Help, junior high school physics, leverage related questions. Problem solving steps are required.

The line of action of the force pulled up at point A is F2, as shown in the figure, AG is the line of gravity of the object, that is, the line of resistance, because the line of power of F2 is just in the same line as the line of resistance, so the power arm is 0, that is, it is not labor-saving.

No, it's like the pressurized water wells before, why don't you mention it, or rely on pressure, that is, it's more labor-saving, it's the lever principle you remember, the formula is not power x power arm is equal to resistance x resistance arm.

I've done this question.,But the only difference is that it's a few numbers less.,What I did is ao=,ob=4,bc=3,So that the landlord knows how to do it, right?!

The connecting AC force is the perpendicular line of the AC, and you can figure it out yourself.

OA is less than the root number (square of ob + square of oc).

By: Power x Power Arm is equal to Resistance x Resistance Arm The force should be vertical oc down.

Friction f r = 1000

f=1000/

f = n = n = 10000n (means that the upward support force at point b is 10000n) according to the principle of leverage equilibrium.

n*ob=f*oa

f=10000*

The resistance to the lever is the pressure and the friction of the brake, f*f=4400n

Let the supporting force at point b be n, the friction force of the flywheel is f, and the moment of the flywheel is mf = n....1

Equilibrium f* f* according to the moment gives f=400n

Friction f*m=1000 is the mechanical efficiency.

f=1000/

f= n= because o is the fulcrum.

So n = 10000n

According to the principle of leverage equilibrium, f*l=f1*l1

N*ob=f*oa can be obtained

f=10000*

So it's 4000n

Solution: Let the center of gravity of the wooden rod be at a distance of l1 from the point of action of the force of 50 Newtons, and the distance from the other end to the other end is l2, and the full length of the wooden rod is l, and the force used to slowly lift the other end is f. According to the lever equilibrium condition, the equation for lifting one end and the other end of the wooden rod respectively, 50n*l=80n*l1 (1).

f*l=80n*l2 (2)

From equation (1), we can get l1 = 5 8l, so l2 = 3 8l, substituting l2 = 3 8l into equation (2) can solve the force used by him to slowly lift the other end.

f=30n

It is assumed that all lever lengths remain the same as the stick is lifted. That is, to keep the stick on the ground at all times.

Set: The length of this stick is L1

The distance between the unlifted end and the center of gravity is l2

The force used to lift the other end is f

The available equation is: 50*L1=80*L2

f*l1=80*(l1-l2)

Get: f = 30 N.

Select D to screw the nut with a wrench, then the wrench can be regarded as a lever, the nut is the fulcrum, then the resistance is at the nut, very close to the fulcrum, and the power is at the wrench handle, so the power arm is greater than the resistance arm, then the lever is a labor-saving lever, A, D option is excluded, and because it is impossible to save power with any machinery, so C is excluded. The magnitude of friction is related to the pressure on the object, under the condition that the roughness of the contact surface of the object is the same and the power is the same, the greater the pressure, the greater the friction. Therefore, in order to avoid abrasions on the hands, gloves should be worn, and D should be selected

You can prove it with practice, it should be C

d, it is impossible for any machinery to save work !!

Set a weight of 10 N then, then because A end of the movable pulley two strands of rope, pull the rope of the fixed pulley downward pull force of 5 N, the fixed pulley does not change the size of the force, so B receives the upward pull force of 5 N, and because of the balance state so B weighs 5 N, so it can be seen that a is heavier than B when balanced, so that when hanging the same weight is the force of the B end should be greater than the A end, so choose B

Question 2: Because the torque generated by A remains unchanged, the center of gravity of L is at the midpoint of L, and in the process of turning 45 degrees, the center of gravity is above the horizontal line, and the projection in the horizontal plane will be shortened. The key is to look at the center of gravity, and the arm is the projection of the line between the center of gravity and the 0 point in the horizontal plane.

Looking at the center of gravity of a uniform cylindrical object, the horizontal distance between the center of gravity and point O gradually increases when rotating at the beginning, because the gravitational force on a homogeneous cylindrical object remains unchanged, so the moment increases, f is always perpendicular to ab, and f increases. When the line between point o and the midpoint of a homogeneous cylindrical object is higher than the horizontal line, the horizontal distance between the center of gravity and point o gradually decreases, and f decreases.