As shown in the figure, the object is in equilibrium under the action of five common forces. If the force F1 is removed and the remaining four forces remain unchanged, please draw the magnitude and direction of the resultant forces of the four forces on the diagram

As shown in the figure, the object is in equilibrium under the action of five common forces. If the force F1 is removed and the remaining four forces remain unchanged, please draw the magnitude and direction of the resultant forces of the four forces on the diagram

According to the equilibrium condition of the common point force, it can be seen that the resultant force of the other four forces is in the opposite direction as F1, so the resultant force of the four forces is opposite to the direction, and the magnitude is also F1;
  
A: as shown in the figure above

When a body is subjected to three common forces, the following four combinations 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=10N、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 10N is 0n, which may be zero with 1n, so D is correct;
Therefore, ad

The balance of the body under the action of a common point force The mass of the object is m. when the long board is inclined to the ground at an angle of a, the object moves at a constant speed and is subject to friction. What is the thrust applied upward in order to make the object slide along the board at a constant speed Q: what's the friction

The downward component of an object moving upward along the inclined plane is twice as large as that of the downward component. Analysis: the downward component is equal to the friction (upward) when sliding at a constant speed; if the sliding upward at a constant speed, both the friction and the component are downward, so the downward component is twice as large

The body is in equilibrium under the action of three common point forces F1, F2 and F3 1. If F3 is removed, what is the direction of the resultant force 2. If F3 is increased to F3 "(direction unchanged), what is the resultant force 3. If F3 is reduced to F3 "(direction unchanged), what is the resultant force 4. if you turn F3 anticlockwise by 90 degrees (the size does not change), what is the magnitude of the resultant force Why not

1. If F3 is removed, what is the direction of resultant force: the direction of resultant force turns 180 degrees along F3, and the size is equal to f3.2. If F3 is increased to F3 "(direction unchanged), what is the resultant force size: the direction of resultant force is F3" direction, and the size is equal to F3 "- f3.3

Physical common point force balance The light spring is a small spring, which still keeps the vertical direction, and the deformation of the spring when the ball contacts the smooth surface Spring a must be stretched B spring may be original length C spring may be compressed d all of the above are possible

D

High common point force balance A smooth ring with radius R is fixed in the vertical plane I didn't do it Two balls with mass of M1 M2 were worn on the ring The two balls are connected with a string The string is placed just against the ring When the two spheres are still, the angles between the position of the ball and the center line of the ring and the vertical diameter are 30 ° and 60 ° respectively Then M1: M2 =? A.2:1 B.1:1 C. 1: radical 3 D. Radical 3:1

The ball is subjected to the elastic force and gravity of the tension ring along the tangent direction of the rope
Force balance
Make vector triangle T: Mg = sin θ
Because the tension is the same
So M1: M2 = sin θ 2: sin θ 1
=sin30:sin60
=1: Radical 3

An object is in equilibrium under the action of 5N, 7n, 9N, 12n, 20n and other common point forces. Now the 9N force is suddenly reversed (the remaining force remains unchanged). In order to maintain balance, how much force should be added? What is the direction of the force?

So, the resultant force of all forces except 9N should be opposite to 9N. When the force of 9N suddenly changes to the opposite direction, the resultant force of those forces is still in the opposite direction

Object a is placed on physical B, and object B is placed on a smooth horizontal surface. Ma = 6kg, MB = 2kg, and the dynamic friction coefficient between AB is u = 0.2. As shown in the figure, if a horizontal right pulling force F is used to act on object a, the following statement is wrong (g takes 10m / S2) A. When the tension f < 12n, a is still; B. When the tensile force F > 12n, a slides relative to B; C. When the tensile force F = 16N, the friction force of B by a is equal to 4N; D. No matter how much tension f is, a is still relative to B I hope we can explain it in detail,

I've done the original question. That's what the teacher said
The answer should be a B D
A. As a whole, the force between AB is internal force, so a and B slide to the right together, but the relative position of AB remains unchanged
B D assumes that there is no relative sliding between a and B. the maximum static friction force between a and B against a is 12n, and the direction is left. Using the isolation method, at this time, a has a right friction force against B, which is 12n. Therefore, at this time, B is only affected by the friction force between a and B, so the acceleration of B is 6 meters per square second, The critical value f of relative sliding of AB should be (6 + 2) × 6 = 48n
So B D is wrong
As for the answer of C, it's very vague. My teacher didn't say it. Let alone him, it's 16-12 = 4
In fact, this problem is to test the integration and isolation
Come on! Happy Spring Festival!

High school physics problems, please help to do a force analysis Suppose the mass of the passenger is 70kg and the speed of the car is 30m / s. The time from the brake to the complete stop of the car is 5 seconds. At this time, what is the force of the seat belt on the passengers?

Vt -- terminal velocity
V0 initial velocity
A -- acceleration
T --- time
Using the formula VT = V0 + at
Find a = - 6m / S ^ 2
Newton's second law f = ma = 70kg * 6m / S ^ 2 = 420N
A: so the force is 420N

As shown in the figure, the mass of the trolley on the smooth and level ground is m, and the mass of the person standing on the horizontal bottom plate of the trolley is m, m ≠ M. people pull the trolley with a rope across the fixed pulley, and the ropes on both sides of the fixed pulley are kept horizontal, regardless of the friction between the rope and the pulley A. One must be subject to friction to the right B. People may be subject to friction to the right C. The greater the pull force, the greater the acceleration D. The greater the pulling force of the rope, the greater the friction force between people and the car

Suppose the rope tension is t and the static friction force between man and workshop is F. suppose that the static friction force between car and man is left and that of man to car is right
T-f=ma
T+f=Ma
T = 1
2(M+m)a… (1)
F=1
2(M-m)a… II.
a=2T
M+m… 3.
According to formula 2, when m > m, that is, the static friction force of vehicle to person is left and that of person to vehicle is to the right; from Formula 3, acceleration is directly proportional to tension. The greater the tension is, the greater the acceleration is. From formula 2, the greater the acceleration is, the greater the friction force is
Therefore, BCD