🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning

Like

Report

Numerade Educator

Like

Report

Problem 37 Medium Difficulty

Two particles are located on the $x$ axis. Particle 1 has a mass $m$ and is at the origin. Particle 2 has a mass 2$m$ and is at $x=+L$ . A third particle is placed between particles 1 and 2. Where on the x axis should the third particle be located so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles? Express your answer in terms of $L$ .

Answer

since the equation is quadratic, we get 2 solutions: $x=L(\sqrt{2}-1)$ and
$x=-L(\sqrt{2}+1),$ but the second solution is easily rejected because we are
given that the third particle lies between 1 and $2 .$ So our only correct
solution is the first one, which is $x=.414 \cdot L$

Discussion

You must be signed in to discuss.

Video Transcript

before adding the third party. Cough the system. The magnitude off the gravitational force that acted of both particles one and two or given by the following remember that the gravitational force had a symmetry. So the magnitude of the force exerted on particle one by particle to is equals to the magnitude off the force exerted on particle to my particle one and both off. This magnitudes are given by G times the mass off particle number one times the mass off particle number two divided by the distance. Between this particles squared so before we had f 12 equals two after one which was equals to two times G and squared divided by else were now after including particle number three. We are adding a new gravitational force on both particles weren't and two and these are given by the following After. Now we have particle tree and we have two new gravitational forces. We have the reputational force exerted on particle one, my particle tree, which is given by G times. The mass off particle number one times the mass off particle number three, Let's say capital M, divided by the distance between particle number one and particle number three squared. Then we have a zoo. A new rev additional force acting on particle number two, which is the gravitational force exerted on particle number two by particle number three, which is given by G times. The mass off particle number two times the mass off particle number three divided by the distance between this particles squared so l minus deep squared. Then we weren't both of these magnitudes to double. So the natural additional force that is being exerted on particle one should double so f one true blood's F one tree should be equals to two times f one truth. So the net gravitational force, after including the third practical, is two times higher than the gravitational force. Before adding that particle and this results in the following F one tree should be then equals two f one truth. At the same time, we have the following f true one plus F 23 should be close to truth times after one so f to tree should be equals two f 21 that you organize it and start solving the question. So if at 13 is equals two f one true and at the same time after treat is equals to after one we have the same. We have the following conclusion. Af one trees is then equals two f true tree. Why? Because af one too easy goes to after one by symmetry then this relation between Afghan three and after three results in the following equation Times M times capped OM divided by D squared Is it Costa g Times True times M time, Scott told him. Divided by l minus the squared We can simplify this common terms on both sides and get the following one divided by d squared Is it Costa? True? Divided by l minus d squared. Now we can send this term to the other side. Fendi's term to the other side to get l minus The squared is he goes to true times the squared Then we can think this square root off both sides So then we have il minus d is equals two plus or minus the square it off true times deep. Then we can send this term to the other side to get the following. So l is Eco's two plus or minus the square it off true times deep plus the then l A is equals to deke times one plus or minus The square it off True before the should be equals two l divided by one blows or minus the square Root off True. And these give us two possible results for deep one is still divided by one plus square it off to on the order is l divided by one minus Square it off. Truth note that the square root of true is bigger than one. So the second result will be smaller than zero and this is not possible because the third particle is in between particles one and two. It means that the is in between zero and l. So it can't be smaller than zero and we don't consider the second result them. The answer is the father This first result, which gives us approximately 0.414 times l and these is our final answer

Brazilian Center for Research in Physics
Top Physics 101 Mechanics Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Aspen F.

University of Sheffield