Find a physics problem about fixed pulleys

Hello, I seem to have seen this:

The upper one is the tensile force of object A and the lower object B B is 6N, and now B moves at a uniform speed on a smooth plane, and the frictional force experienced by A is found.

It is known that the mass of a large wooden block is m, and the mass of a small wooden block is m The friction between a large wooden block and the ground is u The friction between a small wooden block and a large wooden block is u Find 1 How big is the pulling force f when you can pull a large wooden block? 2. What is the pulling force of the fixed pulley against the wall? （g＝1o)

Lifting the weight with a block of pulleys, when the free end of the rope is pulled down 2m, the height of the object is raised, and the mass of the lifted weight is 20kg. If the weight of the pulley and the frictional resistance are not counted, the force of pulling the rope is ( ).

The movable pulley and the fixed pulley are essentially a deformed lever, among them can be regarded as the equal arm lever is ( ) The diameter of the small wheel of a pulley is 6cm, and he is used as a movable pulley, and its power arm is cm, and the resistance arm is cm

Object A is on top of object B. The coefficient of friction between AB is U1, the coefficient of friction between B and the ground is U2, and the mass of AB is M and M respectively.

The gravitational force of the wooden block is 100n pulley weight and 400n to find the tensile force f

There are many classic questions in the physics weekly newspaper, and when I took the high school entrance examination last year, the example questions taught by the teacher were all on them.

I know this question.

The upper object A weighs 200N, the lower object weighs 300N, the dynamic friction factor between AB is, and the dynamic friction factor between B and the ground is, the rope bypasses the fixed pulley and connects with the two objects of AB, and pulls the B object with a tensile force of F=300N, and finds the tensile force of the rope when the object B moves at a constant speed.

Idea: Because the velocity of object B is to be uniform (the two of them are on a line and have the same velocity), the acceleration of both of them is the same, and both are 0You can use the holistic approach.

AB is regarded as a whole, then the force between AB is not counted, B is subjected to the left pull force = 300N, and the right friction force is 240N, then to make the balance, then two ropes should provide 60N, and one rope pull force should provide 30N

So the rope pull is 30n

This problem can also be solved by the isolation method, but it is troublesome, and the whole method is generally used, which is the general way to solve this pulley.

If you have a diagram, you can create your own scenario, set up unknowns, and then use letters to calculate all the quantities that can be calculated, which is much better than doing a problem.

There is a smooth fixed pulley fixed, a light rope has a weight A weighing mg on the left side, and a B weight weighing 3mg on the right side, first hold it with your hand, when you remove the hand, how much tension is the rope?

Object A is placed on object B, and the two objects are connected by a rope around the fixed pulley, and find out how much force can pull object B

Workbench stress: downward workbench weight g1, human pressure g man - 100n = 600n - 100n = 500n, upward force a pulley upward pull f and ground support force on the workbench is 450n

So G1+500N=F+450N, i.e. F=G1+50Na Force on the pulley rope Fa=(Ga+F) 2=(100N+G1+50N) 2=75N+G1 2

b. The force on the pulley rope is 100N=(GB+FA) 2=(50N+75N+G1 2) 2

The solution yields g1=150n

Solution: Let the tensile force of the rope on the movable pulley a be fa, and the tensile force of the rope on the movable pulley b be fb, as can be seen from the figure, fa=2fb, fb=100n, the total weight of the workbench, the movable pulley and the person:

G total = G person + G platform + G wheel, the tension of the workbench, movable pulley and people:

F pull = 3FB + FA = 3FB + 2FB - GB = 3 100n + 2 100N-50N = 450N, G total = F pull + F branch = 450N + 450N = 900N, G set = G total - G wheel - G man = 900n - (100n + 50N) - 600n = 150N

So the answer is: 150

Answer: 150n

Tips: Overall analysis, the tension of the rope in the human hand is 100N, the tension force in the leftmost rope is 150N, and the total upward tension is 450N, and the supporting force of the ground is used to balance the A, B pulleys and the person and the workbench, so the workbench weighs 150 300 450 100 50 600 150N

Solution: (1) The distance traveled by the person is s=vt=;

There are 2 strands of rope on the movable pulley, so the distance of the object moving is l=1 2s=25m;

Friction force f = useful work w = fl = 600n * 25m = 15000 J2) w total = w useful 75% = 15000j 75% = 20000 jw total = fs f = w total s = 20000j 50m = 400n3) p = w total t = 20000j 100s = 200w Answer: (1) The useful work done by a person is 15000 J (2) The pulling force of a person is 400N (3) The power of a person is 200W

There is a diagram showing that the displacement of an object and a person in the same time is the same.

1 useful work = the work done by friction on the object, the cargo movement distance s=vt=, friction f=, so w useful = fs=600*50=30000j

2. The mechanical efficiency is 75%, so the total work done by a person w=w is useful 75%=40000j, fs=w, so f=w s=800n

3. Power p=w t=400w, or p=fv=800*I did it wrong. Lose face...

Because the pulling distance of the free end is 4 times the lifting distance of the object, this pulley group is composed of 4 sections of rope to bear the moving pulley and the heavy object, and the tensile force f can be obtained according to f=(g object + g motion) 4.

