Physics problems, acting on forces! Physics and mechanics problems!

Think of it this way: Suppose that BC has a force F1 on AB, AB must also have a reaction force F2 on BC, and the two are equal in magnitude, opposite in direction and collinear.

Let's consider F1 first: AB is equivalent to a lever with A as the axis of rotation, and its gravity is in balance with F1. Then F1 must be upwards (the upper right side of the oblique bar ab, you can understand), so it can only be A, E, F.

Consider F2 again: BC is equivalent to a lever with C as the axis of rotation, and its gravity and F2 maintain moment balance. Then the moment of F2 must be upward, then F1 as a reaction force must be downward (the crossbar BC is downward), so it can only be C, D, E.

To sum up, only the E direction is possible. The answer is e.

Actually, I think this question is very simple, ab bc is hinged so there is no friction and is connected to one point, so if ab has no action of bc, then it is perpendicular to the upper plane Since you only ask about the force of bc on ab There is Newton's law f=f' So the force of bc on ab = the reaction of bc is the direction e.

The force of AB on BC is in the B direction, and the force of BC on AB is in the E direction, what grade is this? Don't know the torque?

If they are not connected, the AB force is downward, and the BC force is also downward. Since A is on the left and C is on the right, AB is a leftward pull on BC and a rightward pull on BC. At the same time, A is on top and C is down, AB is an upward pull on BC, and BC is a downward pull on AB.

So BC has a downward and rightward pull on AB, and only E is compliant.

It should be e. The resultant force of the tension force is e

The feeling is stationary, and the force analysis of AB, a downward gravity and an oblique upper left pull, using the parallelogram law to decompose the virtual equilibrium force of gravity is the direction where E is located.

There is f=am:

f1=2m，f2=4m；(m is the mass of the object).

So there are two possibilities for the resultant force to occur when both forces act on an object:

1.If the direction of the two forces is the same, then f1+f2=6m, so the acceleration = f combined with m=6m m=6;

2.If the direction of the two forces is opposite, then f1+f2=2m, so the acceleration = f combined with m=2m m=2.

Friction is a principle: f = n, and n is the supporting force.

Since there is no displacement of the object in the vertical direction of the inclined plane in this problem, the supporting force of the inclined plane to the object n=mgcos (the component of gravity in the direction of the vertical inclined plane).

So f= n= mgcos

When the block shudders back and moves upwards at a constant speed, the forces exerted on the block in the vertical direction are: downward gravitational force g, downward frictional force f, and upward tensile force f=. So g+f=f=, the solution is f=

Now the direction of motion of the object changes downwards and the frictional force changes upwards, but because the pressure and coefficient of friction do not change, the magnitude of f does not change. At this time, the forces that leak down are: gravity g, pull force f, and the upward force of the hole finch has friction force f=. So g+f=f=, the solution is f=

The team will answer for you

If the friction force is f, then there is f1=f+g on the leech

The solution is f=downward, and there is a disadvantage f2+g=f

The solution is f2=

Gravity is added when it is up, and gravity is subtracted when it is down, so Zhengda down=up- twice the gravity of Daiga.

Understand, lift imitation vertical.

1.Under the action of F, the object moves in a straight line with uniform acceleration.

The support force on the object n=mg-fsin37°=acceleration a1=(fcos37°- n) m =[fAfter the force is removed, the object moves in a uniform and decelerated linear motion.

Acceleration a2 = g = 5

by a1t1=a2t2

Bring in the above to obtain: 2*(

f = can be solved by standing these 3 equations

2.The object accelerates uniformly from 0 to velocity v, and then decelerates uniformly to 0s=v(t1+t2) 2

v can be simply calculated from the acceleration calculated in the first question: v=at=s=20*(2+4) 2=60m

Hope it helps.

Xiaohong practiced using the spring dynamometer, before using it, she found that the pointer of the spring dynamometer when it was not subjected to tension is as shown in the figure, and then she should adjust the pointer so that it returns to the zero value, and there is nothing hanging The spring dynamometer displays, but the display is not zero, which does not affect the spring dynamometer force.

I wish you progress in your studies and go to the next level! (

There is no weight hanging and there is an indication because the spring dynamometer has not been zeroed, and there are two adjustment methods.

One. When the spring dynamometer is suspended and naturally elongated, slide the scale plate on the spring dynamometer of the bar box up and down to 0 scale and align with the spring.

Two. Toggle the metal plate above, or rotate the spring to reset the indication to zero.

After zeroing, you can hook another spring dynamometer to check if it is adjusted.

The needle of the spring dynamometer is movable, not fixed. Because the 0 scale of the spring may vary under different measurement environments and conditions, it is not possible to fix the 0 scale like a scale.

The spring is old, or there is no zeroing.

Adjust the pointer so that it returns to the zero tick mark.

n pulls the rope to keep itself and the basket still in the air (g=10n kg)225

The analysis topics are as follows:

If the rope is perpendicular to the horizontal plane, the angle between the ball and the splitting bevel is 60 degrees. Because the ball is split at 30 degrees, the rope is 60 degrees upward from the water level!

Because the splitting slope is smooth, the ball is balanced under the elastic force of the wooden block, the pull of the rope and its own gravity

In this equilibrium case: the gravity along the inclined plane is balanced with the rope along the inclined plane; The resultant force of the component perpendicular to the inclined plane and the elastic force of the rope is balanced by the component force of the perpendicular inclined plane of gravity.

mg sin30 = f cos30 so f is equal to three thirds of mg

The second question: Assuming that the splitting is stationary, the system composed of the ball and the splitting is balanced by force

The whole system is subject to static friction, ground elastic force, and rope tension.

The equivalent substitution of the whole system is that the splitting weight is 2m, the force of the rope on the splitting is equal to f, and the direction and level are 60 degrees oblique upward!

The system is perpendicular to the ground and the forces are balanced.

2mg= n +f sin60

Horizontal force balance:

kn=f cos60

Substituting the f-number, K is less than nine and the next three.

I don't know the value of the tree, right, the idea should be right.

f1=3m

f2=5m, when the two forces act at the same time, the resultant force is f1+f2cos, the angle between the two forces is obviously, the two forces are in the same direction, that is, the angle is 0 degrees, when cos = 1, the maximum resultant force is 8m, and the acceleration is 8m per second.

When the two forces are reversed, the angle between the two forces is 180 degrees, and the minimum resultant force is 2m, and the acceleration is 2 meters per second.

These are the two extreme values, and the resultant force and the combined acceleration between them are achievable, so the acceleration may be 2 meters per second, less than or equal to a, less than or equal to 8 meters per second.

f=ma

f1=m*3m/s²=3m

f2=m*5m/s²=5m

When f1 is in the same direction as f2, f(max)=f1+f2=8m a(max)=f(max) m=8m s

When F1 is in the opposite direction to F2, F(Min)=F2-F1=2m a(min)=F(min) m=2m s

When f1 and f2 act on the object at the same time, the magnitude of the acceleration a is between 2m s and 8m s depending on the direction of action of the 2 forces.

The law of vector synthesis. The Law of Triangles.

Between 2 8 m s*s.

Force is a vector, and the sum of vectors obeys the law of parallelograms. In this case, a is a maximum of 8, a minimum is -2, -2 < = a< = 8, and its magnitude is 2---8 meters between seconds and seconds without indicating direction.

The magnitude of acceleration acquired by the object may be 2m s2-8m s2

Force and acceleration are vector quantities with directions, and the addition of 3m s2 and 5m s2 vectors ranges from 2m s2 (when two forces act in opposite directions) to 8m s2 (when two forces act in the same direction).