A pinata of mass M = 1S.48 kg hangs on a rope of negligible mass that is strung
ID: 1437173 • Letter: A
Question
A pinata of mass M = 1S.48 kg hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is D = 2.48 m, the top of the right pole is a vertical distance h = 0.357 m higher than the top of the left pole, and the total length of the rope between the poles is L = 3.36 m. The pinata is attached to a ring, with the rope passing through the center of the ring. The ring is frictionless, so that it can slide freely along the rope until the pinata comes to a point of static equilibrium. Part 1 out of 2 a) Determine the distance from the top of the left (lower) pole to the ring when the pinata is in static equilibrium.Explanation / Answer
Mass of pinata M = 15.48 kg
Horizontal distance between the poles D = 2.48 m
Vertiacal distance between the top of each pole h = 0.357 m
Total lengt of the rope L = 3.36 m
Let the distance from the top of the left pole to the ring be X m
Then the distane of the ring from top of right pole = L - X = (3.36 - X) m
At static equilibrium net force will be zero ié sum of all horizontal forces at that point will be zero , and so will be the sum of vertical forces.
Let the tension in left side of string be T1 and tension in right side be T2 .
Let the angle made by left string from horizontal be X1 and by right string be X2 degree.
Let the horizontal distance of the point from left pole be L1 hence from right pole will be (2.48 - L1 ) m
Now Cos( X1) = L1 / X and Cos( X2) = (2.48 - L1) / (3.36 - X).
(3.36 - X)Sin (X2) - X*Sin(X1) = h = 0.357
Now force balance
Horizontal direction : T1*Cos(X1) = T2*Cos(X2)
Vertical direction : T1*Sin(X1) + T2*Sin(X2) = Mg
Now we have 5 equations and 5 variables , so we easily can find the value of X.
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