Question 4: There is a fruit shop, and the scale used is a hanging rod scale with a range of 10㎏. There is a larger watermelon that exceeds the scale range. Clerk A found another weight, which was exactly the same as this weight, and tied it with the original weight as a weight t

Question 4: There is a fruit shop, and the scale used is a hanging rod scale with a measuring range of 10㎏. There is a larger watermelon that exceeds the scale range. Clerk A found another weight, which was exactly the same as this weight, and tied it with the original weight as a weight to weigh it. When balanced, the double weight was located at 6.5 kg. He multiplied this reading by 2 to get 13 kg, and sold it to the customer as the weight of the watermelon. Clerk B expressed doubts about this weighing result. In order to test, he took another watermelon, weighed it normally with a single weight to get 8 kg, and weighed it using the double weight method of clerk A, and got the reading of 3 kg, multiplied by 2 to get 6 kg, which proved that clerk A's method was unreliable. What is the actual weight of the watermelon that clerk A sells to the customer?

Analysis: 1 Understand the basic composition of the rod scale, as shown in the figure below:

1.1C point is the location of the tick and the position of the fulcrum;

1.2A point is the hanging point of the tick (hook);

1.3O point is the zero scale point, that is, the fixed disk star, it can be on the left of the tick and the side can be on the right.

2 Set the relevant letters refer to each physical quantity

2.1. Suppose the distance between the gear (i.e., the fulcrum) of the rod scale and the hanging point A of the weighing pan is d,

2.2. Suppose the distance between the zero scale O and the fulcrum C is l0,

2.2.1 If point O is on the left of point C, that is, on the same side of the gear as point A, then l0 takes a negative value, as shown in the figure below; if point O is on the right of point C, then l0 takes a positive value, as shown in the figure above;

2.3 Suppose the distance between the scales per kg on the rod scale is K, and the weight of the weight is m0;

2.4 Suppose the weight of the scale rod and the weighing pan is G, and the tension arm of the scale rod is L (C is used as the fulcrum).

3 When there is no object placed in the weighing plate, as shown in the figure:

To balance the scale, the weight should be placed at point O. According to the equilibrium conditions of the lever, there is:

m0gl0= GL①

3 When there is an object with a weight of m in the plate, to balance the lever, the weight should be km away from point O, as shown in the figure below:

According to the equilibrium conditions, there is:

m0g(l0+km)=mgd + GL②

From the two formulas ① and ②, you can get: mgd =m0kmg

i.e.:d=km0③

4 When using a double weight to weigh an object with a mass of m, let the reading is m', and when balanced, there should be:

mgd + GL=2m0g(l0+km')④

will replace ④, i.e.:2m'=m-(m0l0/d)⑤

Therefore, when using 2m' as the weighing result, the difference between its value and the actual value is:

Δm=-m0l0/d⑥

This value has nothing to do with the mass of the object to be measured.

5 From the question, it can be seen that

5.1 Shop B Use a single weight to weigh a west claw normally. It shows that the number is 8 kg, and then use the double weight method of clerk A to weigh the melon. The number is 3kg, multiplied by 2 to get 6 kg. It can be seen that:

Δm =6㎏-8㎏=-2 kg⑦

5.2 Since Δm has nothing to do with the mass of the measured object, it can be seen that the actual mass of the watermelon sold by clerk A to the customer is:

m =2m'- Δm =13 +2 =15(kg)

Summary: The rod scale is a weighing tool made based on the principle of lever balance. The scale value is shaped according to the weight counterweight. When the weight weight is changed, the scale value will change, but it is not a simple proportional relationship.