Where Did the Mysterious '6' Come From in F24 Calculation?

AI Thread Summary
The discussion centers on the derivation of the force equation F24 = 6kq^2/L^2 in a physics problem. Participants clarify that the factor of 6 arises from multiplying the 2 from the r^2 calculation with the 3 from the charge interactions. The forces from charges q1 and q4 are noted to cancel each other out, while the forces from q1, q2, and q3 contribute to the net force acting on charge q4. This results in F24 being the only force affecting q4, leading to the final expression. Understanding these interactions is crucial for grasping the overall force dynamics in the scenario.
Gemy4
Messages
16
Reaction score
0
Hi first post sorry if I didn't follow anyone specific rule. Me and my fiancee are stumped how my physics 2b teacher got 6kq^2/L^2 for F24. We understand that where r= L square root2/2 is squared so the result is L^2. But where in the heck did the 6 come from in 6kq^2.
 

Attachments

  • GetAttachment.aspx.jpg
    GetAttachment.aspx.jpg
    25.8 KB · Views: 449
Physics news on Phys.org
(L\frac{\sqrt{2}}{2})^2=\frac{2L^2}{4}=L^2/2

ehild
 
Yes. As ehild said in his post, the result of r^2 is (L^2)/2. So the 2 multiplies into the 3 on 3q and it becomes 6q^2 (pardon me for my lack of Tex).
 
Beautiful! Also why is it that she used q1 and q4 vs. q3 and q4. I realize it will produce the same answer.
 
Your teacher indicated q1-q4 and q3-q4 because they are two forces of the same magnitude/size pulling in opposite directions. Because these two forces are pulling in opposite directions, their net force cancel to 0.
However, your teacher also killed two birds with one stone because her equations following F14 = F34 not only indicated that the two forces canceled out, it also calculated F24 (as each of the three charges, q1,q2,q3 , provide the same force in different directions). Since F24 does not have any other force cancelling it out, it provides the only force that moves the q4 charge. And that force is 6kq^2/L^2, as you've indicated in your first post.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top