The portion of the cable attached to the spring is at an angle of ? =35? from th
ID: 1404254 • Letter: T
Question
The portion of the cable attached to the spring is at an angle of ?=35? from the ground. The spring has a spring constant of k=14.5 kN/m . The pulley is attached to the ceiling by rod BD that forms an angle ?=62.5? with the ceiling.
1) Find the weight of the hanging mass when the spring, with spring constant 14.5kN/m , has been stretched 12.7cm . The tension in rod BD is3267 N . Recall that the pulley is massless.
Express your answer to three significant figures and include the appropriate units.
2)
Determining the displacement of a spring
The cable segment attached to the spring is at an angle of ?=50.0? from the ground. The spring has a spring constant of k=14.5 kN/m . The pulley is attached to the ceiling by rod BD that forms an angle ?=70.0? with the ceiling. The hanging mass causes a tension of 3461 N in rod BD. What is the displacement of the spring?
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
1) We assume that the tensions in all the rods are distributed in an uniform way. The tension of the rod BD is given, and is 3267 N.
The tension of the rod BC will be equivalent to the force that the spring is applying to the rod, since the system is in equilibrium. This force, according to Hooke's law, is F = k*X, where k = 14.5 kN/m and X = 12.7 cm. We convert all the units to N and m, therefore k = 14500 N/m and X = 0.127 m. With this,
F_{BC} = 14500 N/m * 0.127 m = 1841.5 N.
Finally, we note that at the point B all the tensions of all the rods must be in complete equilibrium. This means that the tension of the rod BA must be equal to the summatory of the vertical component of all the other tensions. In order to find these components, is that we need to project with the angles given. The angle of the segment DB is heta = 62.5, and the projection will be sin(62.5º)*3267 N = 2898 N.
Meanwhile, the vertical projection of the segment BC is sin(35º)*1841.5 N = 1056 N.
Notice that the vertical projection BD is directed upwards, while the BA and BC vertical projections are downwards, therefore
BD = BC + BA,
2898 N = 1056 N + BA,
BA = 1842 N.
This implies that the mass mA is
mA * g = 1842 N,
mA = 1842 N / 9.8 m/s^2 = 187.95 kg.
2) In this part of the problem, all we have is the tension of the rod BD, which is 3461 N. In the point of intersection, we have that the summatory of all the horizontal forces must be in equilibrium. We know that the rod BD has a horizontal projection of the tension of cos(70º)*3461 N = 1184 N. The tension of the BC rod is equivalent to the force produced by the displacement X of the spring, which is 14500 N/m * X, where X is measured in meters. We project this force in its horizontal component, obtaining cos(50º)*14500 N/m*X = 9320 N/m * X. Therefore, since the forces point in opposite directions, it is satisfied the following equality:
9320 N/m * X = 1184 N,
This implies that
X = 0.127 m = 12.7 cm.
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