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A 19000 kg sailboat experiences an eastward force 16400 N due to the tide pushing its hull while the wind pushes the sails with...Asked by Big Poppa Smurf
A 15700 kg sailboat experiences an eastward force 22000 N due to the tide pushing its hull while the wind pushes the sails with a force of 80900 N directed toward the northwest (45◦ westward of North or 45◦ northward
of West).
A) What is the magnitude of the resultant acceleration of the sailboat?Answer in units of m/s^2
B.) What is the direction of the boat’s acceleration? Answer in units of ◦ (N of West)
of West).
A) What is the magnitude of the resultant acceleration of the sailboat?Answer in units of m/s^2
B.) What is the direction of the boat’s acceleration? Answer in units of ◦ (N of West)
Answers
Answered by
Henry
Fr = 22,000N @ 0o+80,900N @ 135o =
Resultant force.
A. X = 22000 + 80900*cos135 = -35,205 N.
Y = 80900*sin135 = 57,205 N.
Fr^2=X^2 + Y^2 = (-35205)^2 + (57205)^2
Fr = 67,170 N = Resultant force.
a = Fr/m = 67170/15700 = 4.3 m/s^2.
B. tanAr = Y/X = 57205/-35205=-1.6249
Ar = -58.4o.
A = 180 - _58.4 = 122o, CCW = 58o N of West = Direction.
Resultant force.
A. X = 22000 + 80900*cos135 = -35,205 N.
Y = 80900*sin135 = 57,205 N.
Fr^2=X^2 + Y^2 = (-35205)^2 + (57205)^2
Fr = 67,170 N = Resultant force.
a = Fr/m = 67170/15700 = 4.3 m/s^2.
B. tanAr = Y/X = 57205/-35205=-1.6249
Ar = -58.4o.
A = 180 - _58.4 = 122o, CCW = 58o N of West = Direction.
Answered by
maddie
where did you get 135 degrees when finding the resultant force?
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