Mechanics and physics problems are solved, and simple physics and mechanics problems are solved in d

Solution: A force situation is shown in the figure. There is no relative slip between a and b, which means that the acceleration of both is the same, both are in the horizontal direction, and the magnitude is aApply Newton's second law to a:

ncosθ+p＝ mg ……1)

f-nsinθ ＝ma…… 2)

Apply Newton's second law to a system consisting of a and b:

f＝(m+m)a ……3)

N 0 ......4)

p≥0 ……5)

a＞0 ……6)

Obtained from (2) and (3).

2f-2nsinθ ＝f

i.e. n f (2sin) 7).

Substitute (7) into (1).

p＝mg-(fctgθ)/2 ……8)

mg-fcosθ/(2sinθ)≥0

f≤2mgtg60°

f≤2√3mg

In order to make a and b accelerate together under the action of horizontal thrust f (there is no relative slippage between the two), f should not exceed 2 3mg

It's been a long time since I've done a physics problem, and I don't know if it's the right thing to do......?

fmax=√3mg

Push the wooden block A with force, analyze the force of A, A is subjected to its own gravity, the support force of the smooth ground, the thrust f, and the force f of the wooden block B on A'。b is subjected to gravity, support, and f'The reaction force f"。

This f'and f"It is a pair of equal reverse forces at an angle of 30 degrees to the horizontal plane, one obliquely upwards and one obliquely downwards. When the force in the vertical direction is equal to the gravitational force mg, the supporting force on the block is 0, and the force decomposed in the horizontal direction is the thrust f, and the thrust is the largest at this time. Solve this triangle:

The opposite side of the 30-degree angle is mg, and the other right-angled side is f, then f = 3mg.

I can't draw pictures on the computer, so I have to write a narrative, I don't know if I can understand it?

The level is limited, don't blame you if you're wrong.

d。First of all, the one below is constantly going to go without saying.

Then, as for why the root above is continuous, it may be too abstract to say that the time is too short and the impact is negligible, you can understand it this way:

The thin wire is regarded as a rope with a large elastic coefficient k, and when it is elongated l>70n k, the rope below is broken, although it reaches 50n, because the action time t can be regarded as infinitely short, the impulse on the weight can also be regarded as infinitely small.

That is, the velocity of the heavy object changes infinitely, that is, the heavy object can be seen as having no change in state, so the rope above is almost not forced.

ps.Although the question is done in this way, whether it will actually be broken depends on the magnitude of the above kt 2.

Due to the short action time, the influence on the upper end of the thin line can be ignored, and the force on the lower end is less than 70N

Hello, I personally think that this question lacks conditions (how much force can the upper thin wire withstand). It is now understood literally as follows:

1. If the thin wire tied to the weight can withstand the force of 50 + 50 = 100N, the result will not break.

The reason is as follows: the upper thin wire is continuous, and the lower thin wire only bears a force of 50N, which does not reach his limit value of 70N, so it is continuous.

2. If the thin wire tied to the heavy object cannot withstand the force of 50 + 50 = 100N, the result is that the upper line will break.

The reason is as follows: the lower thin wire only withstands a force of 50N, and does not reach its limit value of 70N, and it is broken. The upper thin wire could not bear it and broke.

The above is a little personal understanding, please advise.

b is definitely b!! MD If you don't pull the lower line, the upper line is already 50 Nn more gravitational than the lower one! And because the two thin wires can only withstand 70N of tension, when the bottom of the 20N of tension, the top has already borne 70N of tension, so the top is broken first!

1. After analyzing MGSin37-F2 and removing F, MGSin37-UMGCos37=MA was solved to be A= down the inclined plane. After another 3s, the velocity is v=at=

3. There is no displacement at the beginning of 2s, and the displacement s=1 after 3s 2*a*t*t=

1): First find the support force n=mg-fsin37=, so the frictional force f=, the horizontal resultant force is f1=fcos37-f=, and the acceleration is a1=104 40=

For a system composed of ceilings, springs, and objects, the external force received is only f, the direction is horizontal to the right, the force in the vertical direction is unchanged, the pressure on the ground is constant, and the friction factor is constant, so the friction force is constant. The springs, ceilings, ropes, and objects are all internal to the system.

1 From the meaning of the title, it can be seen that the weight of the object g is close to the rough end of the wood is set to x, according to the principle of moment balance, according to the meaning of the title, when the thick end is raised, it is 680

When the thin end is lifted, it is g (

Solution (1) and (2) The center of gravity of the object is separated from the thick end of the object by equation x=m), the weight of the wood is g=680+420=1100 (Newton), and the center of gravity of the wood is spaced at the thick end of the knotted end is x=m).

20n vertically up.

x direction 50 times root number 3

25 times in the y direction (root number 2 minus root number 6).

No. Method 1: According to the formula g=mg, the weightlifter can only lift 200kg, 10n kg=2000n, and the gravity of the steel ingot is 2010n, which is greater than 2000n, so it cannot be lifted. (take g=10n kg).

Method 2: Because the gravity of the ingot is 2010N, according to m=g g, the mass of the ingot is 2010N, 10N kg=201kg, because 201kg is greater than 200kg, so it cannot be lifted.

Question 2: According to g=mg, it can be obtained that the gravity of 120kg of an object on the earth is 120kg 10n kg=1200n, and the person can lift an object with a gravity of 1200n. Then on the moon, it can only lift an object with a gravity of 1200n, but the gravity of an object on the surface of the moon is one-sixth of the gravity it exerts on the surface of the earth (that is to say, the surface of the moon g month = 1 6g ground = 10 6n kg), so the mass of an object with a gravity of 1200n on the surface of the moon is m = g g month = 1200n 10 6n kg = 720kg

If you want to improve your ability to understand the problem, you must listen carefully in class, do the questions carefully, and try to use a variety of solutions to do a question (it is very important to use a variety of solutions to improve the ability to understand the problem), and if you can't afford to do the difficult questions during the exam, don't do it, check the simple questions in front. If you want to learn physics well, you still have to exercise your thinking! Good luck learning physics!