![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Given the magnitude of the two-component force, how to decompose the force? It's about circles Where are the A, B and P points on the first floor?
According to the law of parallelogram (triangle) of force, the resultant force is the diagonal of parallelogram formed by component force
Take the component force F1 with a compass
Take the end point O of resultant force F as the center to make a circle 1
Then use a compass to take the component force F2
Take the point P at the other end of the resultant force F as the center, make circle 2, intersect circle 1 and a
Do ob / / AP, BP / / OA
Then OA and ob are two components of F
Two ends o of resultant force F, P. intersection a of two circles
To decompose a force, if the magnitude of one component and the direction of the other component are known, it may have several groups of solutions
Do you mean there are only two components? If there are only two components, there may be one or two solutions. It's a triangle problem. It depends on the difference between the component and the vertical
A force with definite direction f = 10N is decomposed into two components. It is known that one component has definite direction and forms an angle of 30 ° with F, and the size of the other component is 6N
Then in decomposition () A. there is no array solution B. There are two groups of solutions C. There is a unique solution D
Wrong pull. Should know 2
RELATED INFORMATIONS
- 1. A force with definite direction f = 10N is decomposed into two forces. One component has definite direction and forms an angle of 30 ° with F, and the other component is 6N When it is decomposed A. With or without array solutions B. with two groups of solutions C. with unique solutions D. without solutions
- 2. The object is affected by two common electric forces, one is F1 = 6N, the other is F2 = 8N. The angle between the two forces is 90 degrees. Find their resultant force I am a liberal arts student I'm in the HKCEE Everybody, help
- 3. F1 = 6N, F2 = 8N, find the range of F3 force and the resultant force of F2 and F3
- 4. Three coplanar forces in different directions at the same level are applied to a 2kg body for uniform straight motion. We know that F 1 = 6N, F 2 = 7n, F 3 = 8N, F 1 and F 2, and the force F 1, the body plus a Three coplanar forces in different directions and at the same level are used to make a uniform straight motion on a 2kg body. If we know that F1 = 6N F2 = 7n F3 = 8N F1 and F2 and the force? F1, how much is the body plus a
- 5. If the existing three forces are 4N, 6N and 9N respectively, what is the maximum resultant force and the minimum resultant force
- 6. The resultant force range of 6N and 9N
- 7. An object is in equilibrium under the action of many common point forces, such as 5N, 7n, 9N, 12n, 20n, etc. now the 9N force is reversed for balance How much force and direction should be added?
- 8. If three forces of 4N, 7n and 9N act on one object, then () A. The maximum resultant force of the object is 20nb. The minimum resultant force of the three forces is 2nc. The minimum resultant force of the three forces is zero. D. the resultant force of the three forces may be 5N
- 9. The acceleration of an object with a mass of 1kg may be () A. 5 m/s2B. 7 m/s2C. 8 m/s2D. 9 m/s2
- 10. What is the maximum acceleration and minimum acceleration of a 1kg object under the action of three forces of 2n, 4N and 5N?
- 11. （2/5mn^3-m^2n^2+1/6n^4)/(2/3n^2)=?
- 12. If two hands push the same object with 3N at the same time, is the force on the object 3N or 6N? Same direction
- 13. An object with a mass of 1kg is subjected to two forces of 2n and 3N, Then the maximum and minimum accelerations of the object are -
- 14. Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}
- 15. Let the sum of the first n terms of the sequence {an} be SN. Given A1 = 1, Sn + 1 = 4An + 2, find: (1) let BN = an + 1-2an, prove that the sequence {BN} is an equal ratio sequence 2) Finding the general term of an The second answer is (1) bn＝a(n＋1)－2an＝3•2^(n－1) ∴[a(n＋1)]/[2^(n＋1)]－(an)/(2^n)=3/4 The sequence {(an) / (2 ^ n)} is an arithmetic sequence with 1 / 2 first term and 3 / 4 tolerance ∴(an)/(2^n)＝1/2＋(n－1)3/4＝3/4n－1/4 That is, an = (3n-1) & _; 2 ^ (n-2) (n ∈ n *) Question: how to get [a (n + 1)] / [2 ^ (n + 1)] - (an) / (2 ^ n) = 3 / 4 and why
- 16. In the arithmetic sequence {an}, A1 = - 60, A17 = - 12. (1) find the general term an, (2) find the sum of the first 30 terms of the sequence
- 17. Let the sum of the first n terms of the sequence {an} be Sn = n ^ 2-8n. If TN is the sum of the first n terms of the sequence {an's absolute value}, find TN
- 18. It is known that the sum of the first n terms of the arithmetic sequence is Sn, and S13 < 0, S12 > 0, then the term with the smallest absolute value in the sequence is () A. Item 5 B. item 6 C. item 7 d. item 8
- 19. Given that the first term A1 of the arithmetic sequence {an} is less than 0, the tolerance D ≠ 0, and the absolute value A4 = the absolute value A12, if the first n terms and Sn get the minimum value, then n= I made a mistake!
- 20. In the arithmetic sequence an, a16 + A17 + A18 = A8 = - 36, find TN = | A1 | + | A2 | + · + | an| The answer is TN = {- 3 / 2 times n square + 123 / 2n, (n is less than or equal to 21, n belongs to n *) 3 / 2 times n square - 123 / 2n + 1260 (n is more than 21, n belongs to n *)*