As can be seen from the question, the number of rope segments is n=s h=2m segments).

Excluding the weight of the moving pulley and the frictional resistance, then f=1 4g=1 4mg=1 4*20kg*10n kg=50n

When pulled down 2m, the height of the object is raised, then the number of movable pulleys is (2.

The force of pulling the rope is 20 Newtons).

When an object weighing 10 N is placed in the right disc and an object weighing 6 N is placed in the left pan, m happens to move at a uniform speed to the right, indicating that the leftward friction of the object is 10n-6n=4n

To make the object move to the left at a uniform speed, the direction of friction is to the right, and the pulling force of the rope should be 4n+10n=14n

Therefore, an object of 14n-6n=8n should be added to the left disk.

20 N, because m happens to move at a constant speed to the right, indicating that the friction of the table is exactly equal to 10 N.

Force analysis is simple, you just need to grasp this point: the friction of the table against the object is always opposite to the direction of the object's motion. Therefore, in the front and back cases, the friction force of the table against the object is in different directions.

Spring dynamometer 2 shows 60N;

The frictional force experienced by a is 60N

This question should have: When moving along a horizontal plane (constant velocity) under the action of a pulley, these two words should be added.

Use movable pulleys to save half the effort:

2f1=f2

f2=2*30=60n

According to the motion of the object a along the horizontal plane (constant velocity): a is balanced by force.

Yes: f=f2=60n

2.60 Ox.

3. If it is advancing at a constant speed, then the friction is 60 Newtons.

..6 can be explained:

To maintain a balance, C should regard A and B as a whole, and C and A + B are of the same quality;

Looking at a and b again, in order to balance the two, then ma=mb=3kg, then mc=ma+mb=6kg

3 words. I can't explain it well, it may be that object A is negligible, and on the whole it is necessary to meet mc=ma+mb=0+3=3kg

It's been 4 years since I touched physics.

Keep your balance.

In the first case, A and B are in equilibrium ma=mb=3kg, and in the second case, A and B are in a uniform acceleration state, and A and B accelerate in opposite directions. If MC=3kg, that is, the right side of the pulley.

The tension of the rope is 30N, to balance C, the tension of the rope on the left side of the fixed pulley should also be 30N, and the tension of the rope on both sides of the moving pulley is 15N, and the direction is to pull the pulley downward. That is, the pulling force of the rope on b is 15, and the direction is upward; The pull of the rope on a is 15 and the direction is upward.

And the gravitational force of b is 30, so b accelerates downwards with an acceleration of a1=f m=(30-15) 3=5m s 2

Let the mass of a be m, and a is accelerating upwards, which is the same as the acceleration of b, i.e., a2=a1=515-mg=m*a2

m=1kg

Fixed pulley There are three strands of rope Movable pulley Suction cup Extractor - The minimum force provided by the suction cup is equal to the weight of the glass itself, and the pressure = area x pressure formula, mass = can be used

To equilibrium c, then a and b must be balanced first, a+b=6=c

The rope on the picture can obviously be seen as 3 sections, but the length of each section will change

When B goes down to the hall, the right end of the rope (the distance between the pulley and the ceiling connected to B) increases, and the same goes for the left side, and the direct distance between the two pulleys increases, and the total length is a certain amount because the rope is still the original rope. Those two paragraphs are long together, so the last remaining paragraph will be shorter.

So, the distance a moves on the desktop is.

In this problem, you can first ignore the fixed pulley. The force f of the object carried by pulling the starting pulley only needs 1 2 of the weight g.

Understood in this way: put a heavy object g on the ground, and a worker on the 5-meter-high floor pulls up with the force of f applied by a movable pulley, then the rope length is 10 meters. When G rises 5 meters, the rope is also used up by 10 meters, and G is lifted up 5 meters with a distance of 10 meters.

Set it as the ideal pulley to play the key (pulley that does not do extra work). Then, pulling G up 5 meters directly is equivalent to lifting the rope up 10 meters with a moving Lianghui pulley. Then the force f of the movable pulley must be the force of 1 2

Twice the distance of action does the same work, and the force is definitely halved. f=g 2, therefore, g rises a certain distance, and the worker has to pull up twice the distance.

Going back to the question, imagine A as the worker, and from the above analysis, we can conclude that if B is going to rise by meters, then A is going to pull up, which is equivalent to going backwards, because they are moving the same distance. So B goes down, A goes back.

It's very simple, b, the object is caught by two pieces of rope, gravity is divided into two parts, and the nature of the pulley is; Labor-saving distance Saving half of the effort of the mountain is half of the distance spent twice as much The answer is.

When the object or cavity year b moves, the rope that moves the pulley will move the shirt 2 The rope of the fixed pulley (i.e., the rope that connects the circular object A) also moves, so the object A moves, understand?