A high school physics topic about interaction forces

Let the force between a and b be f because a and b remain relatively stationary at all times, so the acceleration of a and b is the same, let it be a

f-f1=ma*a ①

f2-f=mb*a ②

If a is correct, then f1 f2=ma mb is possible, but not necessarily.

If b is correct, then the direction of acceleration must be to the right, but if f2>f1, then the direction of acceleration is to the left, if c is correct, then it is obtained by a=0, but if f2 is not equal to f1, then a is not equal to 0

When f1=f2, a=0 and f=f1=f2 hold.

A must be equal to 0 and must be false.

Does b have to be less than f1? You think, if the contact surface is smooth and frictionless, then, f1 = f2 = interaction.

c, if there is friction, it is less.

In the absence of friction, the interaction force may be equal to f2, so D is chosen

If there is friction, F1 and F2 may not reach the maximum static friction, then there may be no force between A and B, so BC can be excluded; When there is no friction on the tabletop, there is only one situation where the object can only be kept stationary if the force on both sides is of the same magnitude, and the force will obviously not be zero. So option A can also be excluded.

The answer is as above, I hope it will be helpful to you.

D, the extreme hypothetical method can be used to rule out ABC.

An ant with a mass of about 5mg can lift a small insect that is 20 times heavier than itself, and the mass of the small insect is 100mg=

The gravitational force on small insects g=mg=10 (-4)*10=10 (-3)n=

The support force is equal to the gravity force, both are.

The mass of the insect m=5mg*20=100mg=the gravitational force of the insect g=mg=

Support force and gravity balance.

So the support force f=g=

The support force of the ant to the insect is the gravitational force of the insect so.

f=5*(10^-6)kg*20*10m/s^2

1*10^-3 n

Let the ball of 2n be a ball and the ball of 3n be ball b, and assume that ball b is under.

Force analysis: A ball is gravity 2n downward, rope tension f down, air resistance f upward;

2 + f - f = 0 (constant motion).

f = 2 + f

The ball B is subjected to gravity 3n downward, the tensile force f of the rope is upward, and the air resistance f is upward;

3 - f - f = 0 (constant motion).

f = 3 - f

f = f2 + f = 3 - f

f = physical wolf pack.

The tension of the rope on both sides of the smooth hook is equal, that is, f1=f2=fThe angle between the rope on both sides of the hook and the vertical direction of the round shed is 1, 2, and the horizontal resultant force at the smooth hook is zero, so f1sin 1=f2sin 2, f1=f2, so sin 1=sin 2, so sin 1=sin 2, so 1= 2=

The total length of the rope is 5m, and the span is 4m

Set the smooth hook position to O

oa+ob=5

oasinθ+obsinθ=4

oa+ob)sinθ=4

5sinθ=4

sin=cos=root[1-(sin)2]=

The resultant force of the upward pull force of the rope on both sides of the smooth hook is equal to the gravitational force of the object:

2fcosθ=mg

f=mg/(2cosθ)=12/(2*

Analyze the force on the node o (i.e., where the object hangs on the rope).

The gravity force of 12NOB direction is the pulling force, the OA direction pulling force, the latter two forces are the same magnitude, both are equal to T to divide these two forces into horizontal and vertical trembling directions.

The horizontal or concise directions balance each other, the vertical direction is equal in magnitude, and the sum = 12n, balancing the gravity.

So the vertical direction is 6n

The slopes of oa and ob are equal, and t is along the rope, which can be known according to the Pythagorean theorem.

t=10n

According to the analysis of the imitation force at point A, the tensile force T2 of the AB rope and the component T1SIN of the tensile force T1 of the OA rope are subjected to the horizontal direction

The two are equal, t2=t1sin

The vertical positive direction, the downward gravity g, and the upward component of t1 t1cos, are equal.

g=t1cosθ

According to the above two equations, we can get t2=gtg

In order to ensure that the rope continues, then take the extreme value, just bring it in, assuming that t2 reaches the extreme value of 30n, then =45°, and bring in the solution t1=30 2>20 3 so it does not work quietly, it is t1 that reaches the extreme value first, then according to t1=20 3=g cos, obtain, =30°, and then verify t2<30n

So the answer is =30°

This kind of problem is mainly to do a good analysis of the force and discuss it.

