “Paste as Picture”. This will enable easy re-sizing of results to fit into your document.
1.    In this experiment washers are used for small increments in mass. The total mass of the system is maintained but the difference between the two masses is altered by transfe
ing washers from one mass to the other.
2. The “Part 2 - Atwoods machine” video shows the single pulley Atwood’s machine (Figure 5 in Theory section). The experiment is performed with two different (larger and smaller) values of m1-m2.
3. For the experimental data, the washers were moved one at a time from m1 to m2 until m2 just began to move downwards (accelerate) when released. Sufficient force was then provided to overcome friction. This is your first experimental point.
4.    The time required for m2 to fall a known distance (1.0 m), to the floor was measured using a stop-watch.
Five measurements were taken for each data point and averaged for the times in the Atwood’s Machine Table. You need to calculate the acceleration of the masses from these times (using the equation(s) of motion discussed in the lectures).
Paste the table with the acceleration column filled in here:
5.    Now use Microsoft Excel (or similar) to tabulate results and to plot a graph of acceleration vs. (m2 - m1).
    Use XY (Scatter) chart from the Chart Wizard - show points (markers) only and add a linear trend-line and check the box to Display equation on chart. As well as plotting the line of best fit (linear regression) your plot will give you the gradient and the intercept on the vertical axis.
As shown in the Theory section, the gradient of this plot enables g to be calculated and the intercept on the vertical axis enables an average kinetic frictional force to be extracted from the measurements.
Paste your graph of acceleration vs. m2 - m1 here.
Determine the gradient and intercept from the graph and use Equation (6) to calculate the acceleration due to gravity (g) and the kinetic frictional force. Comment (in your own words) on sources of e
or that may lead to e
or in your value of g.
    g
    
    m/s2
    Kinetic Friction
    
    N
Answered Same Day Apr 28, 2021

Solution

Sannidhya answered on May 03 2021
153 Votes
E1 Mechanics and Forces - Laboratory Report
Experiment report
    Student name
    Student no
    Grade
    
    
    
Preliminary Exercise
Complete the following questions on this sheet, either typed or handwritten.
In these exercises use g = 9.8 m/s2
1:    In the theory document, you will find the experimental set-up for the statics lab (three suspended masses connected by a string). Ignoring friction and taking the special case where m1 = m2 = 0.1 kg and m3 = 0.15 kg.
Find 1 and 2 at the equili
ium position.
· Based on the given masses,
F1 = m1g = (0.1 kg)*(9.8 m/s2) = 0.98N
F2 = m2g = (0.1 kg)*(9.8 m/s2) = 0.98N
F3 = m3g = (0.15 kg)*(9.8 m/s2) = 1.47N
Substituting these values in the equili
ium equations (ignoring friction), we get:
-0.98 sin θ1 + 0.98 sin θ2 = 0         ------(i)
0.98 cos θ1 + 0.98 cos θ2 – 1.47 = 0    ------(ii)
Equation (i) gives us:
sin θ1 = sin θ2 or θ1 = θ2 (as they are acute angles)
Putting θ1 = θ2 in equation (ii) gives us:
1.96 cos θ1 = 1.47
which gives
θ1 = θ2 = 41.410
2:    In Part B: Attwood’s Machine, ignoring friction and the rotational inertia of     the pulley (ie. Assume the pulley is mass-less) calculate the acceleration of m2    (downwards is positive) for the case where m2 = 120 g and m1 = 100 g.
· Ignoring friction and rotational inertia of the pulley, the acceleration of m2 is given by
a = (m2 – m1)g/(m1 + m2)
= (120 g – 100 g)*(9.8 m/s2)/(120 g + 100 g)
= 0.89 m/s2
Hence,
    Acceleration of m2 is 0.89 m/s2 in the downward direction.
3:    In an experiment using m2 = 120 g and m1= 100 g, the measured acceleration     was 0.85 m/s2. In this case assume that there is friction at the pulley but that     the mass of the pulley is negligible. Determine the additional mass in     grams that must be added to m2 in order to overcome this friction.
· Based on the given masses,
Fideal = (m2 – m1)g = (0.12 kg – 0.1 kg)*(9.8 m/s2) = 0.196N
Since friction exists, hence, Newton’s second law of motion gives:
Fideal – f = (m1 + m2)a, where f is the frictional force
Substituting the values, we get:
0.196N – f = (0.1 kg + 0.12 kg)*(0.85 m/s2)
O
f = 0.196N – 0.187N = 0.009N
To overcome this friction, we need some extra force which is provided by adding some additional mass to m2 such that the gravitational force acting on this extra mass equals the frictional force.
Hence, if m is the additional mass added, then
    mg = f
o
    m = f/g = (0.009N)/(9.8 m/s2) = 0.00092 kg = 0.92 g
Therefore, 0.92 g of additional mass must be added to...
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