Question

Solve this:
Q.13. Which of the following pairs of forces cannot be added to give a resultant force of 4 N?
(A) 2 N and 8 N
(B) 2 N and 2 N
(C) 2 N and 6 N
(D) 2 N and 4 N

Answer 1

Dear student,

$$\begin{gathered} \mathrm{From}\mathrm{parallelogram}\mathrm{law}\mathrm{of}\mathrm{vector}\mathrm{addition},\mathrm{the}\mathrm{resultant}\mathrm{between}\mathrm{two}\mathrm{forces}\mathrm{can}\mathrm{be}\mathrm{given}\mathrm{as}, \\ \mathrm{R}=\sqrt{{{\mathrm{F}}_1}^2+{{\mathrm{F}}_2}^2+2{\mathrm{F}}_1{\mathrm{F}}_2\mathrm{cos\theta }} \\ 1.{\mathrm{R}}_1=2\mathrm{N},{\mathrm{R}}_2=2\mathrm{N} \\ \mathrm{Therefore},\mathrm{R}=\sqrt{2^2+2^2+2\times 2\times 2\mathrm{cos\theta }}=\sqrt{8+8\mathrm{cos\theta }} \\ \mathrm{For}\mathrm{cos\theta }=1,\mathrm{R}=\sqrt{8+8}=\sqrt{16}=4\mathrm{N} \\ 2.{\mathrm{R}}_1=2\mathrm{N},{\mathrm{R}}_2=4\mathrm{N} \\ \mathrm{Therefore},\mathrm{R}=\sqrt{2^2+4^2+2\times 2\times 4\mathrm{cos\theta }}=\sqrt{20+16\mathrm{cos\theta }} \\ \mathrm{For}\mathrm{cos\theta }=-1/4,\mathrm{R}=\sqrt{20-4}=\sqrt{16}=4\mathrm{N} \\ 3.{\mathrm{R}}_1=2\mathrm{N},{\mathrm{R}}_2=6\mathrm{N} \\ \mathrm{Therefore},\mathrm{R}=\sqrt{2^2+6^2+2\times 2\times 6\mathrm{cos\theta }}=\sqrt{40+24\mathrm{cos\theta }} \\ \mathrm{For}\mathrm{cos\theta }=-1,\mathrm{R}=\sqrt{40-24}=\sqrt{16}=4\mathrm{N} \\ 4.{\mathrm{R}}_1=2\mathrm{N},{\mathrm{R}}_2=8\mathrm{N} \\ \mathrm{Therefore},\mathrm{R}=\sqrt{2^2+8^2+2\times 2\times 8\mathrm{cos\theta }}=\sqrt{68+32\mathrm{cos\theta }} \\ \mathrm{So}\mathrm{for}\mathrm{any}\mathrm{value}\mathrm{of}\mathrm{cos\theta },\mathrm{R}\ne 4\mathrm{N} \end{gathered}$$
Regards