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If two objects of the same mass are placed on a horizontal table top, the dynamic friction coefficient between them and the table top is the same, because the pulling force on the horizontal direction is different. If a moves at a constant speed and B moves at a variable speed, then the relationship between the sliding friction force and the sliding friction force on them is, write out the reason
The sliding friction force is equal to the product of the positive pressure and the friction coefficient. The positive pressure is numerically equal to gravity, and the friction coefficient is equal, so they are subject to the same amount of friction
The maximum static friction force between the box and the ground is 120N, and the dynamic friction coefficient is 0.25 (1) Push the wooden box with a horizontal force of 70N. What is the friction force on the wooden box? (2) Push the wooden box with 150n horizontal force. What is the friction force on the wooden box?
When a horizontal force of 70N is used to push the wooden box, the wooden box will not move. At this time, the friction force is static friction, which is equal to the horizontal force thrust. Static friction F1 = F1 = 70N;
When the horizontal force of 150n is used to push the wooden box, 150n is greater than 120N, and the object wants to move. At this time, the friction force is sliding friction force, and the sliding friction force is f = μ FN = 0.25 × 400N = 100N;
Answer: (1) push the wooden box with a horizontal force of 70N, and the friction force on the wooden box is 70N,
(2) Push the wooden box with a horizontal force of 150n, and the friction force on the wooden box is 100N
[physics of senior high school] a problem of solving friction force There are pictures, A cylinder with uniform thickness and weight of G = 100N is placed on a V-shaped groove with an opening angle of 60 degrees. The friction coefficient between the contact surfaces on both sides of the V-shaped groove is u = 0.25. First, the cylinder should be pulled out at a uniform speed to calculate the horizontal tension in the axial direction of the cylinder The answer is 50N. I calculated 25N. I calculated the force of the ball on the inclined plane perpendicular to the inclined plane, bisecting the angle of 60 degrees below, to get the force of the ball on two inclined planes to be 25N respectively, and then through F = UN = 0.25 * 25N = 12.5n, then * 2, equal to 25N?
50N, so the force on the inclined plane of the cylinder is 1002 * 25, so it should be supported by the oblique plane of the cylinder
There's something wrong with your stress analysis!
When pushing the bicycle forward, the friction forces of the front and rear wheels under the action of the ground are F1 and F2 respectively; when riding the bicycle, the friction forces of the front and rear wheels under the action of the ground are F3 and F4 respectively. Which direction is forward? Why?
When riding, the rear wheel is the driving wheel, so the friction direction F4 is the same as the forward direction
When pushing the cart, the rear wheel is driven wheel, and the friction direction F2 is opposite to the forward direction
Whether riding a bicycle or pushing a cart, the front wheels are driven wheels, and the direction of friction is opposite to the direction of advance
High school physics about power problem! Urgent! Online and so on! When a vehicle with a rated power of 30kW is driven at constant power on a horizontal highway, the maximum possible speed is 108km / h. assuming that the resistance is constant and the mass of the vehicle is 500kg, what is the acceleration when the speed is 36km / h and 72km / h? Process!
The process is as follows:
In the first step, according to the conditions, when driving at constant power on the horizontal highway, the maximum possible speed is 108km / h, and the traction force F of automobile engine can be obtained
30kW=108km/h*f
Where f = 600N is the resistance at the maximum speed on the horizontal road, which is also equal to the traction force F of the automobile engine
In the second step, the resistance is proportional to the speed of the vehicle
When the speed is 36km / h, f = 200N
When the speed is 72 km / h, f = 400 n
The second step is to calculate the acceleration according to the traction force of the engine and the resistance when driving at various speeds
When the speed is 36km / h, a = 0.8m/s2
When the speed is 72 km / h, a = 0.4 m / S2
There is a fountain in the park. The height of water spray is 5M and the power of water pump is 8kw. How much water is sprayed in the air at any time
After a long time, two water columns with 5m height are formed in the air, so m / 2 V ^ 2 = Pt (at this time, t is the moving time of the upper and lower water column) and then v = 0.32m ^ 3 is obtained from the formula of density!
A high school physics problem about force A piece of wood stick with uneven thickness is balanced at a certain point. If the stick is divided into two sections, the weight of the two sections is () The answer is that the thin section is heavy and the thick end is light. Who can explain why it is not the same weight? Can you explain it without torque? Or to be more specific, we didn't learn about torque
The answer is wrong, but it is not equal, but the thin section is light and the thick section is heavy torque. There are two elements: ① force, f is the length of force arm, l is the product of these two elements, and M is used to express it. Therefore, torque is m = Fl
As shown in the figure, the three forces acting on point o are balanced. Let one of the forces be F1. Along the - Y direction, the angle between the unknown force F2 and the + X direction is θ A. Force F3 can only be in the second quadrant B. The smaller the angle between force F3 and F2 is, the smaller is F3 and F2 C. The minimum value of F3 is f1cos θ D. The force F3 may be in any range in the third quadrant
A. According to the balance condition, the force F3 may be in the second quadrant or in the third quadrant. Therefore, a is wrong. B. according to the triangle rule, three force composition diagrams are made. Because the resultant force of the three forces is zero, the three forces form a closed triangle with the end connected
High school physics movement and force Suppose a firefighter with a mass of 60kg is holding a vertical pole 18m above the ground and glides in the shortest time. The mass of the rod is 200kg. The speed of the firefighter landing on the ground should not be greater than 6m / s. The maximum pressure of hands and legs on the pole is 1800n. The dynamic friction factor is 0.5. The local gravity acceleration is 10m / s 1. Maximum speed of firefighters during descent 2. The maximum pressure of the rod to the ground during the sliding process 3. The shortest time of decline (detailed explanation is required)
According to Newton's second law, a is equal to the ratio of friction force to mass, and the friction force is equal to the product of maximum pressure and friction coefficient, which is 900N. Therefore, a = 5m / S2
There are: GT1 ^ 2 / 2 + (GT1 + 6) T2 / 2 = 18
gt1-at2=6
T1 = (2 √ 10) / 5 s T2 = (4 √ 10-6) / 5 S
The maximum velocity Vmax = GT1 = 4 √ 10m / s
The maximum pressure of the rod to the ground is f = Mg + μ FN = 3800n
The shortest time of descent t = T1 + T2 = = (6 √ 10-6) / 5 S
As shown in the figure, an object of mass m is suspended on a light support, the angle between the inclined beam OB and the vertical direction is θ, and the elastic forces of horizontal beam OA and inclined beam ob acting on o point are F1 and F2 respectively A. F1=mgsinθ B. F1=mg sinθ C. F2=mgcosθ D. F2=mg cosθ
The stress analysis of node o is shown in the figure. According to the equilibrium condition, f can be obtained
One
=mgtanθ，F
Two
=mg
cosθ
Therefore, D
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