A piñata of mass M=12.0 kg hangs on a rope of negligible...

A piñata of mass $M=12.0 \mathrm{~kg}$ hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is $D=2.00 \mathrm{~m},$ the top of the right pole is a vertical distance $h=0.500 \mathrm{~m}$ higher than the top of the left pole, and the total length of the rope between the poles is $L=3.00 \mathrm{~m} .$ The piñata 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 piñata comes to a point of static equilibrium. a) Determine the distance from the top of the left (lower) pole to the ring when the pinata is in static equilibrium. b) What is the tension in the rope when the pinata is at this point of static equilibrium?

A piñata of mass $M=12.0 \mathrm{~kg}$ hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is $D=2.00 \mathrm{~m},$ the top of the right pole is a vertical distance $h=0.500 \mathrm{~m}$ higher than the top of the left pole, and the total length of the rope between the poles is $L=3.00 \mathrm{~m} .$ The piñata 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 piñata comes to a point of static equilibrium.
a) Determine the distance from the top of the left (lower) pole to the ring when the pinata is in static equilibrium.
b) What is the tension in the rope when the pinata is at this point of static equilibrium?

Key Concepts

Geometric Constraints in Equilibrium Problems

In problems where connections occur via ropes or cables of fixed length, geometry plays a fundamental role. The physical setup often involves relationships between the rope length, the positions of fixed points, and the angles formed by the rope with horizontal or vertical axes. This geometric relationship is key to writing equations that relate forces (through their directional components) to the constraints imposed by the system.

Frictionless Constraints

A frictionless constraint, such as that provided by a sliding ring or pulley, means that there is no resistive force opposing the relative motion between the object and the constraint. This allows for the tension to remain consistent along the rope and ensures that only gravitational and tension forces need to be considered in the analysis. The absence of friction simplifies the equilibrium conditions considerably.

Static Equilibrium

This concept involves the condition where an object at rest has zero net force acting on it. In problems like this, all vector forces, including gravitational force and tension from the rope, are balanced. This ensures that there is no acceleration, allowing the object (or the ring in this case) to remain at a constant position.

Tension in Ropes

Tension is the force transmitted through a rope, string, or cable when it is pulled tight by forces acting from opposite ends. In systems with massless ropes and frictionless pulleys or rings, the tension is uniform throughout each segment of the rope, which simplifies analysis as the same magnitude of force is involved in balancing the weight.

Transcript

00:01 So for this problem, we are given the situation that is shown in this figure with a piñata of mass capital m and that is equal to 12 kilograms and the distance between the poles is a distance capital d equal to two meters the difference between the poles is a distance capital d equal to two meters the difference between the pole height, this difference right here, h, is equal to 0 .5 meters, and the rope length is equal to 3 meters.

01:09 Now, the piñata is in equilibrium.

01:16 So we know that the forces, the net forces in the eds and the white company must be equal to zero.

01:25 So for part a of this problem, what we need to determine is the distance from the top of the left, of the left lower pole to the ring between the piñata is in static egg inhibition.

01:42 So we need to find the, um, distance d, this distance d in here, we call this distance d, then if this is the distance d, then this distance because the whole rope measures l, then this remaining sediment in here is d minus and with that we also are going to have that this angle, we have that angle in there in this other angle in here and with that said we're going to call this the angle theta 2 and this theta 1 now trigonometry can be used to find the value of d because that's the one and we want to find and of course oh of course we need that and that distance but also we need to calculate the angle so that we can calculate that distance d.

03:08 Now, in here we will have three forces that are opting on this, and those are the tension, attention to the right, we're going to call that t, and attention to the left.

03:33 And we also have the weight of the piñara.

03:37 Which is the mass times acceleration due to gravity.

03:43 So what we can do is to apply newton's second law to the forces acting on the x component.

03:55 So as you can see, the only ones that are contributing to that are the tensions.

04:02 So in this case, we are going to set that the one that is to the right is positive.

04:09 So we have the tension cosine of theta 1 because that will give us the x component of that tension minus the one that is directed to the left, which is the tension cosine of theta 2.

04:26 And this is equal to 0.

04:32 Now from here is easy to obtain that if we pass this to the other side and cancel the temptions, will obtain that tension 1 is equal to tension 2 that we are just going to call theta for both.

04:50 So with that set we can now use trigonometry in order to calculate the distance d.

05:03 So from that we are going to have the following.

05:11 Let me just see in here in the picture so we have this total distance d so we can write a total distance d as the cosine of theta plus this other segment in here plus l plus well minus d if i'm correct yes cosine of teta so from here we obtain you can see that these cancel with this one so we obtain that d is equal to alt cosine of theta and from there we can obtain the angle.

06:14 So the angle theta is equal to cosine of minus 1 of thee over alt.

06:26 From there we can obtain the angle so we can substitute the values for that.

06:32 Remember that the distance d is given and that distance is two meters.

06:40 At the distance l is also given, that distance is three meters.

06:45 So from there we obtain a value of 48 .19 degrees.

07:00 So now we can use another trigonometry relation in here...

Please give Ace some feedback