The mast of a sailboat is supported at the bow and stern by a stainless steel wi
ID: 1529902 • Letter: T
Question
The mast of a sailboat is supported at the bow and stern by a stainless steel wires, the forestay and backstay, anchored 10 m apart. The 12.0m long must weighs 800 N and stands vertically on the deck of the boat. The mast is positioned 3.74 m behind where the forestay is attached. The tension in the forestay is 584 N. Find the tension in the backstay and the force that the mast exerts on the deck __________ N ____________KN Consider the ideal Alwood's machine below. When N washers are transferred from the left side to the right side, the right side descends 24.5 cm in 0.25 s. Find N.Explanation / Answer
a).
Let be the angle between the mast and the forestay.
= arctan(3.74/12) = 17.3105°
Fx = 584sin17.315° = 173.812 N
Fy = 584cos17.315° = 557.534 N
let be the angle between the mast and the backstay.
= arctan(10.0 - 3.74)/12) = 27.549°
The horizontal force exerted on the mast by the forestay must be balanced by the force exerted on the mast by the backstay:
Fx = 173.812 N
Fy = 557.534 N/tan(27.549°) = 1068.78 N
The tension in the backstay, F is
F = 173.812/sin(27.549°) = 375.804 N
b.
To calculate the acceleration based on time and distance use d = 1/2(a)(t)^2. a = 2d/t^2.
To calculate the acceleration based on the weight difference. a = 2*N*mg/10m.
Set these two expressions for a to be equal. 2(.245)/(0.25^2) = 2*N*9.8/10
N = (.245/.0625)* 10/9.8 = 4 Washers transfered.
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