Question

If the positions of two like parallel forces on a light rod are interchanged, their resultant shifts by one -fourth of the distance them then the ratio of their magnitude is :

• A. $1:2$
• B. $2:3$
• C. $3:4$
• D. $3:5$

Answer 1

The correct option is B $2:3$

Let force be $F_1$ and $F_2$ and situation is explained in diagram.

Since there is no rotation so $\tau =0$

for case 1:

$\tau =F_1{x}_1-F_2{x}_2=0$; assuming anticlockwise direction +

$F_1{x}_1=F_2{x}_2$

for case 2:

$\tau =F_2(x_2+(x_1+x_2\left)/4\right)-F_1(x_2-(x_1+x_2\left)/4\right)=0$; assuming anticlockwise direction +.

$F_2(x_2+(x_1+x_2\left)/4\right)=F_1(x_2-(x_1+x_2\left)/4\right)$

putting values from above we get

$2x_1{F}_2=3F_1{x}_1$

$F_1/F_2=2/3$

Hence Option B is correct.