Draw a Force Diagram to the right of the above drawing, using the provided coord W friction, f. & applied force, F.

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Answer to #3 force diagram force table etc.
335
kg
III. Data Tables & Calculations:
Data Table - Part 1 - Horizontal Ramp:
Lab05
Data Table -- Part 1
335
Normal
force
mass of wood block =
total
mass
kg
13,45
3.55
cargo
mass
kg
0.100
0.200
0.300
0.400
0.500 3.85
0.600
3.95
best linear fit slope, ₁ =
N
33.81
34.49
13.65
35047
36-75
kg
1
120
150
150
200
37.73
38.47 240
Coefficient of Friction
2
130
150
140
200
240
critical
hanging
trial #
3
1200
130
150
200
240
mass
kg
4
1205
120
200
240
f max
avg
avg
hanging hanging
mass
force
N
5
kg
e12
120
130
134
50
240
1.150
•24
=
11.2
1.31
1.47
11.96
2.352
X9.8
19=X=1000
X. 9.X NF
Calculations - Part 1 - Horizontál Ramp:
1. Record the mass of the wood block. Add its mass to the specified cargo masses in the 1st column to
get the total mass on the ramp that is recorded in the 2nd column. Using MKS units, multiply by g to
convert this total mass on the ramp into the Normal force, N, & record in the 3rd column.
2. For each specified added cargo mass, gradually increase the hanging mass until the block just slip
& continues to move "slowly" across the ramp. A mm/sec is too slow, a cm/sec is too fast. Record thi
critical hanging mass for 5 trials for each of the 6 block set-ups. For each new trial, remove part of the
hanging mass, then gradually add it back onto the hanger until it once again starts sliding.
Transcribed Image Text:335 kg III. Data Tables & Calculations: Data Table - Part 1 - Horizontal Ramp: Lab05 Data Table -- Part 1 335 Normal force mass of wood block = total mass kg 13,45 3.55 cargo mass kg 0.100 0.200 0.300 0.400 0.500 3.85 0.600 3.95 best linear fit slope, ₁ = N 33.81 34.49 13.65 35047 36-75 kg 1 120 150 150 200 37.73 38.47 240 Coefficient of Friction 2 130 150 140 200 240 critical hanging trial # 3 1200 130 150 200 240 mass kg 4 1205 120 200 240 f max avg avg hanging hanging mass force N 5 kg e12 120 130 134 50 240 1.150 •24 = 11.2 1.31 1.47 11.96 2.352 X9.8 19=X=1000 X. 9.X NF Calculations - Part 1 - Horizontál Ramp: 1. Record the mass of the wood block. Add its mass to the specified cargo masses in the 1st column to get the total mass on the ramp that is recorded in the 2nd column. Using MKS units, multiply by g to convert this total mass on the ramp into the Normal force, N, & record in the 3rd column. 2. For each specified added cargo mass, gradually increase the hanging mass until the block just slip & continues to move "slowly" across the ramp. A mm/sec is too slow, a cm/sec is too fast. Record thi critical hanging mass for 5 trials for each of the 6 block set-ups. For each new trial, remove part of the hanging mass, then gradually add it back onto the hanger until it once again starts sliding.
3. Crooked String Problem: Assume the string in Part 1 of the lab is tilted at some angle 0 above the
horizontal.
N
b. Fill in the
(M)
F
W
a. Draw a Force Diagram to the right of the above drawing, using the provided coordinate axes.
orce Table below for the normal force, N, weight, W, friction, f, & applied force, F.
List of Forces
y-components
x-components
+ N
N
0
- W
W
0
f
- f
F
+ F cos()
(sum the component columns
+ F cos() - f = 0
& the sum = ?)
X
0
+F sin()
+F sin()+N-W = 0
c. Use vertical equilibrium to derive the equation for the normal force, N, on the block when the force,
F, is tilted by an arbitrary angle, 0.
insert your solution for Nom(d, Aa
d. Qualitatively, would you expect the experimental determination of the coefficient of friction be the
same if the string is tilted instead of being level? (Yes or No)
Transcribed Image Text:3. Crooked String Problem: Assume the string in Part 1 of the lab is tilted at some angle 0 above the horizontal. N b. Fill in the (M) F W a. Draw a Force Diagram to the right of the above drawing, using the provided coordinate axes. orce Table below for the normal force, N, weight, W, friction, f, & applied force, F. List of Forces y-components x-components + N N 0 - W W 0 f - f F + F cos() (sum the component columns + F cos() - f = 0 & the sum = ?) X 0 +F sin() +F sin()+N-W = 0 c. Use vertical equilibrium to derive the equation for the normal force, N, on the block when the force, F, is tilted by an arbitrary angle, 0. insert your solution for Nom(d, Aa d. Qualitatively, would you expect the experimental determination of the coefficient of friction be the same if the string is tilted instead of being level? (Yes or No)
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