Pulley statics equilibrium

Frictionless pulley system. What's the tension in each rope and the force P, such that the system is in equilibrium? Side question. Why is the tension in the rope around the pulley constant? From an assignment a while ago. Always sucked at pulley questions, but I think I should probably know how to solve them.

Help, hints, directions to some helpful resources - all welcome. Thanks a bunch! I always sucked at pulley problems too, Maybe I still do. But I think I can muddle through this one. First, your side question: good question!

It isn't intuitive to me either. But think about it: at what point would the tension dissipate from the rope? I think you can generalize that as long as there are no "corners" in the rope that is, nothing other than smooth curves, as around a pulley that the tension is constant.

Maybe you just have to put it in the "faith" column.

pulley statics equilibrium

Let's call the pulleys A, B and C from left to right. Given a force of P, the tension in the rope supporting pulley C must be 2P.

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P down on the right, therefore P down on the left, therefore 2P up on the pulley. Therefore the tension in the rope looping over pulley B and supporting pulley A is 2P. The rope on either side of pulley A still has tension P, meaning that there is a total of 4P holding up pulley A, which in turn holds up N. I am an Electronics Engineer and my daughter a civil engineer but the math in AE is an order of magnitude past what we had to do.

Because AE is truly rocket science. This is the main reason they get paid more than most other engineers. Vii Lv 4.

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Answer Save. Woodsman Lv 7. Favorite Answer. Statics Pulley Problems.Hot Threads. Featured Threads. Log in Register. Search titles only. Search Advanced search…. Log in. Support PF! Buy your school textbooks, materials and every day products Here!

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An Asymmetrical Static Equilibrium Physics Problem Involving Pulleys and Hanging Masses

Thread starter Violagirl Start date Oct 28, Homework Statement Hi, I am confused by this physics question and I was wondering if someone could help me out. I have a system with 2 pulleys and 3 weights.

They are all held together by one string, the two weights on the outside of the two pulleys, both weighing the same, hold up the weight in the center. The system has a central object B with mass M which is suspended half way between two pulleys by a string. The whole system is in equilibrium. The objects A and C, which are on the outside of the pulleys, have the same mass of m. What I'm trying to figure out, what is an equation for the vertical displacement of the central object B in terms of the horizontal distance between the two pulleys Lthe mass of object B Mand the mass m of objects A and C?

Make a drawing first. Show the data given. Then draw the forces acting on each weight. The system is in equilibrium so the vector sum of forces at each weight has to be zero. When it comes to determining how the angle changes, that's what I find to be difficult in putting it together.

Why does anything change when the system is in equilibrium? Nothing moves. Can you show your picture, or should I show it? Oh yes, that is true, so then the angle would remain consistent. Attached is my picture for it along with the forces that are significant. Lab 3 Problem 4. Do you need to find how d is related to the masses and to L? Draw the forces at each weight. The tension in the rope is the same everywhere.

Do you know its magnitude in terms of m? Some more hint: See attachment. Thanks for your drawing! Violagirl said:.An interesting phenomenon in physics, and physics education, is the simplicity of symmetric situations, compared to the complexity of similar situations which are, instead, asymmetrical. Students generally learn the symmetrical versions first, such as this static equilibrium problem, with the hanging masses on both left and right equal.

Here is the setup, using physical objects, rather than a diagram. The masses on the left and right are each g, or 0. Specifically, downward tension in the strings must be balanced by upward tension, and the same is true of tension forces to the left and to the right.

In the symmetrical situation, all that is really needed to solve the problem is the fact that the vertical forces are in balance. These answers can then be checked against the physical apparatus. The error here could be caused by several factors, such as the mass of the string itself neglected in the calculations abovefriction in the pulleys, or possibly the fact that the pulleys did not hang straight down from the hooks which held them, but hung instead at a slight diagonal, as can be seen in the second image in this post.

This is testable, of course, by using thinner, less massive string, as well as rigidly-fixed, lower-friction pulleys. However, reducing the error in a lab experiment is not my objective here — it is, rather, use of a simple change to turn a relatively easy problem into one which is much more challenging to solve.

How to calculate tension in a multiple pulley system

In this case, the simple change I am choosing is to add 50 grams to the g already on the right side, while leaving the central and left sides unchanged. This causes the angles where the strings meet to change, until the situation is once more in static equilibrium, with both horizontal and vertical forces balanced.

With the mass on the left remaining at 0.

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As can be seen above, these derived values are close to demonstrated experimental values. Like Liked by 1 person. Like Like. You are commenting using your WordPress. You are commenting using your Google account.

Statics - Rod and pulley

You are commenting using your Twitter account. You are commenting using your Facebook account. Notify me of new comments via email. Notify me of new posts via email. Like this: Like Loading In calculus we may have used it to simplify integrands. Leave a Reply Cancel reply Enter your comment hereThe system is in static equilibrium.

The tension of the string and the gravitational force the weight act on the mass m 2. The mass of the pulley is negligible, so we will not use it to solve the problem.

The rod is subject to the weight, the tension of the rope and the two components of the reaction for the joint that we will calculate separately. To calculate the tension of the string we consider only the right part of the figure, that is, the forces acting on the mass m 2. Since the pulley has a negligible mass, the tension of the rope that holds the rod will have the same magnitude.

Taking into account the external forces represented in the previous figure that act on the rod, the first condition is:. We now project on the x and y axes taking into account the positive directions defined in the figure.

We have represented the angles necessary to calculate the projections of the forces on the axes in the figure below:.

We therefore need a third equation to solve it, which comes from imposing the second condition of static equilibrium on the rod. This condition must be fulfilled for any point that we take as the origin of torques. To solve this problem we will use the point A represented in the figure.

And where r is a vector that goes from the origin of moments to the point of application of the force.

Statics: A pulley system in equilibrium

The first two terms of the previous expression are null because the forces R x and R y are applied at point A and, therefore, the vector r is null for both. We have represented both vectors in the following figure. And using the right-hand rulewe determine that the direction of the cross product is perpendicular to the screen and inward -k. The weight has been moved to point A in the figure in pink so that it is easier to see the direction of the cross product.

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And using the right-hand rulewe determine that the direction of the cross product is perpendicular to the screen and outward k. The equations 12 and 3 will allow us to solve the problem.

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For convenience we will write the three equations together here:. The minus sign indicates that R y has a sign opposite to the one represented in the figure. Determine: the tension of the rope. Ad blocker detected Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics.Soccer matches as a result soccer tips. Consequently free fixed matches tips. Gambling either paid www betting expert tips matches.

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Statics: Pulley system in equilibrium?

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pulley statics equilibrium

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pulley statics equilibrium

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pulley statics equilibrium

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