Ask a high school moment problem, high school physics moment problem

Children, when they go to college, they will know that this kind of connecting rod is called a two-force rod. It is itself balanced, so its force must be in the direction of the rod (explained in the diagram below). According to Newton's third law, the force of the crankshaft point B on the connecting rod is along the direction of the connecting rod, then the force of Liangan on the crankshaft point B is also in the direction of the stem and opposite to the upper force of the same magnitude.

So the force of the piston on the O point must be along the direction of the AB rod.

If, according to you, the force of the piston on point o is perpendicular to the piston, then as shown in the figure below:

Therefore, the force of the piston on the O point must be along the direction of the connecting rod, not perpendicular to the piston, nor along the other direction.

<>f=ps=

The arm is L, according to the knowledge of the machine.

l/ob=ao/ab

l = moment m = fl = 300*

Keeping the gas temperature constant, when the crankshaft turns 90 clockwise to the OBA in a straight line, the piston moves inward for a distance of x=ab+bo-ao=8cm

v1=400ml,v2=v1-xs=240mlp1=

There is by Boyle's law.

p1v1=p2v2p2=

The same principle as the wrench, the moment refers to the force (n) you exert into the upper force arm (m), so it is nm **Reference: The force applied n force arm m So the moment is n*m To resist this force at the red dot (force arm m), you have to use the force of n because it is the same fulcrum.

The same object So n*m = n*m The farther away from the fulcrum at the force application the larger the force arm is calculated with the above diagram I apply a force of 3n to rotate the force arm 5m at the red dot The force arm 3m You have to fight with a force of 5n to resist the rotation More details zh *wiki/%e5%8a%9b%e7%9f%a9

Reference: I&*

Think about it, if you put a force on the fulcrum, it will move without any other external force, but it won't transcode and tease, right?

Why nm?Because moment = force applied x force arm The unit of force is n The unit of force sold by the force arm is m, so the unit of moment is nm

i) Moment: Change angular momentum Force: Change linear momentum Obviously, these two are different things, angular momentum refers to the momentum possessed by an object in motion Linear momentum refers to the momentum possessed by an object in motion (ii) The greater the moment, the greater the angular acceleration given to the object (iii) The definition of moment is:

Moment = Force x (cross) Arm cross means vertical effective so the vertical distance between the computing power and the fulcrum 2012-02-06 18:23:49 Supplement:

To put it simply: even if the amount of force applied is the same, the effect will be different at different distances from the support point. Something like this.

Think of the whole system as a lever, with o as the fulcrum.

The arm of G1 is L1=DCCOS -OCSIN The arm of G2 is L2=OCSIN

where oc=lcos30°

dc=l/4

It can be seen from the principle of leverage equilibrium.

g1(l/4cosθ-lcos30°sinθ)=g2lcos30°sinθ

The solution yields tan = G1L4 (G2LCOs30°+G1LCOs30°) = 3G1 (6G1+6G2).

The specific value is related to gravity.

Set the angle a = angle b = a

Moment balance: A: G1Cos L 4 + G2Cos L 2 = TBSinal

B: G1SIN 3L 4+G2SIN L 2 = Tasinal Force Balance along the Rope Direction: Tacosa = TBCOSA + G1SIN +G2Sin

Vertical direction: tasin(a+ )tbsin(90+ -a)=g1+g2

Which province is so difficult in the first year of high school!!

Make a plumb line through point O, intersect AB at point E, let de=x, ce=y be balanced by moment: g1 * x=g2 * y

Let the side length of the triangle be l, and from the geometric relationship, x+y=l 4 is solved by the above two equations.

y=g1l/4(g1+g2)

Since OC is perpendicular to AB, it can be judged by geometric relations = angle EOC, so tan = EC OC = y ( 3L 2) = G1 [2 3 (G1 + G2)].

oa, ob is a rope of equal rod length meaning is oa ob equal to rod length?

It's zero degree, the AB is analyzed by force, OA and OB are orthogonally decomposed, the horizontal force does not care, the sum of the vertical force is always equal to the sum of G1G2, if there is an angle, then there must always be a rope that is not forced, because there is no horizontal force. Do the test and you can understand it.

Let the ab length be L

Because AB is a homogeneous rod of uniform thickness, AC=L*M1 (M1+M2).

cb=l*m2/(m1+m2)

g are omitted. m3*ac+m1*ac 2=m2*cd 2: 2m1*m3+m1 2=m2 2

m1+m3) 2=m2 2+m3 2>m2 2: m1+m3>m2

Let the b angle be atao*ac*cosa-m3*l 2=tbo*cb*sina to get: tao*ac*cosa>tbo*cb*sina due to tana=oc cb=ac oc

tana=[ (m1*m2)] (m1+m2)tao tbo> (m2 m1).

Because m2 > m1

tao/tbo>1

Then: tao>tbo

This question is explained with a torque:

First of all, the rod is in equilibrium, and the moment is 0, otherwise it will rotate.

The overall divide: the force is also balanced, then the intersection of the three forces of the type hail car (dotted line intersection) consideration: Note: The A below the Bu Shi represents the acute angle in the figure, because I can't hit it. Selling...

Along the rod direction: NCOSA=GSINA Perpendicular to the rod direction: T+NSINA=GCOSA

Then consider the moment at point d: Let db length: x

nsina*（l+x）+t*x= gcosa*[（l+x）/2]

Convert: Bring t=gcosa-nsina into the above formula: simplify.

nsina*l+ 1 2gcosa*x=1 2gcosa*l, and then bring ncosa=gsina into the above formula: simplify:

l*( Corner is required here... I'll think about how to ask for horns. After knowing the horn, you can find it. ）

Set the length of the wood l, and the center of gravity is x from the left end of the stick

When lifting from the right end, the point where the left end touches the ground is considered as the fulcrum, and the gravitational moment is balanced according to the counterclockwise moment and the clockwise moment, that is, the moment of the vertical upward force. The equation is obtained: 480L=GX

When lifted from the left end, the same analogy is given: 650L=G(L-X) and the above two equations are divided to obtain: X=(48 113)L

Take this to the first equation and get g=1130n

You can do it without using torque.

The left and right sides are lifted at the same time, with 480 at the right end and 650 at the left end, just balanced.

In fact, its weight is 1130N

The first floor is very skillful, but I advocate you to use the second floor method, the first floor technique is extracted from the second floor method, you should learn a type of problem from such a problem rather than just know how to do this problem.

Recommend the second floor with a clear idea. I used the first floor for the exam.

The answer is d

First of all, according to the force acting is equal in magnitude and opposite in direction. You can perform an individual analysis of the CD rod. As long as the moment of the external force is the same as the moment of gravity itself, the direction is opposite.

For the force T1, then the Cd rod is subjected to the same force as T1 but in the opposite direction. At this time, its moment is counterclockwise, and the moment of gravity of the CD rod itself is also counterclockwise, so it cannot be balanced. Therefore a false.

B is the same as option A.

For C, this T3 force just passes through the rotation point C, and the force arm is 0, so its moment is also 0, and the gravitational moment cannot be balanced, so it is not true.

For D, the analysis method is the same, and it can be found that for the force on the CD rod (which is the same size as T4, in the opposite direction), its moment is clockwise and can be balanced with the gravitational moment.