One to the physics topic about leverage .

Let the length of the ruler be L, and the weight will be G ox, there is:

5 N * 1 4L + g 4 N (protruding foot weight) * 1 8L (protruding center of gravity) = 3 4g N (table foot weight) * 3 8L (table ruler center of gravity).

The solution is g = 5 newn, so the weight of the ruler is about kilograms.

Solution: set the ruler weight m kg, the gravitational acceleration takes g = 10n kg, the ruler length is 1m (meter ruler) and the pressure of the ruler body on the desktop is 0, indicating that the edge of the table is used as the fulcrum, the lever is balanced, considering the weight of the meter ruler, the center of gravity at the end of the meter ruler desktop 3 4 is at 3 8, which is the length of the left lever arm, the extension end of the meter ruler 1 4 the center of gravity is at 1 8, which is the length of the right lever arm, the weight is hung at the tip of 1 4, the lever arm is 1 4 long, and the column balance equation is as follows.

3 4*m*g)*3 8=(1 4*m*g)*1 8+5*1 4 m=

To be concise, the equilibrium formula for the direct list of forces: f1*l1=f2*l2f1=g=mg l1=1 2-1 4=1 4 f2=5n l2=1 4

Therefore mg=5n

When g = 10 n kg, m = 5 10 =

When g=,m=5

It's still a bit messy, be patient.

Set the length of the ruler l, according to the lever theorem, 5n*1 4l=3 4mg*3 4l*1 2 (center of gravity).

l approximate, mg = 40 9n

Ah, I answered wrong, ashamed, in pursuit of speed, I didn't think clearly ......

The moment is generated by friction, and the arm is the radius r, therefore.

f (Mo) = 1000

The pressure at point b f(b) = f (motor) u = 10000n according to the principle of lever f * oa = f (b) * ob

f=10000*

Sorry, in addition to f(b)*ob, there should be a drag moment where the friction of the wheel to the right of the lever is multiplied by d

So there is f*oa=f(b)*ob+f(mo)*d to get f=(10000*

It's supposed to be this, I'm really sorry just now.

(1) F1 = 3000 root number 2 is an isosceles right triangle gravity is a right angle side Force analysis diagram drawn by yourself.

f2 = 3000 is an equilateral triangle Force analysis diagram draw your own f1: f2 = root number 2 to 1

2) Because the center of gravity is higher by meters, the negative work done by gravity is 3000 * The negative work done by gravity is equal to the work done by people.

Power = total work Time = 7500 30 = 250

I read your link, you actually have a question about how much the center of gravity has increased, right? The work done by gravity depends on the change of the center of gravity, so what is the change of the center of gravity related to? Is it related to the distance the center of gravity moves?

No, the change in the center of gravity is the change in the relative height of the object's center of gravity from the ground, that is, the displacement of the center of gravity in the vertical direction. In this case, the center of gravity is raised in the vertical direction, that is, the vertical distance from the midpoint (center of gravity) of the OA to the surface of the water. In this case, it is easy to understand below.

The answer upstairs is correct, but the method is incorrect, the heavy posture sedan force of the X segment should be counted on the right instead of the left, and the correct solution is as follows:

Set the movement x, in order to facilitate the calculation, you may wish to set the length of the wire 1, weight 11 2 (1 2-x) (1 2-x) = 1 2 (1 8 + x) + 1 2xx1 8-1 2x+1 2x =1 16 + 1 2x + 1 2x x = 1 16

Then the filial piety slag moves 1 16L

At the beginning, it is the midpoint balance. The weight of the dust section on each side is g, and the arm is 1 4L (in addition, the fine weight is 2g, and the length is L).

After folding the socks in half, the weight of one side did not change, it was g, and the strength arm became half, which was 1 8L, and the other side, there was no folding, and the weight was like a school of arguments, and the strength arm was 2L

To balance, move x.

1/4*l-x)g=(1/8l+x)g

x=1/16*l

Mobile 1 16l

When the person sits and leans back, the size of G remains the same, the force arm OE becomes larger, and the size of the force arm OH does not change, so the pulling force F becomes larger, so the positive option B and the negative option C;

For option D: "The wide bag is locked on the calf to enlarge" is obviously to increase the force area of the calf, to reduce the pressure on the calf from the outside world, and to make people feel more comfortable, which is the same as sitting on the sofa instead of sitting on the bench.