Question

Tension in rod of length L and mass M at a distance y from F, when the rod is acted on by two unequal forces F₁ and F₂ where (F₂< F₁) at its ends is

• A. F₁(1 -y/L) + F₂ (y/L)
• B. F₂(1 -y/L) + F₁(y/L)
• C. F₁(1 + y/L) + F₂(y/L)
• D. F₂(l +y/L) + F₁(y/L)

Answer 1

Answer: (A)

F₁(1 -y/L) + F₂ (y/L)

Answer 2

Acceleration along F₁,

a = $\frac { \mathrm { F } _ { 1 } - \mathrm { F } _ { 2 } } { \mathrm { M } }$

Mass of length y = m = $\frac { \mathrm { M } } { \mathrm { L } }$ y

If T is tension in this length, then

F₁ – T = ma = $\left( \frac { M } { L } y \right) \left( \frac { F _ { 1 } - F _ { 2 } } { M } \right)$ = (F₁ – F₂)y/L

∴ T = F₁ – (F₁ – F₂)y/L

or T = F₁(1 – y/L) + F₂(y/L)