It is very important to keep the balance of the carrying bell in weightlifting. As shown in the figure, if an athlete lifts a barbell of 1800n and keeps his arms at 106 ° angle, what is the force on both arms (sin53 ° = 0.8, cos53 ° = 0.6)?

This paper takes barbell as the research object, analyzes the force situation: gravity mg and the force F1 and F2 of the athletes' arms on the barbell, and makes the diagram as shown in the figure
2Fcos53°=mg
The result is: F = 5
6G＝5
6×1800N＝1500N
Answer: at this time, the strength of the two arms of the athlete is equal, which is 1500N

As shown in the figure, the object a on the horizontal ground moves in a straight line at a uniform speed to the right under the action of the oblique upward tension F, then () A. Object a may not be subject to ground support B. Body a may be affected by three forces C. The value of sliding friction force on object a is fcos θ D. The supporting force of horizontal ground to a is fsin θ

A. The body must be affected by friction in the horizontal direction, so the body is in balance by gravity, tension, supporting force and friction force. It is known that the body must be affected by four forces
C. According to the equilibrium of common point force, f = fcos θ. Therefore, C is correct
D. The resultant force of the supporting force of body a and the component force of F in the vertical direction is equal to gravity, i.e. n = mg fsin θ
Therefore, C

If five blocks of the same mass are placed on a smooth horizontal surface, and the horizontal force F acts on the first block, what is the force of the third block on the fourth block?

First use the holistic approach
For the whole composed of 5 wood blocks: F = (5m) * a
And then the isolation method (four and five are regarded as a small whole with a mass of 2m)
Then: three to four force: F three to four = (2m) * a
By combining the above two formulas, we can get: F three to four = (2 / 5) F

The balance problem of body under the action of common point force in senior one physics When goods are transported by sled on the snow, the total weight of the goods and sled is 1 ton, the dynamic friction coefficient between the sled and the ground is 0.5, and the angle between the tow bar and the horizontal plane is 37 degrees. What is the pulling force F in the direction of the draw bar to move forward at a constant speed?

Orthogonal decomposition
y:Fsin37+N=mg
x:Fcos37=μN
F. Cos37, m, G, sin37, μ all know
It is enough to solve the system of bivariate linear equations

As shown in the figure, the diameter of smooth ball a is 10cm, the length of suspension line is 50cm, the thickness of object B is 20cm, and the weight is 12n (1) B what is the pressure on the wall? (2) How much does a ball weigh?

According to the known sin θ = 20 + 550 = 12, so θ = 30 ° slider B is subjected to gravity, pressure of ball, pressure of wall on ball and upward friction force; in vertical direction, according to the equilibrium condition of common point force, f = mg = 12n; from F = μ f, f = f μ BC = 120.2 = 60N. The force analysis of the ball is shown in the figure: g = ftan, θ = 603

Body equilibrium under the action of common point force A constant force F with an angle to the horizontal direction is used to pull the slider with mass m to move in a straight line at a uniform speed on the rough horizontal plane

Let the angle be α
Ground pressure FN = mg fsin α
Because it's a uniform linear motion in the horizontal direction, so the friction is equal to the resistance, so there is
f=Fx=Fcosα

A physics problem. On the balance of bodies under the action of a common point force The lower end of the light rod is hinged on the ground, and the upper end is pulled by a constant horizontal force F, and the upper end of the light rod is pulled by the iron wire obliquely. The rod is just vertical. Now move the iron wire to the direction close to the pole, and the length of the wire will also change, while the rod is still in the original position. How will the tension in the wire change? How will the pressure on the rod change?

Did not look at the picture, I do not know if it is inclined to pull down the iron wire, so that the node of the iron wire moves downward, and the wire is correspondingly shortened?
If so, then
When the tension in the wire increases, the pressure on the rod remains unchanged
Because f is constant, in order to keep the balance, the moment of the force pulling the iron wire on the hinge remains unchanged (which can be understood as lever balance), and the node moves down and R decreases (the arm of force decreases), then the force perpendicular to the rod increases (inverse relationship), M = F1 * r
Let the angle between the iron wire and the pole be α, and then let d = R * sin α
Finally, the force F = F1 / sin α = m / (d * cos α) can be obtained by simplification
If the node moves down, α increases, then f increases
The pressure n = m / D is a constant

As shown in the figure, there are two pieces of wood 1 and 2 with mass of M1 and M2 respectively on a rough horizontal surface, which are connected by a light spring with original length of L and stiffness coefficient of K. the sliding friction coefficient between the wood block and the ground is μ A. L+μ km1g B. L+μ k（m1+m2）g C. L+μ km2g D. L+μ k（m1m2 m1+m2）g

Block 1 is studied. Block 1 is subjected to gravity, spring tension, ground support and friction
According to the balance condition, the spring force F = μ m1g
According to Hooke's law, the elongation length of spring x = F
k=μm1g
K
So the distance between two blocks is s = L + x = L + μ
km1g．
So choose a

Two objects superposition problem The maximum static friction force between AB and B is 20n, and the dynamic friction coefficient between B and horizontal plane is 0.1, If the current horizontal force F acts on B to accelerate the motion of AB, then the force f satisfies the condition? 2 ＞ if the horizontal force F acts on a to accelerate the motion of AB, then the force f satisfies the condition? (g = 10m / S2) Tell me what the maximum static friction does

【1】 A = (f-0.1 * 150) / 15 for AB and a '= f / 5 (f) for a

A fixed smooth vertical rod is covered with a small ball with mass of M. the ball can slide freely on the rod. Now a light rope is used to pull the ball to make the ball in a static state. The angle between the rope and the rod is a and the acceleration of gravity is g. the pressure of the ball on the rod is calculated

F tension * cosa = mg
So f tension = mg / cosa
F pressure = f tension * Sina = (mg / COSA) * Sina = mg * Tana
(you can draw the schematic diagram of force analysis, and then decompose the tension of the rope into horizontal and vertical directions. According to the force balance in each direction, you can solve the problem.)

It is very important to keep the balance of the carrying bell in weightlifting. As shown in the figure, if an athlete lifts a barbell of 1800n and keeps his arms at 106 ° angle, what is the force on both arms (sin53 ° = 0.8, cos53 ° = 0.6)?