Radial force is calculated by circular force

1.

1. The calculation of the circumferential force of the belt pulley

first use the conversion ratio of the belt pulley speed and the belt pulley diameter, speed ratio = output speed: input speed = pitch circle diameter of the load belt pulley: pitch circle diameter of the motor belt pulley. The circular force and the reference force are the same, diameter - 2H = pitch circle diameter, h is the groove depth on the reference line, different types of V-belt h are different, yzab CDE, the circular force on the reference line are h = 1.62 2.753.54.88.19.6 respectively

2. Calculation of belt pulley radial force:

the radial force is the theoretical force of belt pulley pitch line position, which is generally expressed by PD, and the outer circle is generally expressed by OD. The conversion formula of pitch circle and outer circle is different for different groove types. Generally, it is easy to measure the outer circle of pulley, and calculate pitch circle according to the formula. SPZ:OD=PD+4; SPA:OD=PD+5.5; SPB:OD=PD+7; SPC:OD=PD+9.6

The axial force of the pulley is set as D1, N1 for the diameter and speed of the motor pulley (driving pulley) and D2, N2 for the diameter and speed of the driven pulley; That is, D2 = D1 * (N1 / N2). The minimum outer diameter of pulley a or spa is 80mm, SPZ belt, and the small pulley is not less than 63mm

extended data:

the recommended degree of pulley groove angle for different types of pulleys in different diameter ranges

1. The recommended degree of pulley groove angle for O-type pulleys is 34 degrees when the pulley diameter ranges from 50 mm to 71 mm; 36 degrees in the range of 71mm to 90mm, & gt; It is 38 degrees at 90mm

2. The diameter of A-type pulley is 34 degrees when the pulley diameter ranges from 71mm to 100mm, and 36 degrees when the pulley diameter ranges from 100mm to 125mm& gt; 38 degrees at 125 mm; When the diameter of B-type pulley ranges from 125mm to 160mm, it is 34 degrees; 36 degrees at 160 MM-200 mm, & gt; When 200 mm, it is 38 degrees

3. C-type pulley is 34 degrees when the pulley diameter ranges from 200 mm to 250 mm and 36 degrees when the pulley diameter ranges from 250 mm to 315 mm; 38 degrees at 315 mm

When the diameter of D-type pulley is 355mm-450mm, it is 36 degrees; At 450 mm, it is 38 degrees; E type is 36 degrees from 500mm to 630mm; It is 38 degrees when 630mm

2.

The solution is as follows:

If Z is used to represent the number of teeth of the gear, then: the circumference of the dividing circle= π D = ZP, that is, d = ZP/ π Make p/ π= m. Then d = MZ where. It's called molus. Because the pitch P of two gears must be equal, so the molus is equal

For the convenience of gear design and machining, the molus value has been standardized. The higher the molus is, the higher the height and thickness of the gear teeth are, and the greater the load they bear. Under the same conditions, the larger the molus is, the larger the gear is

extended data

Properties of circular force:

1. In any case, the resultant force of the moment center is zero (that is, the moment center is a fixed point, which should have the conditions to balance the circular force)

The circular force can be divided into several parts or couples without changing the effect on the figure

The translation theorem is not completely applicable to circular force

3. Bending radius generally depends on the material, space, design capacity requirements and other factors, so it is considered in many aspects.

4.

What is perpendicular to the diameter on the dividing circle is the circumferential force

The direction of the circular force is opposite to the motion direction on the driving wheel and the same as the motion direction on the driven wheel; The direction of the radial force points to the axis of the two gears; The direction of the axial force is determined by the helical direction of the tooth and the rotation direction of the gear

for the driving wheel, the left and right hand rule can be used to judge: the left screw uses the left hand, the right screw uses the right hand, the thumb is straight and parallel to the axis, and the other four fingers hold the axis along the rotation direction, then the direction of the thumb is the axial force direction of the driving wheel, and the axial force direction of the driven wheel is opposite to that of the driving wheel

extended data:

circular force can be decomposed into several parts (circular) force or couple without changing the effect on the figure

The

translation theorem is not completely applicable to circular force

rotation --- the movement of a circular body around the center of moment under the action of circular force is called rotation, also known as fixed axis rotation. The moment center line perpendicular to the action surface of the circumferential force is called the

rotation axis; It is usually specified that the counterclockwise rotation is positive. The direction of the circular force at a certain point on the circle points to the rotating side, and changes with the rotation

around the circle

5.

