As shown in the figure, it is a long enough inclined plane with a dip angle of θ = 37 ° and a mass of M = 10kg. Under the action of a force of F = 100 N along the inclined plane at the bottom of the inclined plane, the body moves from rest to 4m in 2S, and the force F is removed at the end of 2S, (sin 37 ° = 0.6, cos 37 ° = 0.8, 2) find: (1) the dynamic friction coefficient μ between the object and the inclined plane; (2) the velocity V of the object at the end of 1.5 s from the removal force F; (3) when the object passes 4.5 m away from the bottom of the slope

As shown in the figure, it is a long enough inclined plane with a dip angle of θ = 37 ° and a mass of M = 10kg. Under the action of a force of F = 100 N along the inclined plane at the bottom of the inclined plane, the body moves from rest to 4m in 2S, and the force F is removed at the end of 2S, (sin 37 ° = 0.6, cos 37 ° = 0.8, 2) find: (1) the dynamic friction coefficient μ between the object and the inclined plane; (2) the velocity V of the object at the end of 1.5 s from the removal force F; (3) when the object passes 4.5 m away from the bottom of the slope

U = 0.2, and then calculate the acceleration from the fall. Notice that he has the initial velocity

FN = f * sin (angle a) f=p*Fn F * cos (angle a) = F + G Take the angle a, G, P as the known condition and find F

Isn't that easy
Put the first two into the third
We get p * f * sin (angle a) + G = f * cos (angle a)
So f = g / (COS ∠ a - p * sin ∠ a)
Mm-hmm

The block is placed on the inclined plane with an inclination angle of 45 degrees. The weight of the object is 40 n. The maximum static friction force with the inclined plane is equal to the sliding friction force, and the dynamic friction coefficient with the inclined plane is three thirds of the root sign. When the object is pushed by horizontal external force, what is the range of f that can make the object rest on the slope? The maximum value of F can be calculated. How to find the minimum value?

It can be thought that if the thrust is very small, n = fsin α + mgcos α, the sliding component of gravity is greater than that of thrust along the slope, so there is F + fcos α = mgsin α
Due to the static friction, the direction is upward along the slope
Now f increases, f decreases, and finally increases in reverse, so the maximum value of F can be obtained
Because the body is not pushed, the body is sliding. In order to make the object stationary, there must be fcos α + μ (fsin α + mgcos α) = mgsin α, and the minimum value of F should be calculated

On the equilibrium of common forces Three forces acting on the same body are in equilibrium. One of them is 4.0n, northward, and the other is 3.0N, eastward?

The third force is 5N and its direction is 37 ° south by West (or arcsin3 / 5)

A body is subjected to the action of three common point forces. The combination of the following four groups of forces may make the body in equilibrium A. F1=7N，F2=8N，F3=9N B. F1=8N，F2=2N，F3=11N C. F1=7N，F2=1N，F3=5N D. F1=10N，F2=8N，F3=1N

A. When the resultant force of 7n and 8N is equal to 9N and the direction is opposite, the resultant force is zero, so a is correct;
B. The maximum resultant force of 8N and 2n is 10N, so the resultant force of 8N and 2n cannot be zero, so B is wrong;
C. The minimum resultant force of 7n and 1n is 6N, and the resultant force of 7n and 1n cannot be zero, so C is wrong;
D. The minimum resultant force of 10N and 8N is 2n, and the resultant force of 10N and 8N cannot be zero, so D is wrong;
Therefore, a

The equilibrium condition of a body under the action of a common point force is______ .

The equilibrium condition of the body under the action of the common point force is: the resultant force of the external force on the object is zero;
So the answer is: the resultant force of the external force on the object is zero

Among the following groups of forces, the common point force acting on the body makes it possible for the body to be balanced A.3N,4N,8N B.10n.30n,50n c.7n,7n,15n D.10n,10n,10n

D. 10N, 10N, 10N. Other forms are not trigonometric

The equilibrium condition of a body under the action of a common point force is______ .

The equilibrium condition of the body under the action of the common point force is: the resultant force of the external force on the object is zero;
So the answer is: the resultant force of the external force on the object is zero

When a body is subjected to three common forces, the following four groups can make the body in equilibrium A. F1=10N，F2=7N，F3=1N B. F1=8N，F2=2N，F3=1N C. F1=7N，F2=1N，F3=5N D. F1=7N，F2=8N，F3=9N

A. If the resultant force range of F1 and F2 is 3N ≤ f ≤ 17N, F3 = 1n and the resultant force of F1 and F2 cannot be equal, then the resultant force of the three forces cannot be zero and the object can not be in equilibrium. Therefore, a is wrong
B. If the resultant force range of F1 and F2 is 6N ≤ f ≤ 10N, F3 = 1n and the resultant force of F1 and F2 cannot be equal, then the resultant force of the three forces can not be zero, and the object can not be in equilibrium
C. If the resultant force range of F1 and F2 is 6N ≤ f ≤ 8N, F3 = 5N and the resultant force of F1 and F2 cannot be equal, then the resultant force of the three forces cannot be zero and the object can not be in equilibrium. Therefore, C is wrong
D. If the resultant force range of F1 and F2 is 1n ≤ f ≤ 15N, F3 = 9N and the resultant force of F1 and F2 may be equal, then the resultant force of the three forces may be zero, which can make the body in equilibrium. Therefore, D is correct
Therefore, D

A problem about the equilibrium of common point forces in Physics A particle with mass m is in equilibrium under the action of several common point forces at the same time. When one of the forces f is constant, the direction of rotation is 180 degrees, and the rest of the forces remain unchanged, the acceleration magnitude and direction of the particle are () A f / m, same as f original direction B F / m, opposite to the original direction of F C 2F / m in the same direction as f D 2F / m, opposite to the original direction of F Choose D but don't know why. Who knows why?

At the beginning, the resultant force is 0. If you change one of the f directions and turn 180, the resultant force of the other forces is f in the same direction as f. the resultant force is 2F,
A = 2F / m, opposite to the original direction of F