1. The figure below shows an overhead view of a 50 g disk and two of the three h
ID: 1418245 • Letter: 1
Question
1. The figure below shows an overhead view of a 50 g disk and two of the three horizontal forces that act on it as it is on a frictionless table.
(Please explain how you did it. I don't understand this problem)
1. The figure below shows an overhead view ofa 50 g disk and two of the three horizontal forces the indicated angles are ,-30° and 2-45°. In theunit-vector notation, what is the third force if the disk (a) has constant velocity u=(13 1-14 j) m/s, and (b) has the varying velocity u= (13t i-14t j) m/s, where t is time?Explanation / Answer
(a)
As the disk has constant velocity, which means acceleration = 0.
This means Net Force on the particle is 0.
F1 = - 6 * cos(30) i^ + 6 * sin(30) j^
F1 = - 5.19 i^ + 3 j^
F2 = 7 * cos(45) i^ - 7 * sin(45) j^
F2 = 4.95 i^ - 4.95 j^
F1 + F2 + F3 = 0
- 5.19 i^ + 3 j^ + 4.95 i^ - 4.95 j^ + F3 = 0
F3 = 0.24 i^ + 1.95 j^
Third force in unit vector notation, F3 = 0.24 i^ + 1.95 j^
(b)
v = a*t
a = v/t
a = 13 i^ - 14 j^
F = m*a
F = 50*10^-3 * ( 13 i^ - 14 j^)
- 5.19 i^ + 3 j^ + 4.95 i^ - 4.95 j^ + F3 = 50*10^-3 * ( 13 i^ - 14 j^)
- 0.24 i^ - 1.95 j^ + F3 = 0.65 i^ - 0.7 j^
F3 = 0.89 i^ + 1.25 j^
Third force in unit vector notation, F3 = 0.89 i^ + 1.25 j^
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