A Physics Pulley Body, Junior High School, Junior High School Physics A Question About Pulleys?

1) The pulling force of the pulley on object B is equal to the pressure of the object on the ground.

Therefore: f pull = ps = 2 * 10 4 *

2) Fb*ob=Fa*OA is obtained from the principle of lever balance (FB and FA represent the tensile force at the B end and the tensile force at the A end, and A, B should be reduced).

And because fa=ga=10n, fb=2*10 1=20n, then analyze the pulley on the left, f left + f middle + fb = g pulley + (g b - f pull) and f left = f middle = fb

8g pulley = g B.

So g pulley = 10n

g B = 80n

Now add another C under A, and the pressure is 0, then push back 3FB=G pulley + G B=90, so FB=30N

Then from the principle of leverage equilibrium, fb*ob=fa*oa, then fa=30 2=15n

So g C = fa-g A = 15-10 = 5n

1) The pressure is equal to the pressure divided by the surface area p=f s

The pressure on the ground is p, and the area is s, then the ground has a support force of 3*10 to the 4th power. （3*104）

Let the weight of g be 1 8g by the principle of lever balance to obtain fb*ob=fa*oa (fb, fa is to represent the pull force of the b end, the pull force of the a end) the pulling force of a to a is 10n, then the pulling force of b to the rope is 20n, and the pulling force of the three ropes to the pulley is 60n

9/8)g=60n+30n=90n

Launch g = 80 n g pulley = 10 n

At this time, the tensile force f of the movable pulley on object B is 80N-30N=50N, and the pressure is exactly zero, and the tensile force without the rope should be 90N, 3 =30N, then FB2=30N, Fa2=5N, so GC, is equal to 5N

1) From the gravity of A, the tensile force of the rope at the end of B is known to be 20N, and the moving pulley and the object B as a whole are analyzed, the gravity of the 9g wheel (B and the movable pulley) and the tensile force of the three-strand rope supporting force are 60N, and the supporting force is equal to ps=3 10 4N (suspect that there is a problem with the data), so as to calculate the G wheel = 3340N (guess that the most likely is 10N, the following is calculated with 10N), and then analyze the movable pulley separately, and the gravity, three tensile forces and B's tensile force on it, list the balance formula, and calculate f=50N

2) List the equilibrium formula 3 (G C + G A) = G B + G wheel (G B = 8G wheel = 80N), so G C = 20N

Pulley and lever principle.

All ropes on the pulley are equal in magnitude, and the rest depends on other people's answers.

As shown in the figure, the movable pulley is directly connected to it by two sections of rope, the middle section of the rope and the right section of the rope are upward on the moving pulley, and the upward pull force of each section of rope on the weight (pulley weight is not counted) is f, so 2f=g.

Note: It is the upward pull of the rope experienced by the movable pulley, not the downward pull of the rope on the fixed pulley. You draw upward arrows on the ropes on either side of the moving pulley and you're good to go.

The two sections of rope in contact with the movable pulley bear the weight of the object and the pulley together, and of course the friction, and theoretically the free end of the rope is equal to one and a half of the above.

The force on the same rope is equal everywhere, and the force on different ropes is not necessarily equal.

In the first diagram, there is only one rope, which means that the tensile force of each rope is f. If the weight is in the air, then there is a G slip + G = 2F balance, if the figure is water, but also take into account the buoyancy of the heavy object, it should be G slip + G = 2F + F float to balance.

In the second figure, there are two ropes, the one on the right is fb, the one on the left is 2fb+g wheel, and the two ropes hang a weight together, so there is 3fb+g slip = g object.

It is obtained by g=mg=vg.

g Fe = 7900*

g aluminium = 2700*

The pulling force of each strand of rope is equal to the gravitational force of aluminum27n

For movable pulleys and iron blocks, the upward force is 27N*2=54N, and the downward force is 10N+79N=89N

The net force is 89n-54n=35n

So the pressure of the iron block on the ground is 35n

The area is 10cm*10cm=100 square centimeters = square meters pressure p = f s = 35n square meters = 3500n square meters 3500pac The answer is that the unit is a unit of density, not a unit of pressure.

Please click to adopt, if you are not sure, you can ask questions.

Aluminum block gravity =

The upward pulling force received by the pulley is 27*2=54n

The upward pulling force received by the iron block is 54-10=44n

Gravity of the iron =

The pressure received on the ground is 78-44=34N

Pressure = 34n

v = 10 = 1000cm = 1 dm , iron = decim , aluminum = decimeter g aluminum = m aluminum g = aluminum vg =, g iron = 78n, g wheel = 10n rope tension t = g aluminum = 27n

For movable pulleys: 2t = F iron + G wheel.