The radius of pipe elbow is calculated as follows:

radius = perimeter ÷ two ÷ π 3.14)

based on: PI

PI is the ratio of circumference to diameter of a circle, usually in Greek letters π Read as P à i) Means, π Is a constant (approximately equal to 3.141592654), representing the ratio of circumference to diameter. It is an irrational number, that is, infinite non cyclic decimal. In daily life, 3.14 is usually used to represent PI for approximate calculation

extended data

formula related to circle:

1, circle area: s= π r²， S= π( d/2)² D is the diameter and R is the radius)

The area of semicircle: s semicircle= π r^2)/2 R is the radius)

3. Circle area: s big circle - s small circle= π( R ^ 2-r ^ 2) (R is the radius of large circle, R is the radius of small circle)

The circumference of the circle: C = 2 π R or C= π d D is the diameter and R is the radius)

5. Circumference of semicircle: D+ π d) / 2 or D+ π r D is the diameter and R is the radius)

6. Divide the area of the circle where the sector is located by 360 and multiply by the angle n of the center angle of the sector circle, as follows:

s = n / 360 ×π r²

S= π r² × L/2 π R = LR / 2 (L is arc length, R is sector radius)

6. Unknown_Error

7.

Attention unit

8. Unknown_Error

9. The horizontal disc rotates uniformly around the vertical central axis, a small block of wood is placed on the disc and rotates with the disc, and the block remains stationary relative to the disc, as shown in the figure; A. The greater the mass of the block, the easier it is to slide on the disc
C. The greater the distance between the block and the rotating shaft, the easier it is to slide on the disc
D. the smaller the rotation period of the disc, The more difficult it is for the wooden block to slide on the disk
homework helps users with physics 2017-10-20
do you want to get the secret script of fast score raising? A. when the wood block moves in a uniform circular motion, the friction force in the motion provides centripetal force, so its direction is perpendicular to the linear velocity direction, so a is wrong
B. the critical state of relative sliding of wood block on the turntable is: mg μ= m ω 2R, the mass is eliminated, so B is wrong
C. The greater the distance from the block to the shaft, the greater the centripetal force required, and the easier it is to slide, so C is correct
D. when relative sliding occurs, the maximum static friction provides centripetal force, and the value is: mg μ= m ω Therefore, the smaller the period, i.e. the angular velocity ω As shown in the figure, there is a horizontal disc which can rotate around the vertical central axis, on which a spring with stiffness coefficient K is placed. One end of the spring is fixed on the shaft o, and the other end is connected with a small block a (which can be regarded as a particle) with mass M. the dynamic friction coefficient between the block and the disc is 0 μ， Open
as shown in Figure 2, a disk rotates uniformly around the vertical axis passing through the center in the horizontal plane, and a small object on the disk is stationary relative to the disk and moves with the disk. As for the centripetal force on the object, the following statement is correct: a
the horizontal disk rotates uniformly around the central axis, and... Two small objects a and B are placed on the disk, which rotate with the disk, Their masses are M1 and M2 respectively, and the distances from the rotating axis are R1 and R2 respectively, and R1 > R2. The friction coefficients between the two objects and the disc are the same, and the static friction forces they are subjected to are F1 and F2 respectively. The following statement is correct:
as shown in the figure, a disc can rotate around a vertical axis passing through the center of the disc and perpendicular to the surface of the disc, The wood block moves with the disc, then () A. the wood block is subject to the friction of the disc, the direction
the object with a mass of 1.0kg is placed on a horizontal disc which can rotate around the vertical axis, and the object is connected with the rotating axis by a light spring. The maximum static friction between the object and the disc is 0.1 times of the gravity, the stiffness coefficient of the spring is 600N / m, and the original length is 4cm,
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problem analysis
when the wood block moves in a uniform circular motion on the horizontal turntable, the static friction provides centripetal force. Since the static friction has a maximum value, it is necessary to keep relatively static ring rotation, There is a maximum value of rotation speed or period.
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examination site of this question:
centripetal force.
examination site comments:

10. Circular force: ft = 2T / D = 2 * 100 * 1000 / 50 = 4000N
Radial Force: FR = ft * Tan a = 4000 * tan20 °= 1455.9n
a-engagement angle, for standard gears. A = 20 degrees