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Tilted pivoted rods with various loads force to hold rods

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A curtain rod is a decorative item used for hanging curtains in your house. While curtain rails are very much hidden from view, curtain rods are visible and need to be attached properly in order to be aesthetically pleasing. If you are interested in attaching a curtain rod to a wooden window ledge or door frame, go through the steps given below.

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Since you will be working from a ladder, make sure that it is deployed on a level surface. Remember, a wobbly ladder can injure you badly.

Make sure that the drills are battery powered and not powered directly from the power outlets. This makes working with these tools much more convenient and eradicates danger from power cables.

Drilling into walls disperses a lot of dust and can cause an allergic reaction. A face mask can protect you from such dangers and should be worn at all times along with gloves for protection.

With these precautions in mind, move on to the attaching the curtain rod to your wall. This is a very crucial step because it determines the finish of your work. Before using the drill in any way, you need to carefully go through this step. Double check all measurements and use a permanent marker to mark where the screws for the brackets will be forced in. An electric drill is hard to handle, especially when being used while standing on a ladder.

Wear gloves to obtain a proper grip on the machine, and drill pilot holes into the marked points made in the previous step. Once you are done with drilling, use a hammer to force in anchors into these holes.

Make sure that you have with you screws that correspond to the diameter of the screw holes present in the brackets. Hold these brackets up to the wall and use an electric screw drill to fasten them. Once these are in place, attaching the curtain rod is a very simple process. With the curtain rod in one hand, climb the ladder and slide the rod into the groves of the bracket. You have now successfully attached a curtain rod to the wall. Hang curtains to complete the effect that you worked for.

We welcome your comments and suggestions. All information is provided "AS IS. All rights reserved. You may freely link to this site, and use it for non-commercial use subject to our terms of use. View our Privacy Policy here. Toggle navigation Login Register How-Tos. How to Attach a Curtain Rod to Wood.

Written by Doityourself Staff. What You'll Need. Curtain rod. Electric screw drill. Permanent marker. Measuring tape.Six identical 2-meter long massless rods are supporting identical Newton masses. In each case, a vertical force F is holding the rods and the masses at rest. The left end of each rod is held in place by a frictionless pin. The rods are marked at half-meter intervals.

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Rank the magnitude of the vertical force F applied to the rods for the six cases. Explain how you determined your ranking. The drawing shows two identical systems of objects; each consists of the same three small balls connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at Solve Prob. It swings down to a vertical position and A slender rod AB is released from rest in the position shown.

It swings down to a vertical position and strikes a second and identical rod CD which is resting on a frictionless surface. Assuming that Questions Courses. Aug 01 AM. Kriti V answered on June 02, Hello Find the Do you need an answer to a question different from the above?

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How to Attach a Curtain Rod to Wood

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tilted pivoted rods with various loads force to hold rods

Statistical Mechanics Question. Posted 7 years ago. Posted 3 years ago.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up. A uniform rod is 2.

tilted pivoted rods with various loads force to hold rods

The rod is pivoted about a horizontal, frictionless pin through one end. What is the angular acceleration of the rod at the instant it is released? Where 9. But I got none of the answers in the multiple choice Also I haven't used the radius. Any suggestions would be helpful It all follows from there. Here is a Free Body Diagram I made for you. Sum of the forces on body equals mass times acceleration at the center of gravity.

Sum of torques about center of gravity equals moment of inertia times angular acceleration. Acceleration of point A must be zero. Finally use 2. For the moment of inertia, in your book there are some lists of moments of inertia for various circumstances, find the one that corresponds to a rod being rotated about one end, or look it up on wikipedia. As you've been given an uniform rod of some length and an angle of drop, it behaves in a similar way to that of a compound pendulum.

Have a look at Wiki for an enormous drop angles relative to horizontal axis. As it's a frictionless pendulum which keeps on swinging forever unless there's air frictionthis problem wouldn't be a problem Sign up to join this community.Hot Threads. Featured Threads. Log in Register. Search titles only.

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For a better experience, please enable JavaScript in your browser before proceeding. Torque of a uniform rod being pivoted. Thread starter monkeysmine Start date Apr 21, Homework Statement A uniform rod 1. Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction.

I think realized that sin might not be valid to use due to the nature of the direction of application of the force and tried 7.

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I then plugged in the values and came up with. PhanthomJay Science Advisor. Homework Helper. Gold Member. You don't need to get into angular accelerations and moment of inertias per attempt 2.

The problem is just asking for the torque as a function of theta, per your first attempt equation. You have a couple of errors. I only tried 2 because 1 did not work. I crossed. Well I double checked my work on a calculator and I'm positive that my cross product was right. I just used my last "guesses" on this problem though, so I will talk to my teacher tomorrow about it. WebAssign is glitchy anyways, and often requires strange sig.

Thank you for the prompt help! PhanthomJay said:. Well, sig figures aside, none of your answers are correct. I inadvertently gave you the correct answer, so i've deleted it.If you apply force at an angle instead of parallel to the direction of motion, you have to supply more force to perform the same amount of work.

You can use physics to calculate how much work is required, for example, when you drag an object using a tow rope, as the figure shows. To find the work in this case, all you have to do is find the component of the force along the direction of motion or displacement. Work properly defined is the force along the direction of displacement multiplied by the magnitude of the displacement, s :.

