Question

In the figure shown below, a horizontal force $F$ is applied on $4\text{kg}$ block towards left. If the coefficient of friction between the surfaces are $0.5$ and $0.4$ as shown in the figure. The value of Tension in the rope and force required just to slide the $4\text{kg}$ block under $2\text{kg}$ block is
(rope is massless and inextensible)

• A. $T=90.9\text{N}$ and $F=29\text{N}$
• B. $T=9.09\text{N}$ and $F=14.5\text{N}$
• C. $T=9.09\text{N}$ and $F=29\text{N}$
• D. $T=90.9\text{N}$ and $F=14.5\text{N}$

Answer 1

The correct option is C $T=9.09\text{N}$ and $F=29\text{N}$

The FBDs of the blocks are as shown below
$R_1$ and $R_2$ are the normal forces acting on the respective blocks.

The condition is just to slide, it means in the question it is talking about limiting condition.
So, from FBD of $2\text{kg}$ we have,
$R_1+T\sin {37}^{\circ }=2g=20$
$R_1=20-\frac{3T}{5}$ ... (1)
$f_1=T\cos {37}^{\circ }$
$\Rightarrow f_1=\frac{4T}{5}$ ... (2)
$\because f_1=\mu R_1$
$\Rightarrow f_1=0.5(20-\frac{3T}{5})=10-\frac{3T}{10}$... (3)

From (3) and (2) we get
$\frac{4T}{5}=10-\frac{3T}{10}$
$\frac{4T}{5}+\frac{3T}{10}=10$
$\frac{11T}{10}=10\Rightarrow T=\frac{100}{11}\text{N}$ ... (4)

From (4) in (1) we get
$R_1=20-\frac{3\times 100}{5\times 11}$
$=20-5.454$
$R_1=14.546\text{N}$

Now, from the FBD of $4\text{kg}$ we get,
At equilibrium,
$R_2=R_1+4g=14.546+40=54.546\text{N}$
$F=f_1+f_2$
$\because f_2={\mu }_2{R}_2=0.4\times 54.546=21.8\text{N}$
and from equation (3)
$f_1=10-\frac{3\times 100}{10\times 11}=7.27$
$\Rightarrow F=21.8+7.3=29.1\text{N}$