Question 8
3 pts
Newton’s second law states that:
The force of gravity is proportional to the inverse
square of the distance between centers
The acceleration is proportional to the force exerted
An object will continue at a constant velocity unless
acted upon by an external force
For every action there is an equal and opposite
reaction
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Question 20
2 pts
Question 21
3 pts
Question 19
2 pts
You are investigating forces applied onto a box.
What does the dotted line indicate in the following
graph? Select all that apply.
What does the vertical dashed line on the graph
represent? Select all that apply
Force 1 = 2637N
=
200
1500
С
Force 2 = 250N
=
C
150
1000
100
Biceps insertion angle = 72.34°
SO
s/800
500
V
0
Angularvelocity (deg/s)
0
10
30
Am
99
loo
-50
What is the net force acting in the
x direction? Round your answer to the nearest
whole number.
-100
-500
Time (%)
Figure 2: Shoulder angular velocity
– 150
-200
-250
†
-300
Xof the start phone
f
A positive acceleration
Change of direction
Peak angular velocity
O Anatomical O
F
O Anatomical 0
Zero Velocity
Peak displacement
Peak negative velocity
Change of direction
Peak displacement
O A negative acceleration
Peak positive velocity
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Question 23
4 pts
Question 24
3 pts
Question 25
Search documents and file names for text
You are investigating forces at the elbow during a
preacher curl exercise. The forearm is parallel to
the ground.
=
You are designing a new weight machine for the
gym but want to assess it for stability based on the
mass of the 3 parts and the location of their
individual center of masses.
The estimated center of mass in the y direction for
the following schematic is at y = 9. Given this y
position, what would you conclude about its
stability?
Force in biceps brachii (F1) = 2000N
=
Force of the dumbbell (F2) = 150N
=
Calculate the center of mass position in
the x direction. You can omit the units in your
=
answer
Biceps insertion angle = 60°
Using the free body diagram, determine the Net
force acting at the elbow joint
cm
10kg (6, 15)
F
4kg (5.5,9)
cm
1
0
10
15
20
25
cm
8kg (6; 2)
رک
11
10
15
20
25
cm
No, the COM is too high
No, the y position does not say if the COM is located
within the object.
1732N
The mid point makes it stable
2131N
Need more information to determine stability
IVE
O 2150N
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1872N
O 1929N
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