As shown in the figure, two objects a and B are stacked on a smooth horizontal plane. It is known that Ma = 6kg, MB = 2kg, and the dynamic friction factor between a and B is μ = 0.2. A thin line is tied on object a, and the maximum tensile force that the thin line can bear is 20n. Now the thin line is pulled horizontally to the right, g = 10m / S2 Find (1) if the tension is 20n, then the friction between ab? (2) In order to make AB relatively static, the value range of horizontal tension f is calculated

As shown in the figure, two objects a and B are stacked on a smooth horizontal plane. It is known that Ma = 6kg, MB = 2kg, and the dynamic friction factor between a and B is μ = 0.2. A thin line is tied on object a, and the maximum tensile force that the thin line can bear is 20n. Now the thin line is pulled horizontally to the right, g = 10m / S2 Find (1) if the tension is 20n, then the friction between ab? (2) In order to make AB relatively static, the value range of horizontal tension f is calculated

If B is above, then Fab = umbg = 4N
The maximum acceleration of B is a = Fab / MB = 2m / S ^ 2
If f = 20n,
aA=(F-f)/mA=(20-4)/6=8/3m/s^2
There is relative sliding between AB and ab
(2) AB is relatively static, then a'a

When Ma = 4kg, MB = 1kg, the dynamic friction coefficient between a and the table is μ = 0.2, the distance between B and the ground is s = 0.8m, and a and B are static, the velocity of B falling to the ground is_
In addition to conservation of energy, can acceleration be used to solve it

Building lord, this problem can use acceleration to calculate
According to the meaning of the title, I think your picture should be a on the desktop, B hanging in the air 0.8 meters above the ground. Right?
If so, this problem first analyzes the force on object B and its acceleration
analysis:
1. Body B is pulled by body a, and body B is pulled by its own gravity. The final resultant force of body B is gravity G-F pull
2. The pulling force F is related to the object a. the pulling force F is equal to the friction force on the object A. f = ma * g * μ = 4 * 10 * 0.2 = 8 N (in general, G is 10 N / kg)
3. In conclusion, the resultant force of object B is MB * G-F pull = 1 * 10-8 = 2 N, and the direction is vertical and downward. Therefore, the acceleration of object B is FB resultant force / MB = 2 m / s;
4. Because the object is stationary and the initial velocity is 0, according to the formula U & # 178; (last) - U & # 178; (first) = 2As, the root of the final velocity of object B is 3.2m/s

The three objects are connected by a string, Ma = 2kg, MB = 3kg, MC = 1kg. The dynamic friction coefficient between AC and horizontal table is 0.25?

For B: MB G - F2 = MB a
For C: F1 - μ MC g = MC a
For a: F2 - F1 - μ Ma g = ma a
The solution is: when G is 10m / S2, F1 = 6.25n, F2 = 18.75n, a = 3.75m/s2
When G is 9.8m/s2, F1 = 6.125n, F2 = 18.375n, a = 3.675m/s2

A. The friction coefficient between a and B is 0.2,
The sliding friction coefficient between B and table is 0.1. When the horizontal right force F = 5N acts on a, select ()
A body a is subjected to four forces
B object B is subjected to four forces
The friction force of C table top to object B is 5N, and the direction is left
The friction force between D and B is 4N, and the direction is left
Note: A is above B. you don't even know the object under force and the object under force

B. They are the pressure of a to B, the support of B to a, the support of desktop to B and the pressure of B to desktop