Physical Integral Isolation Method, Urgent !!!

1) Firstly, the acceleration of the slider is calculated according to the displacement time formula, and the force of the slider is analyzed, and then the friction and support force are solved according to Newton's second law, and finally the force of the inclined surface is analyzed, and the friction force and direction of the ground facing the inclined plane are solved according to the common point force balance conditional column

2) Firstly, the acceleration of the slider is calculated according to the displacement time formula, and the force of the slider is analyzed, and then the friction and support force are solved according to Newton's second law column, and finally the force of the inclined surface is analyzed, and the support force of the ground on the inclined plane is solved according to the common point force equilibrium conditional column.

1. The supporting force of the inclined wooden block n=mgcos37=16n.

If the wooden block slides at a uniform speed, the friction force of the inclined face of the wooden block f=mgsin37=12n2, taking the whole as the research object, the system has no horizontal acceleration, then the horizontal resultant force is zero. Then the ground friction on the inclined plane is equal to zero.

If there is no acceleration in the vertical direction of the whole system, then the net force in the vertical direction is zero, and the ground support force is equal to the gravitational force, n=(m+m)g=70n

1) Slide at a constant speed and balance the force. F support = mgcos 37 ° = 16 n f friction = mgsin 37 ° = 12 n

2) The inclined plane is stationary. The inclined plane is vertically downward by its own gravity, and the pressure of the wooden block is the gravity vertically downward. Support force = mg + mg = 70n

Friction 0

In uniform motion, the sliding force is equal to the frictional force, and the supporting force on the inclined side of the wooden block is equal to the gravitational force of the wooden block. There is no friction between the ground and the inclined plane, and the supporting force is equal to the sum of the gravity of the block and the inclined plane.

1. The blocks m and m of the same material are connected with a light rope, and a constant force f is applied on m to make the two blocks move in a straight line with uniform acceleration, 1) the ground is smooth, and the common acceleration a = f (m+m) is first obtained by the overall balance method

Then use the isolation method to analyze m: t=ma=fm (m+m).

2) The ground is rough, first use the overall Fa Que Zheng to find the common acceleration a = f (m+m)- g

Then the isolation method was used to analyze the m: t- mg = ma = fm (m+m)- mg

t = fm/(m+m)

3) Vertical acceleration upward, first use the global method to find the common acceleration a = f (m+m)-g

Then analyze m with the isolation method: t-mg=m=fm (m+m)-mg

t = fm/(m+m)

Second, use the overall method:

The rope pulls t t upwards on the chairlift;

The rope rubs against the athlete's hands t, upward;

athlete gravity mg, downward;

Lift gravity mg, downward.

The resultant force produces a common acceleration: t+t-(m+m)g = m+m)a

t = m+m)(a+g)/2 = 65+15)*(10+1)/2 = 440n

The force of the athlete pulling the rope vertically downwards is 440N

Analysis of the chairlift by the isolation method:

Athlete pressure f, downward;

rope pull t, upward;

Lift gravity mg, downward.

The resultant force produces acceleration: t-mg-f=ma

Athlete pressure to chairlift pressure f = t-m(g+a) =440 - 15*(10+1) =275n

The main thing is Newton's second law!

The maximum static friction between the two wooden blocks on the right side is mg, and under the action of this static friction force, a wooden block on the upper right side and two wooden blocks on the left side move together according to the acceleration a, a = mg (m + m + 2m) = g 4

The tension of the string on the two wooden blocks on the left:

2m+m)a=3m* g 4=3 mg 4 The static friction between the two wooden blocks on the left: 2ma=2 mg 4< mg, which meets the requirements, so the tensile force of the string on the left two wooden blocks 3 mg 4 meets the requirements of common acceleration motion.

So b is correct.

Resources.

Let the maximum tensile force of the rope be f1, and let the friction between objects be f

Analyze the two objects on the right f-f1=3ma f1-f=ma analyze the force on the large object on the left f=2ma, then f1=3ma, when the static friction force is the largest, the tensile force is the largest, and analyze the force on the small object on the right mg-f1=ma=f1 3

then mg=4f1 3

F1=3 mg4 so choose B

Let the density of the rope line be x(=m l).

dp=dm*v=xvdl=xwldl

p=1/2*xw *l^2=1/2mwl

dek=1 2dm*v 2=1 2x(wl) 2*dlek=1 6x*w 2*l 3=1 6*m*(wl) 2dl(angular momentum, the previous l is the length)=dp*l= xw*l 2dl angular momentum=1 3*xw*l 3=1 3*mw*l 2 is the answer?

Solution: This is a simple problem for integrals.

Take a short segment of dx far from the endpoint x on the rod, then the velocity is x, the mass is mdx l, then the momentum of the rod p = m x ldx = 1 2m l, and the upper and lower limits of the integration are l and 0 respectively;

The kinetic energy is e=1 2 m x ldx=1 6m x ;

Angular momentum l= m x ldx=1 3m l.

o(∩_o~

The radius of action of different points of the rod body is not the same, j= m(r 2) is the moment of inertia of a particle, and this rod is composed of countless particles, so it is integrated j=m(l 2) 3 ! j=s m/l(r^2) dr=m(l^2)/3！Do you understand?

a=15/2=

s=15*2-1/2*

h=s*1/2=

mgh=1*10*

The kinetic energy of the slider is finally converted into gravitational potential energy and the work done to overcome the frictional force, i.e., w=1 2*15*15=mgh+fs=75+un*15, and the solution is u=