A 1.2 kN horizontal force is applied acting horizontally to the right at end \"A
ID: 1899885 • Letter: A
Question
A 1.2 kN horizontal force is applied acting horizontally to the right at end "A" of a lever 600 mm long which is attached to a shaft at the origin denoted as "O." The angle formed between the horizontal (x) axis and lever OA is 60degree. End "A" is to the upper right of "O." Determine: the moment of the 1.2 kN force about O; use both scalar and vector formulation and confirm that the same result is produced either way; the magnitude of the vertically downward force applied at A which would create the same moment at (about) O; the smallest force applied at A which creates the same moment about O; how far from the shaft (origin) a 2.4 kN horizontal force must act to create the same moment about O; and whether any one of the forces obtained in parts b, c, and d is equivalent to the original force. A 1.2 kN horizontal force is applied acting horizontally to the right at end "A" of a lever 600 mm long which is attached to a shaft at the origin denoted as "O." The angle formed between the horizontal (x) axis and lever OA is 60degree. End "A" is to the upper right of "O." Determine: the moment of the 1.2 kN force about O; use both scalar and vector formulation and confirm that the same result is produced either way; the magnitude of the vertically downward force applied at A which would create the same moment at (about) O; the smallest force applied at A which creates the same moment about O; how far from the shaft (origin) a 2.4 kN horizontal force must act to create the same moment about O; and whether any one of the forces obtained in parts b, c, and d is equivalent to the original force.Explanation / Answer
a)
m = r × F = r F sin60 = 600e-3 * 1.2e3 * 0.86603 = 623.54 624 N.m
b)
m = r × F = r F cos60
623.54 = 600e-3 * F * 0.5
F = 2078 N
c)
m = r × F = r F sin90
623.54 = 600e-3 * F * 1
F = 1039 N
d)
m = r × F = r F sin60
623.54 = r * 2.4e3 * 0.86603
r = 0.300 m
d)
No
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