A physics problem of Fangshan s pulley group, and a physics problem of pulley gong

Analysis: G 30N, Sen Carry N 2 Section, T 10S, G Wu Zhichun Hui 200N, H 5m

Tensile force: f (g matter g moving) n (200n 30n) 2 115n;

Tensile work: W FS rope FNH 115N 2 5M 1150J;

Pulling power: P T 1150J 10S 115W.

Since there are 3 ropes.

And the total head macro mass is 90 2 20 60

The pull of the rope by the master worker is 30N

Each rope bears 30n).

w＝mgh＝90x10x10＝9x10＾3j

Solution: Because there are two ropes on the movable pulley, it is said that the two ropes bear the total weight of the car.

Because it is stationary at 10m, F pull = 1 2g total = 1 2m total g = 1 2 (2kg + 8kg + 20kg + 60kg) * 10n kg = 450n

Because there is a force of 1 2 left, it takes 2 times the distance, so the S rope = 2H = 2 * (20m-10m) = 20m

W=F Pull S rope = 450N*20m=9*10 3J Answer: When the workbench is stopped at a height of 10m from the ground, the pull force of the worker's master on the rope is 450N, if the workbench rises from 10m to 20m, the worker's master will dig the hole and do less Gongna sail core 9*10 3J

The rope that bears the weight of the object and the moving pulley in Figure B is 3 sections.

According to the formula f=g n (g is the sum of the weight of the body and the weight of the movable pulley, n is the number of segments of the rope that bears the weight of the object and the weight of the moving pulley).

1. The tensile force f=100N, the weight of the object is 260N, and it is estimated that F=(G + G) N gets G=3F-G = 3x100 N - 260 N = 40 N

2.When lifting an object weighing 500 N, the pulling force f = (500 N + 40 N ) 3 = 180 N

3. When the object is lifted, the distance of the free end of the rope is s=

A and B should be different, and the first one is A.

1. (Look at Figure B) From the 3 sections of rope, we can know that 3*100 300N, minus the weight of 260N, 300N-260N=40N

2. The weight of the object plus the weight of the pulley = 500n + 40n = 540n, divided by the number of rope segments, it can be obtained, 540n 3 = 180n

3. Because there are 3 sections of rope, the object is lifted, so the free end movement distance is 3

It's time to do a comprehensive multiple-choice question.

Since the right side is a movable pulley, A moves, B only moves meters, and similarly, the force is only half of MB!

So f=147n, the velocity is, so p=f*v= is troublesome to explain, don't explain.

The truth will help you see, but I forgot everything in junior high school, and I can't help it.

The idea of this question tells you.

Use mechanical efficiency = g (g + g) = 150n (150n + g motion) = 60% to find the power of the moving pulley, g motion shortcode = 100n. then you can ask for it.

When hanging a 250N weight, mechanical efficiency = 250N (250N + G moving) = 250 (250 + 100) =

w has =gh=250*2=500j

w total = 500

w has this wild work = 250 * 2 = 500;

Mechanical efficiency = 60%;

Therefore, w total power = w useful power Mechanical efficiency = 500 60% = 833

The picture is so small, besides, I'm lucky!

f=30n..

Regard the fixed and moving pulleys as one with the weight and the rope... And then on this whole afterburner 30N

So the ceiling there rally 110n

First take the moving pulley as the research object, the resultant force of the line on both sides of the movable pulley is the weight and the pulley weight is 60N, and then the fixed pulley is the research object, you look at the figure in reverse, it is a moving pulley, obviously for the same section of rope on the wrestler is the same size, it is also 60N, then the force on the ceiling is the resultant force plus the gravity of the pulley, that is, 120N + 20N = 140N

1.The research object is people, baskets, and movable pulleys as a whole.

Three ropes pulled them up.

f=(g man+g basket+g wheel) 3=(500+100+0) 3=200 N2Take people as the object of study.

f+n=gn=g-f=500-200=300 N.

First of all, what about the figure?

Secondly. Solution (Huai shed 1) side Ming noisy useful work = weight of the object * height of the object being pulled up = 90000n * 1m = 90000j

Total work = work with transport cover Mechanical efficiency = 90000j 80 = 112500j tensile force = total work The distance the rope is pulled = 112500j 3m = 37500n2) speed = tensile force The power of the tensile force = 37500n 2500w = 15s and finally ends.