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Question: Three identical masses are at the three corners of the triangle, connected by massless identical spr...

Three identical masses are at the three corners of the triangle, connected by massless identical springs (rest length 10) forming an isosceles right angle triangle. If the two sides of equal length (of length 210) lie along positive x-axis and positive y-axis, then the force on the mass that is not at the origin but on the x-axis is given by ax^+by^a\hat{x} + b\hat{y} with

Answer

(-410+5\sqrt{2}),\hat{x}+(210-5\sqrt{2}),\hat{y}

Explanation

Solution

Solution:

We have three masses located at the vertices of an isosceles right triangle with legs of length 210 along the positive x‐axis and y‐axis. Thus the coordinates are:

  • A = (0,0)
  • B = (210,0)  → (mass on the x‑axis; our point of interest)
  • C = (0,210)

They are connected by identical massless springs with rest length L0=10L_0=10 (and let spring constant be kk, which we may take as 1 if not specified).

The three springs are:

  1. Between A and B: current length =210=210. Extension =210−10=200=210-10=200.
  2. Between A and C: current length =210=210. (Not directly acting on mass at B.)
  3. Between B and C: current length =(210−0)2+(0−210)2=2102=\sqrt{(210-0)^2+(0-210)^2} = 210\sqrt{2}. Extension =2102−10=210\sqrt{2}-10.

Now, the mass at B (at (210,0)) is attached to two springs: one connecting B with A and one connecting B with C.

Force from spring AB (between B and A):

  • Direction: from B toward A. The displacement r⃗BA=(0−210, 0−0)=(−210,0)\vec{r}_{BA} = (0-210, \,0-0) = (-210,0) has unit vector r^BA=(−1,0)\hat{r}_{BA} = (-1,0).
  • Magnitude: k(210−10)=200kk(210-10)=200k.

Thus,

F⃗BA=−200k i^.\vec{F}_{BA}= -200k\,\hat{i}.

Force from spring BC (between B and C):

  • Displacement from B to C: r⃗BC=(0−210, 210−0)=(−210,210)\vec{r}_{BC} = (0-210, \,210-0) = (-210,210).
  • Its magnitude is 2102210\sqrt{2} so the unit vector is
r^BC=(−210,210)2102=(−12, 12).\hat{r}_{BC}=\frac{(-210,210)}{210\sqrt{2}} =\left(-\frac{1}{\sqrt2},\,\frac{1}{\sqrt2}\right).
  • Magnitude of force: k(2102−10)k(210\sqrt{2}-10).

Thus,

F⃗BC=k(2102−10)(−12 i^+12 j^)=(−k(2102−10)2,  k(2102−10)2).\vec{F}_{BC}= k(210\sqrt{2}-10)\left(-\frac{1}{\sqrt2}\,\hat{i}+\frac{1}{\sqrt2}\,\hat{j}\right)=\left(-\frac{k(210\sqrt{2}-10)}{\sqrt2},\;\frac{k(210\sqrt{2}-10)}{\sqrt2}\right).

Net force on mass at B (at (210,0)):

Add the contributions:

F⃗B=F⃗BA+F⃗BC=(−200k−k(2102−10)2)i^+(k(2102−10)2)j^.\begin{aligned} \vec{F}_B &= \vec{F}_{BA}+\vec{F}_{BC} \\ &=\left(-200k-\frac{k(210\sqrt{2}-10)}{\sqrt2}\right)\hat{i} + \left(\frac{k(210\sqrt{2}-10)}{\sqrt2}\right)\hat{j}. \end{aligned}

Assuming k=1k=1 (or the answer expressed in terms of kk), we have:

a=−200−2102−102,b=2102−102.a = -200 - \frac{210\sqrt{2}-10}{\sqrt2},\quad b = \frac{210\sqrt{2}-10}{\sqrt2}.

Let us simplify these if desired:

For aa:

a=−200−21022+102=−200−210+102=−410+102=−410+52.a = -200 - \frac{210\sqrt{2}}{\sqrt2}+\frac{10}{\sqrt2} = -200 -210+\frac{10}{\sqrt2} = -410 + \frac{10}{\sqrt2} = -410+5\sqrt2.

For bb:

b=2102−102=210−102=210−52.b = \frac{210\sqrt{2}-10}{\sqrt2}=210-\frac{10}{\sqrt2} = 210-5\sqrt2.

Thus the net force on the mass at B is:

F⃗B=(−410+52)i^+(210−52)j^.\vec{F}_B=\left(-410+5\sqrt{2}\right)\hat{i}+\left(210-5\sqrt{2}\right)\hat{j}.

Explanation (minimal core):

  1. The mass at (210,0) is connected to (0,0) with a spring extended by 200 (210–10) giving force −200 x^-200\,\hat{x}.

  2. It is also connected to (0,210) with a spring extended by 2102−10210\sqrt2-10 acting along (−12,12)\left(-\frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right), giving force (−2102−102,2102−102)\left(-\frac{210\sqrt2-10}{\sqrt2}, \frac{210\sqrt2-10}{\sqrt2}\right).

  3. Summing these forces gives:

    a=−200−2102−102andb=2102−102,a = -200-\frac{210\sqrt{2}-10}{\sqrt{2}} \quad \text{and} \quad b = \frac{210\sqrt{2}-10}{\sqrt{2}},

    which simplifies (with k=1k=1) to a=−410+52a=-410+5\sqrt2 and b=210−52b=210-5\sqrt2.