For the analysis of the force on M2, it is concluded that the effect of the external force on him is to make him move to the right with the acceleration of a. Therefore the force at the level is for GTAN. So the horizontal stress m2gtan .

This force is provided by the friction between the carriage and the m2. So the frictional force received is m2g. (Action vs. Reaction Force).

Action and reaction force, the support force is the pressure of m2 on the carriage. M2G-F pulls for pressure. The f-pull force is cos, so the support force is m2g-cos.

When the force is decomposed, the line in the oblique direction according to m1 is the x-axis.

The rope tension force f=m1g cos, so the support force is m2g-f, the friction force = m2a, for m1, m1a=fsin, a can be obtained

Simple, leave a label to have time to write.

m1xa=f1

m1xg=f2

The square of f1 + the square of f2 is the square of the force on the rope.

The support force is m2g-f

The frictional force is m2xa

The answer is: the force of C balances weight C from the resultant force of the longitudinal component of ao and bo, and the mass of weight c remains unchanged, the component force of ao and bo in the vertical direction remains unchanged, and the direction of ao remains unchanged, the force of ao remains unchanged, and the component force of ao in the vertical direction does not change, then, the component force of bo in the vertical direction must not change, and the direction of bo is also unchanged, so the tensile force f in the bo direction is also unchanged.

The simple question is that if three forces are balanced, and the magnitude and direction of two of them are constant, then the magnitude and direction of the third force must not change!

Sliding slowly, the resultant force of the force on the three ropes of a b c is always zero, since ao is always horizontal, indicating that the angle of the three forces is unchanged, so any of the three forces is unchanged, choose c,

AC Rally F1, BC Rally F2:

Horizontal force balance: f1sin30°=f2sin60°, that is: f1 2=f2 root number 3 2, f1 = f2 root number 3

Vertical force balance: F1COS30°+F2COS60°=50, that is, F1 root number 3 2 + F2 2=50

AC pull: F1 = 25 root number 3

BC pull: F2 = 25N

f1 = f2 root number 3, i.e. f1 f2 root number 3

f1max=180n, f2max=100n, f1max=root number 3

Therefore, f2max should be used as the basis for calculating the bearing:

f2 100, f1 = f2 root number 3 100 root number 3

Maximum lifting weight: g = f1cos30 ° + f2cos60 ° = f1 root number 3 2 + f2 2 100 root number 3 * root number 3 2 + 100 2 = 200n

h=，l=，m=50kg，μ=

sinθ=h/l=

cos = root number (

mgsinθ-μmgcosθ=ma

a=g(sinθ-μcosθ)=10*(

l=1/2at^2

The speed at which the person slides to the ground: v=at=2*2=4m s

The maximum friction f between A and B is U1m A g = and the maximum acceleration of A A is a = U1g = 4m s 2

A and B remain relatively stationary, and B's maximum acceleration is also 4m s 2 B is subjected to ground friction f'=u2(mA+mB)g=f-f-f'=ma

f = 4 * 1 + 21 + 80 = 105n, this is the maximum value of the tensile force f, b error f if it decreases, according to the overall analysis of the acceleration of a and b also becomes smaller, the acceleration of a is provided by friction, so the friction of a a is also gradually decreasing c correct.

If the force f is applied to A, and the friction force of A to B is provided to B for acceleration, and the mass of B is very small, and the maximum static friction of A and B remains unchanged, then the acceleration provided is much larger than before, and the range of values of f is also much larger, and it can maintain a relatively static motion, and D is also correct.