Friction is difficult, friction is a problem.

It's not hard to figure this out.

In the case of straight pulling, the tensile force f1=umg=f used to reach the critical state, where u is the critical static friction coefficient, m is the cement mass, g is the gravitational acceleration, and f is the maximum static friction.

When pulling one end to rotate, assume that the end point to which the pulling force is applied is on the right side of the axis of rotation, and the distance between the axis and the end point is d1 and the distance from the other end point is d2. If the tensile force f2 makes it have a tendency to rotate clockwise, then the friction force must make it have a tendency to rotate counterclockwise. We consider that the cement block is divided into two parts from the position of the rotating shaft, the projected area on the right side of the rotating shaft is subjected to positive pressure of m1g, the maximum static friction force f1 = um1g, and the direction is opposite to the direction of tensile force (the moment it produces is opposite to the moment generated by tensile force); The projected area on the left side of the shaft is subjected to positive pressure m2g, and the maximum static friction force f2 = um2g, which is the same direction as the tensile force (the moment it produces is also opposite to the moment generated by the tensile force).

From static equilibrium and moment equilibrium, two equations can be listed:

f2+f2=f1 (1)

f1* (2)

And m1+m2=m, so there is f1+f2=f (3) and obviously there is d1 d2=m1 m2=f1 f2 (4) Solving this system of equations yields:

f2 = (root number 2 1) * f=

d1 = (root number 2 2) * d where d is the length of the whole block of cement.

The implication of this solution is that the force f2 that pulls one end to reach the critical rotation is much smaller than the force f1 that reaches the critical translation of the straight pull, and that once this force is reached, the cement block will start to rotate on its axis from the tensile end.

You can also think about what will happen after the rotation once it starts.

If you ask which mile is bigger, the straight pull is bigger, because the friction required to reach the critical movement is the maximum static friction, which is greater than the friction force. The tensile force that makes it critically rotate is equal to the frictional force, so the straight pull is large.

Although I don't know what you're asking, one thing is certain: the friction required for critical movement = the pulling force for critical rotation.

It should be a big straight pull. If it is simplified into a rod, the straight pull is un, and the axis with any point on the lever is the sum of two forces. Cement slabs have to be integrated with weight, which is not discussed here.

Did the landlord ask which of the two forces was bigger?

The straight pull force is high.

When it rotates, the force has a lever amplification effect on the cement slab, and the length of the "lever" is the distance from the action point of the force to the corresponding points on the cement slab.

Among them, the leverage effect from the point of force application to the point of rotation is the largest.

Tell me some conditions in general, right?

How big is cement? What is the coefficient of friction?

The conditions are enough, it's so difficult.

Solution: Let the speed of the block before landing be v1, and the speed of the car at this time is v

Before the small object hits the ground, it is regarded as a whole with the car, which can be known from the kinetic energy theorem

fd=1/2*(m+m)*v^2

Let's take a look at the blocks and carts separately

The block is only subjected to the rightward kinetic friction f between the block and the trolley, while the trolley is subjected to two forces, one is the horizontal constant force to the right, and the other is the kinetic friction force given to it by the left-facing block, so the resultant force is f-f, again using the kinetic energy theorem:

f(d-1)=1/2*m*v1^2

f-f)d=1/2*m*v^2

f=mg can be obtained by bringing in the relevant data, v=2 2m s, v1=2m s, f=300n, f=100n

After the block falls from the trolley, it does a parabolic motion with the initial velocity v1=2m s from the place of height h=, while the trolley only accelerates to the right by the constant force of f=300n, the initial velocity is v=2 2m s, and then the object landing time is t, the parabolic motion distance of the block is s1, the distance of the trolley continues to move after the block lands is s2, and the trolley velocity is v2 when the trolley block lands, then the following relationship can be obtained:

h=1/2*gt^2

s1=v1*t

ft=mv2-mv

f*s2=1/2*m*v2^2-1/2*m*v^2

s=s2-s1

As for the direction of f and f, as the first floor said, I hope it can help you.

Because the horizontal force F to the right is applied to the trolley m, the trolley will have a tendency to move to the right, and at this time M has no tendency to move to the right, it will prevent the trolley from moving to the right, and it can also be understood that the trolley will move to the right with animal blocks. Then the friction on the trolley is to the left, and the block M is only subjected to the force f (friction) to the right

To give you the explanation of the national standard GB T 17754-1999 on the face of friction and friction:

Friction: The phenomenon of resisting the tangential relative motion of the contact surfaces of two objects under the action of external force.

Friction: The tangential force that occurs on the contact surface of two objects that hinders the relative motion of two objects when one is in relative motion relative to the other or has a tendency to move relative to each other.

In fact, it is generally said that the increase in friction refers to the increase in friction, and I agree with the views given by the 3rd floor.

This question itself is not rigorous, but as far as this question is concerned, I think it is necessary to score, but it is necessary to see whether the questioner and the marking are not given, and some questions can only be dumb and eat coptis, and there are hardships, and everyone comes as students.

Hehe, it should be friction, increase the roughness, increase the friction.