Say that you use a rope to drag a gold ingot, and the rope is at an angle of 10 degrees from the ground instead of parallel. If you want to apply the same amount of force parallel to the ground as before, then you would need the component of your force that is in the direction of the displacement to be the same as if you were applying a parallel force — in this case, 2, newtons.

This means that. If you pull at a degree angle, you have to supply about 40 extra newtons of force. But will you end up doing the same amount of work? You would if you pulled with this force. The coefficient of friction is the same as if you are pushing the ingot, but now the normal force with the ground is given by the weight of the ingot minus the upward component of the force you supply. Therefore, the force of friction is given by. The force of friction must be smaller than if you were pulling the ingot parallel to the ground because the normal force is smaller — you can already see that you need to do less work to move the ingot.

Because you want to do the least amount of work, you want to drag the ingot across the ground with the smallest force needed to overcome friction. So set the horizontal component of your force equal to the force of friction:. If you rearrange this equation to solve for F pullyou can find the magnitude of the force you need to apply:. Now, say that you want to move the ingot to your house, which is 3 kilometers away. If you pulled the ingot straight on, you would use 7.

More force is required to do the same amount of work if you pull at a larger angle.Hot Threads. Featured Threads.

Determine the smallest angle u for equilibrium

Log in Register. Search titles only. Search Advanced search…. Log in. Forums Physics Other Physics Topics. JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding. Lifting one end of a beam. Thread starter osuaaron Start date Jan 15, Hello, This is my first question, I've been pondering for awhile now.

Example: I have a beam, it is 21' in length and weighs I am about to lift one end of this beam. I would like to know how to figure the weight I am lifting on this one end. Assume the weight is constant throughout the beam.

Thank you for any assistance. Hootenanny Staff Emeritus. Science Advisor. Gold Member. Welcome to the Forums, If this is a homework question, could I please remind you that homework questions should be posted in the homework forums.

With respect to the question, tell me what you know of torques. Well, I work for a major oil company, and I hurt by back attempting to do this. As a result of the injury, I am working on a presentation for my peers that explains how much weight we are lifting in this way.

I know what torque isaside from that not very much. I'm sort of a grunt, you could say. I would like to learn this universally so I can apply it to other beams as well, to give the guys a good idea of what they're lifting. I shouldn't be contacting my lawyer before advising you should I? Well, if you want to, that's your perogative :. I'm just looking for some help to figure this out, it's been a long time since I've done this kind of thing.

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Basically it looks like I'm trying to convert Torque into weight. My head hurts, I didn't realize it was this involved. If you lift one end of a beam that has a uniform mass distribution, you are lifting half the weight. What if there where no one to carry it on the other side?When a beam is simply supported at each end, all the downward forces are balanced by equal and opposite upward forces and the beam is said to be held in Equilibrium i.

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What do you think the reading on each would be if you place the 2kg mass just a quarter of the way along the beam i. Imagine a builder's plank resting on two pieces of scaffolding as shown The loads acting on a beam can be represented in a diagram as a single force acting on the centre of the beam - ' F'.

This total downward force must be resisted by upward forces which together are equal and opposite to it. In the diagram these are shown as Reactions R1 and R2. If either of the two reactions were to be removed, the beam can be imagined to pivot on the one remaining.

It follows that each Reaction must be providing a turning force or Torque to keep the beam in position. And we know from using a simple lever that it is not just the size of the force or effort that matters but also its distance from the pivot or fulcrum - longer levers need less effort.

The force created by the load or the effort multiplied by its distance from the pivot or fulcrum is known as the Moment of Force. Torque and Moments are not quite the same thing but both are measured in newtons x metres Nm. In the case of a Second Class Lever as shown. The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.

If the wheelbarrow is held steady i. This gives a way of calculating how much force a bridge support or Reaction has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!

The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up. The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example.

tilted pivoted rods with various loads force to hold rods

But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated. Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams or bridges have several loads acting upon them. In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.

The forces in the loaded beam example are not only important with regard to their size or magnitude but also their direction. F1 is not only 60kg but is also acting downwards R1 is not only 40kg but is also acting upwards. It is also established that if the beam is to stay in equilibrium then the sum of all the downward forces must equal the sum of all the upward forces. Forces are known as vector quantities because they not only have magnitude but also direction.

Vectors can be represented by drawing arrows. The length is equal to its magnitude and the arrow points in the direction the force is acting. The two loads on the beam can be drawn as vertical lines pointing downwards scaled such that their lengths equal F1 and F2.

Similarly the two reactions can be drawn as vertical lines pointing upwards scaled such that their lengths equal R1 and R2. The total length of the downward pointing line is equal to the sum of all the downward forces and is known as the Resultant.

The total length of the upward pointing line is equal to the sum of all the upward forces and is known as the Equilibrant - because it is the force required to balance the two loads and maintain balance or equilibrium.

This is known as Bow's Notation and it is applied simply by working around the structure clockwise and labelling the space between each force in capital letters. This is known as a Space Diagram. The direction is known from the Space Diagram and magnitudes are calculated where possible using Moments - R2 was the only force not calculated. The two lines can be merged together to produce a single Force Diagram representing all the forces on the beam and, if drawn to scale, the value of R2 can be measured.

As before, the Resultant ' ac' can be measured downwards and the Equilibrant ' ca' measured upwards note that small letters are now used to name the vectors in Force Diagrams. We also know from experience that if we walk along a builder's plank, for example, it will wobble up and down, or deflect, and we know that it is easier to bend a rule when it is held flat rather than on edge.


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