Question

Mass of block A is $100\text{kg}$ and that of block B is $200\text{kg}$. As shown in figure, the block A is attached to a string tied to the wall. The coefficient of static friction between blocks A and B is $({\mu }_s{)}_A=0.2$ and the coefficient of static friction between block B and floor is $({\mu }_s{)}_B=0.3$. Then calculate the minimum value of force $F$ required to move the block B. $(g=10{\text{m/s}}^2)$

• A. $980\text{N}$
• B. $1000\text{N}$
• C. $1100\text{N}$
• D. $1200\text{N}$

Answer 1

The correct option is C $1100\text{N}$


From the FBD of A:
$T=f_1$........(1)
From FBD of B:
$F=f_1+f_2$......(2)

Since we want the minimum value of $F$ (for just slipping condition), we will take the limiting values of friction.
$f_1={\left({\mu }_s\right)}_A{m}_{A}g$ .........(3)
and $f_2=\left({\mu }_s{)}_B\right(m_A+m_B)g$ ......(4)

Substituting (3) and (4) in (2),
$F=f_1+f_2$
$=\left(0.2\right)\left(100g\right)+\left(0.3\right)\left(300g\right)=110g=1100\text{N}$