Ask for help with engineering mechanics statics problems, thank you

This kind of static problem is a test of carefulness and basic knowledge, and the answers are analyzed as follows:

1) Find AE length.

2) Taking the AB member as the research object, point A is the center of moment, and the balance is based on the moment, and FE is found.

3) Taking the cylinder as the research object, the tension of the rope is obtained according to the plane force balance. Finished.

Take the distance from the left endpoint: f*3a-m=right end support reaction force*2a, solve, right end support reaction, and then list the balance equation of the vertical direction force to solve the left end support reaction.

Let's ask for the first one.

Balance of x-axis force: fa+fb=2000

B-point bending moment balance: fa*3a=2000*a

Result: FA = 2000 3 N

fb=4000/3 n

The later Lu Sleepy Bird will solve it for you, which is very simple, that is, it is more troublesome to enter.

The second: Formula:

fa*3a=2000*2a+2000*a

Result: FA = 2000 N

fb=4000 n

The third: set up the support A, accompany the early B support reaction force is vertically upward.

fa+fb=2000

1500+fb*2a=f*3a

Result: fa=-625 n

fb=2625 n

When the thin plate is stationary, there are three force arms acting, with the vertex of the lower right corner of the thin plate as the fulcrumThe direction of friction coincides with the ground, and the force arm is 0, the column force arm equation has:

2a·f=p·(a/2)

where a is the side length of the sheet.

Solution: f= 2p 4=

Solution: Let the upward force of the wall to the wheel be x, and the force of the ground to the wheel upward is y, and the maximum weight of a is z x=0 x y z 210 y=0 x

m=0 100z 200 x solve the system of equations.

x＝30y＝480

Z 300300N is what you want.

Let the ground support force for the wheel be n1 (direction upward), the wall support force for the wheel is n2 (direction to the right), the friction force of the ground on the wheel is f1 (direction to the left), and the friction force of the wall on the wheel is f2 (direction upward), which is obtained by x=0: n2-f1=0

is obtained by y=0: n1+f2-g-f=0

is obtained by m=0: f1*r2+f2*r2-g*r1=0f1=f*n1

f2=f*n2

The above five formulas can be substituted into the number to obtain: g = 300n

From the meaning of the title, the hook on both sides of the rope tension filial piety ruler size is equal, set to f, and the angle between the rope on both sides and the horizontal direction is also coincidentally high, set this angle as, set the hook on the left side of the rope length of y meters, then the right length is (5 y) meters.

Then the geometric relation is obtained by y*cos (5 y)*cos 4 , that is, 5*cos 4 , cos

From the relation of force, 2*f*sin g is obtained, and the desired tensile force is f g (2*sin ) g (2* 2*Newton.