Mechanics good come in, please, help

First of all, the direction of friction and motion are opposite (note!). At the beginning, use a force of 10 N to "push the cart" to make it uniform forward Indicating that the magnitude of the friction force is 10 N backward (opposite to the direction of movement of the car), and then the friction force is still backward (because the car is still moving forward) when decelerating, then the force used by the person to "pull" the car is backward, and the direction of the two forces is the same, so the resultant force is the sum of the two forces.

The magnitude of friction is only related to the mass of the car and the friction coefficient of the ground and not to the magnitude of the force provided by the person, and I have reminded you that the direction of friction is opposite to the direction of movement, you think about whether the car is still moving forward when decelerating, since the car is moving forward, the friction force should be opposite to the direction of movement of the car, is the direction of friction should be backward.

When the trolley starts to move in a straight line at a uniform speed, that is, the force is balanced, then the friction force = thrust = 10n In the process of deceleration movement, the trolley is still moving forward, then he is still subjected to the friction force of 10n backwards.

In order to stop the trolley, Xiao Ming changed the forward thrust of 10N to a backward pull of 15N.

Then the net force experienced by the trolley in the horizontal direction is the backward force of 25N.

The reason why the car is not affected by the forward force is that it is impossible for Xiao Ming to push the car forward and pull the car backward at the same time.

As long as the object is in motion and the contact surface is not smooth, then the object will be subjected to dynamic friction in the opposite direction of motion!

Your teacher calculates it like this, Xiao Ming pushes a trolley at a uniform speed with a force of 10n to maintain a uniform motion, so the external force on the trolley is 0, that is, the ground friction force is 10n (the direction is opposite to the direction of pushing), (excluding air resistance).

When Ken suddenly pulled the cart instead of pushing, the cart was still moving forward at this moment, so the cart still had friction, but it slowly disappeared with the feed of speed.

Let's assume that the trolley moves from left to right.

From the title to know that the trolley moves at a uniform speed under the action of 10n to the right thrust, that is, the force.

Balance. Then we can know the sliding friction of the trolley f = 10n , direction.

It should be left-facing;

When you slow down and brake, the direction of the pull you use should be the same as the one you applied earlier.

The direction of thrust is reversed, i.e., to the left, which is 15n.

At this point, let's analyze the direction of friction: when braking, the car slows down.

Movement, the speed is constantly decreasing, but the trolley does not change the direction of movement at this time.

direction, that is, the direction of speed remains to the right, because the direction of friction is the same as that of the trolley.

The direction of motion is always opposite, so at this point the frictional force f = 10n and the direction remains.

Left. The two components of Nama's car at this time are both to the left, then in the horizontal direction.

The resultant force is 15 + 10 =25n and the direction of the resultant force is to the left.

Note*** The thrust of 10n to the right when braking has been undone, no.

There is a force in the right direction!

At the beginning of the trolley, the force is flat, and the force is flat at a constant speed, and a trolley is pushed at a uniform speed with a force of 10N.

Then at this time the trolley is subjected to 10n forward thrust and 10n backward friction, because suddenly he found that there is an old man in front, so immediately use the force of 15n horizontal direction to pull the trolley, so that the trolley slows down, then the direction of friction is opposite to the relative direction of motion, that is, it is still backward, then the direction of its force direction is the same as that of 15n, then the net force of the trolley is 25n

The teacher is right, Xiao Ming pushes the trolley with a force of 10N, which means that the trolley is also balanced by the friction force of 10N, so that it moves at a uniform speed, and when it decelerates, it is not subject to the forward force The friction force is opposite to the direction of motion.

Because the resultant force is the sum of force and friction, the pulling force is 15N, the frictional force is 10N, and the direction is backward, and the resultant force is 25N

In the process of deceleration motion, the trolley is also subjected to a backward friction force of 10N.

When Xiao Ming pulls the trolley in the opposite direction with the horizontal force of 15N, because the trolley is moving forward, it is also affected by the frictional force f=10n that is opposite to its relative motion, and this force is the same as the direction of the tensile force of 15N, so the resultant force in the direction of the tensile force is 25N.

The direction of friction is opposite to the direction of the object's motion or the direction of the object's tendency to move, not to the force.

So the frictional force experienced by the trolley is in the direction of tensile force and is 25N.

Look at it. Math formulas are not easy to type.

Seek internal force first. The first.

The second. <>

The third shear force is 0

If x=,y= is the center line of the object, then the length of these two line segments is not a sum, and the coordinates of the end points of the verse are (,,30) and (20). The center point of the two line segments is (,,,,,

1. It may be a force.

After simplifying A and B, the principal moment is zero, the principal loss is not zero, and both A and B pass the principal loss;

It can't be a couple.

If it is a force couple, it is a force couple after simplifying any point, which does not match the conditions of the title;

Probably balance.

For points A and B, the simplified principal moment is zero, and the main loss is also zero, isn't it balanced?

2. Balance. Suppose there are three points: a, b, and c. It is known that the principal moment of point A is zero, and if the principal loss is also zero, it must be balanced; If the main loss is not zero, the main loss must be over point A.

The principal loss (at this time the principal moment is zero) is simplified to points b and c, because the three points a, b, and c are not collinear, and the principal moment of point b and point c must have a non-zero, which contradicts the topic and the assumption is not valid. Therefore, the force system is a balanced force system.

3. No hypothesis, the earth revolves around the sun, only revolves, not rotates (just hypothesis). At this time, every point on the earth is moving in a circle, but this is not called fixed-axis rotation, this is called translational (parallel movement).

The first one is the acceleration at point D, and the second is the acceleration at point E, both of which are written in tangential and normal form.