Question

For the equilibrium of the system of a toy shown the relation between force F₁and F₂when the members are arranged to form three identical rhombuses is

• A. F₁ = F₂
• B. F₂ = $\frac { \mathrm { F } _ { 1 } } { 3 }$
• C. F₂ = $\frac { \mathrm { F } _ { 1 } } { 2 }$
• D. F₁ = $\frac { \mathrm { F } _ { 2 } } { 3 }$

Answer 1

Answer: (D)

F₁ = $\frac { \mathrm { F } _ { 2 } } { 3 }$

Answer 2

Let AB = BC = CD = l unit and end A moves slightly towards right such that l is increased to (l +δl).

Let the total work be zero, we get

F₁3(δl) – F₂(δl) = 0

( displacement of A = 3δl and that of C is δl)

or F₁ = $\frac { \mathrm { F } _ { 2 } } { 3 }$