A physics problem The principle of leverage, the principle of physical leverage

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c Pair. Because in the process of lifting the object, the force arm of the object's gravity gradually increases, so f also increases.

Select C, you can draw the picture, and the repercussions extend the power f to find that the power arm is reduced. According to the principle of leverage equilibrium, f should be increased.

According to the lever theorem: f1l1=f2l, the effective distance of the heavy object becomes longer in the process of the weight rising, and the effective distance of the pulling force remains unchanged.

(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.

To be honest, this question is difficult, and the quality of the hanging plate should be considered.

1) Assuming that the quality of the hanging plate is negligible, the position of the hand pull ring on the scale rod is O, and the position corresponding to the hanging wire of the hanging plate is Aoa|=a, the position corresponding to the hanging wire of the weighing stone is b, and the position is |ob|=b, let the weight of the scale weight be m, for the single scale, the object of weighing 1kg satisfies a = MB, in the same way, the object of weighing 2kg satisfies 2a = 2MB, that is, the position of the hanging wire corresponding to the weighing stone is exactly 2 times that of the hand pull ring when weighing 1kg, that is, 2b, so the double scale stone method weighs 2kg of small watermelon, then xa = 2b * 2m, gets x = 4kg, and uses a single scale to weigh 5kg, obviously the mass is not equal, so the quality of the hanging plate can not be ignored, 2) set the quality of the hanging plate is k, for a single scale, the object of 1kg is called to meet (1+k) a = mb, in the same way, the object of 2kg is to meet (2 + k) a = mb*, so b * = (2 + k) a m, because the mass weighed by the single scale method is the exact value, so the mass of the small watermelon is 5kg, and the double scale method weighs the small watermelon is 2kg, and the lever balance principle obtains (5 + k) a = 2mb*, that is, (5 + k) a = 2m (2 + k) a m, and the solution is k =1, that is, the mass of the hanging plate is 1kg, the mass of the large watermelon is x, and then the larger watermelon is 8kg by the double scale method, and the distance b from the point of 8kg from the position o of the hand ring is calculated first, and the formula (8 + 1) a = mb is obtained b = 9a m, then.

x+1)a=2MB, that is, (x+1)a=2m9a m, and the solution is x=17

You can assume that the number of views is 1kg = 1cm, the number of views is uniform, and the double scale method is 1cm = 2kg, but it has a hanging plate, so the zero scale line is set to a certain distance from the fulcrum, and the distance between the hanging plate point and the fulcrum is l, and the weight of the scale is m, and the equation can be listed:

m（5+a）=5l ； 2m(2+a)=5l ；

The solution is a=1cm. So you use the double scale method to move towards the fulcrum with zero scale. Double scale method. In this way, the result obtained by the double scale method is to multiply the number of sights by 2 plus 1.

So 8*2+1=17.

Analysis: To solve this problem, we only need to use the principle of four rods, which is recorded as From daily life experience, we know that the scale of the rod scale is uniform, the zero line does not coincide with the fulcrum, and the reading is proportional to the distance from the point of action of the scale force to the zero line.

Solution: We set the weight of the weighing pan as m, the weight of the weighing pan as m0, the mass of small watermelon as m1, the mass of large watermelon as m2, the corresponding arm of the hanging pan as l, the arm corresponding to 0kg as l0, the weight arm corresponding to 2kg as 2l+l0, 5kg as 5l+l0, and 8kg as 8l+l0.

Small watermelon measured by double scale method: 2m (2L+L0)=(M1+M0)L

Measuring small watermelons by single scale method: m (5L + L0) = (M1 + M0) L

From the above two equations, l0=l is obtained.

Measuring large watermelon by double scale method: 2m (8L+L0)=(M2+M0)L

When there is no load: ml0=m0l, get ml=m0l

Substituting L0=L into the formula, we get 6ml=(M1+M0)L, and the formula, we get M1=5M0, so M0=1kg.

Substitute L0=L into the formula to obtain 18ML=(M2+M0)L, and substitute the formula, to obtain M2=17M0=17KG.

There is a problem with the landlord's question.

If the power is doubled by the double scale, the power arm should be reduced to half of the original size. The resistance and resistance arm are constant, so the smaller watermelon should be 4kg on a single scale. How is 5kg?

First, calculate that f is a downward discrete, and secondly, it can be determined by the equilibrium principle f1l1=f2l2.

Calculate the downward force at point b, and this force plus buoyancy is gravity.

The force f is at an angle of 30 degrees to the lever, f=3n, so the downward component is , fl=fl, so f=3n.

Because g = f + f buoyancy and f buoyancy = 100 1 , g = f + f buoyancy =

The downward tensile force at the end of a is f=fsin30°=3n*1 2=the principle of lever equilibrium, f*ao=f*aobo, f-pull=f*aobo=the object is buoyantized, f-float=pvg=1000kg, m 3*100*10, -6m3*

Weight g = f pull + f float = 3n + choose b

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.