Question

plzz give me answer of 8th...question....plz​

Answer 1

Answer:

Explanation:

Refer to Fig. 6.42=(AB2+BC2)1/2

=(l2+l2)1/2=l2–√

Gravitational pull on body at A due to body at B is

F1=Gm×2ml2=2Gm2l2alongAB.

Gravitational pull on body at A due to body at C is

F2=Gm×2m(l2–√)2=2Gm2l2alongAC.

As, AB=AC and ∠ABC=90∘ , so ∠BAC=45∘.

i.e., angule θ between F1−→ and F2−→ is 45∘

Using parelogram law of forces, the resultant gravitaitonal force on body at A is

F=F21+F22+2F1F2cos45∘−−−−−−−−−−−−−−−−−−−−√

=[(2Gm2l2)+(2Gm2l2)2+2×2Gm2l2×2Gm2l2×12–√]1/2

=−2Gm2l2(1+1+2–√)1/2=3.696Gm2l2