F iron = 2t-g wheel = 2 * 27-10 = 44n

For the iron block:

F pull = f iron = 44n

Pressure on the ground f=g, iron-fpull=78-44=34n, compacted area, s=10*10=100cm=

Ground pressure p=f s=34

Object A moves up a distance and the size of the corner remains the same!

Solution: Cut to the chase and talk about the corners first!

Force Analysis: When the tensile force of the rope is f, the weight of object A is ga, and the weight of object B is GBP point (first equilibrium) can be written.

ga = 2 * f * sinβ

f = gb

Derive the equilibrium relationship between GA and GB.

ga = gb * 2 * sinβ

That is, when the upper test is satisfied, the system in the diagram is in equilibrium!

When the rope goes from point P to point Q!

The pulley system will readjust to equilibrium, but there is no variable in the equilibrium test, and the value in the test is the value at the time of the first equilibrium!

In summary, it remains the same.

The height of object A In your question, there is no standard for measuring height, so I will compare the height of object A with the height of object A in the first balance and the second time, and let the distance from object A to the ceiling in the first equilibrium be ha, and the distance from the ceiling in the second equilibrium is ha'

The distance from point p to pulley is hp

The distance from point q to pulley is HQ

Then there are equations. ha = 1 2 * hp * tan at first equilibration ha'= 1 2 * hq * tan hq how much does it rise specifically Just do it badly!

Unchanged, only in this way, the force of a remains the same, and then the balance is maintained.

According to the principle of force and reaction force, the Fa and FB of the object in the original diagram A must be equal in order to be balanced, such as slowly moving the fixed point P at one end of the rope to the right to the Q point. When the whole system is rebalanced, the height of object A increases, otherwise, the reaction force of the object FA will change and its equilibrium will be broken. Only the angle direction of the rope between the object fa pulleys can be balanced at the same time, so the height of object a and the angle of the rope between the two pulleys in the horizontal direction will not change.

The included angle does not change.

Because of the balance of the three forces, the rope at both ends of object A and the rope pulling B experience equal pulling force.

Answer A is correct.

A pulley group composed of a movable pulley and a fixed pulley, a two-section rope, a three-section rope, the gravity of the movable pulley is the same, the same object is raised to the same height at a uniform speed, w has = gh, equal, and the same movable pulley is also lifted to the same height, the extra work done is equal, the total work is equal to the useful work plus the extra work, the total work is equal, so the mechanical efficiency is also equal. w A and w B indicate that the work done by the pulling force is the total work.

The answer should be helpful to you.

Picture in**? The answer may also be wrong, ask the teacher, and the teacher will definitely give you more accurate.

- Amitabha, the Boss... Where's your picture???

If a person stands on the ground, it is two. Because then only the two ropes that hold the movable pulley carry all the weight.

If a person stands in a hanging basket, it is 3. Because at this time, the two ropes that hang the movable pulley and the rope in the human hand bear all the heavy loads.

There are two pulleys in total, a moving pulley and a fixed pulley, first look at the moving pulley, the triangle is regarded as an object weighing 600N, the rope in the middle and the left hand side bear 300 forces each, and then look at the fixed pulley, the rope on the left hand side and the rope on the right hand side bear 300 forces each because it is an equal arm lever.

The title equates "strip" with "section", which is a rope that is divided into three sections, each of which is obviously subjected to force, that is, to carry a heavy load.

The basket hangs on a moving pulley and moves with it.

First analyze the force on object A Obviously, a gravity force and a pull force are balanced, so the downward pull force at point A is 10N, then point B should also be subjected to a downward pull force of 20N to balance the rod. Then the force analysis of object B, the vertical downward gravity is set to 8mg, the vertical ground upward support force n and the pressure of B on the ground are the action force and reaction force, so it should be n branch = p * s = 2 * 103 *, and object B should also be pulled by the moving pulley on him f pull, f pull + n branch = 8mg (1). Then the force analysis of the movable pulley, the movable pulley should be balanced, first of all, the rope on the 2 sides of the movable pulley should be equal, so it is subjected to the upward pull force of 20n per rope of 3 ropes, and the downward gravity of the moving pulley is mg, so mg + f pull = 60n (2) comprehensive (1) (2) can solve m and f pull.

Personally, I think the question data is difficult to calculate, so I didn't do it. The second question will not be discussed in detail.

W total = 600n * 2m = 1200j

w amount = w total - w has = 1200j-960j = 240j

Mechanical efficiency 960J 1200J*100%=80%.