Mechanisms and machines
In this subject we are going to study certain mechanical systems (mechanisms) that make possible the man to reduce the effort, as for example the handle, the pulley, the wheel, etc. The machine can be defined as a system of diverse elements next to mechanisms that carry out a task determined with the presence of force, movement and work.
A Mechanism is a set of mechanical elements that carry out a specific task within a machine We put an example.
Resistance X BR = Power X BP
Handle of 2º class. Now we have the resistance in means of the other protagonists. It gives equal as it is looked and therefore, is just like the power is to the left that to the right but always the resistance in means. Example the nutcrackers
Handle of 3º class. In this case the one is the power that is in center. Example the opener of bottles
It goes, that mess I am going away to do with the classes, certainly forgets me in the examination! Good there is a way to remember them.
The word remembers ARTICHOKE and ten in account that there is in center of the handle.
ALCACHO F A →
The first letter (1º class) is the A of support, 2º is the C of load and the F (force) 3º of 3º class.
1º In the Roman of the left, the distance to the plate from the strongpoint (BR) is 10 cm and the distance of the support to the weight is 50 cm. If we want to raise a resistance R of 4kg, to calculate the value of Power P. Solution: 800 grams 2º a nut needs a force 10 kg so that the rind is broken. If the Bp is 20 cm and BR of 5 cm, to calculate the force that we must apply for being able to break it.
|Solution: 2.5 kg
It calculates whichever 100 weights of gr., 50 gr., 10 5 gr. and gr. we needed to balance the balance if in the plate we have 675 grams 4º
135 kg 5º
A cane to fish has a length of 170 cm and the distance between the two hands that hold it is 25 cm. To calculate the force that I must apply if the fish weighs 10 kg (Obvious he is not the one of the photo). Solution: 68 kg 6º In this last example we have BR = 30 cm and BP is 20 cm. To calculate the value of R if the power is 20 kg.
Solution: 13.3 kg
Pulleys and blocks and tackle.
The pulley is composed by a wheel with a crack in the external part through which it passes the cord or the strap.
Its basic function is to facilitate the work carried out and/or to reduce the effort that there is to make to raise a load.
In this case the armor of the pulley serves to support the resistance and one of the ends of the cord is tied to a fixed point. As we can see in the figure, the pulley can raise or lower and with her the load.
If we analyzed the balance of forces, in balance, that is so that the resistance does not fall or raises, one is fulfilled that the forces that throw upwards are equal to that they throw downwards (something similar to the game of the cord, where a group throws to the right and another one to the left).
Downwards we have only the resistance but upwards it throws the tied cord but the cord where we applied the effort, therefore Resistance = T + P As the force in the two equal cords is Resistance = 2 P → P = Resistance/2
In a movable pulley, the power is midad that the resistance, P = Resistance/2
We see other cases in systems of movable pulleys whereas clause the work and equilibrium of forces (W = Force X Distance)
Case 1. We have a pulley fixes where the force of the resistance is 100 Nw. The force of the cord also is 100 Nw. If I move the cord from the end in 10 cm, the load also raises to 10 cm, being fulfilled the law of the work by which Force X Distance is always the same, that is F X d = Constant. In this case, 100 xs 10 = 100 xs 10.
Case 2. We have one determines but a mobile. The two cords of the mobile each strip with 50 is distributed to the effort upwards and therefore Nw. As the force in all the cord is 50 Nw, the force in the end also is 50 Nw. To the smaller being the force, I must move but the cord to raise the 10 cm, in this case: 100 Nw X 10 50 cm = Nw X d => the distance is 20 cm Case 3. We add a pulley but and therefore when looking over the pulley that bears the burden, we have three cords that throw upwards of the load. the force in each is 33 Nw, that is the same that in the end of the cord. The distance that there is to cross to raise the load is: 100 Nw X 10 33 cm = Nw X d => the distance is 30 cm Case 4. Another pulley is added and in this case the 25 force is now Nw and the distance 40 cm We finished seeing the blocks and tackle that are not another thing that a combination of fixed and movable pulleys that crosses them a single cord with one of its ends tied to a fixed point.
To calculate the force that there is to apply for each one of the blocks and tackle
* *Soluciones: 1º - > 100 kg, 2º - > 20 kg 3º - > 50 Nw. Important: In order to make these problems it is not enough with dividing force between pulleys because it can give error. It is necessary to explain what it happens in each pulley and to make a diagram of forces in each plane
Example of resolution of problem 2.
In this two case we have fixed and one mobile. If we divided 60 between 3 pulleys provide the solution to us but with an erroneous procedure because the pulleys of the same form cannot be considered. We must consider a plane superior that cuts to the 3 cords more or less of where they are in favor you shoot with an arrow of it forces.
The winch allows us to elevate great loads by means of a cord that is coiled in a drum when driving a crank. In addition it is a system that turns a linear revolving motion into one continuous one.
There is nothing no new and the laws of the physics it must them fulfill, therefore. The work that I do in the crank is the same that is used to raise the load.
The work is force by distance. In the part of the crank the force I do it in the crank and the distance is the arc that it crosses when I move it to the left (it observes the system of fixation of the pawl) the distance that it crosses is the length of the circumference, that is 2πR = πD If I give a complete return, the work in the crank is F X π X DF the roller also turns a return and therefore the work that is realised is R x π x Dr
As the work is the same, equaling the two expressions we have F X π X DF = R x π x Dr ==>> Force X Diameter of the handle = Resistance X Diameter of the drum. Also we can use the radius when dividing the expressions by 2 and have P X BP = R X BR That is, the same expression that before.