Asked by thaokieu
Before starting this problem, review Conceptual Example 7. The force vector A has a magnitude of 79.0 newtons (N) and points due east. The force vector B has a magnitude of 107 N and points 75.0 ° north of east. Find the (a) magnitude and (b) direction of A - B. (Give the direction as a positive angle with respect to due east). Find the (c) magnitude and (d) direction of B - A. (Give the direction as a positive angle with respect to due west).
Answers
Answered by
Henry
Given: A = 79N, East; B = 107N @ 75deg.
a. X = hor. = 79 - 107cos75,
X = 79 - 27.7 = 51.3N.
Y = ver. = 0 - 107sin75,
Y = 0 - 103.4 = -103.4N.
M^2 = X^2 + Y^2,
M^2 = (51.3)^2 + (-103.4)^2,
M^2 = 2631.7 + 10691.6 = 13323,
M = 115.4N Magnitude of A-B.
b. tanA = Y/X = -103.4 / 51.3 = -2.015,
A = -63.6Deg. = 63.6deg South of East = Direction of A-B.
Use the above procedure for B-A.
Mag.= X/cosA = 106.7 / cos44.1 = 148.6N.
b. A = 79N,
B = 107N@75deg. = 107cos75 + i107sin75,
B = 27.7 + i103.4.
A - B = A to B = (27.7 + i103.4) -79,
A to B = -51.3 + i103.4
tan(theta) = 103.4 / -51.3 = -2.0156,
Theta = -63.6deg = 63.6deg South of East.
c.
a. X = hor. = 79 - 107cos75,
X = 79 - 27.7 = 51.3N.
Y = ver. = 0 - 107sin75,
Y = 0 - 103.4 = -103.4N.
M^2 = X^2 + Y^2,
M^2 = (51.3)^2 + (-103.4)^2,
M^2 = 2631.7 + 10691.6 = 13323,
M = 115.4N Magnitude of A-B.
b. tanA = Y/X = -103.4 / 51.3 = -2.015,
A = -63.6Deg. = 63.6deg South of East = Direction of A-B.
Use the above procedure for B-A.
Mag.= X/cosA = 106.7 / cos44.1 = 148.6N.
b. A = 79N,
B = 107N@75deg. = 107cos75 + i107sin75,
B = 27.7 + i103.4.
A - B = A to B = (27.7 + i103.4) -79,
A to B = -51.3 + i103.4
tan(theta) = 103.4 / -51.3 = -2.0156,
Theta = -63.6deg = 63.6deg South of East.
c.
Answered by
Henry
OOPS!!
Ignore data shown after part b.
It is incorrect.
Ignore data shown after part b.
It is incorrect.
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