Force Exerted by Roads of Length x & Separation x - Physics Problem

  • Thread starter Thread starter jingu
  • Start date Start date
  • Tags Tags
    Physics
Click For Summary
SUMMARY

The discussion centers on calculating the gravitational force exerted by two parallel roads of length x, separated by distance x, with masses k and p. The primary equation used is Newton's law of gravitation, expressed as F = G(m1*m2)/r^2. Participants emphasize the importance of correctly defining mass elements and integration variables, suggesting the use of distinct symbols for clarity. The final consensus indicates that the approximate force is F = (Gm1m2)/4x^2, contingent on proper calculus application for precise results.

PREREQUISITES
  • Understanding of Newton's law of gravitation (F = G(m1*m2)/r^2)
  • Basic knowledge of calculus, particularly integration techniques
  • Familiarity with mass distribution in continuous bodies
  • Concept of gravitational potential and force between point masses
NEXT STEPS
  • Study the application of double integrals in gravitational force calculations
  • Learn about mass distribution and its impact on gravitational force
  • Explore the concept of gravitational potential energy in continuous systems
  • Investigate the differences between point masses and extended bodies in physics
USEFUL FOR

Students studying physics, particularly those in high school or early college, who are learning about gravitational forces and calculus applications in physics problems.

  • #31
Good. Now that you know F(u) acting on a material point at x = u of mass m, let there be another rod between x = 2d and x = 3d, of mass m2. What is the net force?
 
Physics news on Phys.org
  • #32
Net force on point mass m? It will depend on the positions of m2 and m bcoz gravitational force is always attractive so there will be two cases ...please tell whether u > 3d...or d<u<2d
 
  • #33
Net force on the second rod from the first rod.

You know the location of the first rod: 0 ≤ x ≤ d. You know the location of the second rod: 2d ≤ x ≤ 3d.

You know the force from the first rod on a point at x = u.
 
  • #34
I solved it I came up with this answer F= (GMm ln(4/3))/d^2...please verify...
 
  • #35
I get the same answer. Very good!
 
  • #36
thank you very much...Are u sure that the answer is the same...?
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
Replies
5
Views
2K
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
13
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K