Question

In the figure shown below, a horizontal force $F$ is applied on $5kg$ block towards left. If the coefficient of friction between the surfaces are $0.8$ and $0.8$ as shown in the figure., find the value of tension in the rope and force required just to slide the $5kg$ block under the $10kg$ block. Take $g=10m/s^2$.

• A. $T=50\text{N},F=120\text{N}$
• B. $T=62.5\text{N},F=140\text{N}$
• C. $T=75\text{N},F=160\text{N}$
• D. $T=87.5\text{N},F=140\text{N}$

Answer 1

The correct option is B $T=62.5\text{N},F=140\text{N}$

The FBDs of the blocks are as shown~.
As per the given question, the block has to just slide which means the condition of friction will be limiting.


From the FBD of $10kg$ block, we have
$\sum F_y=0$
$\Rightarrow N_1=10g-T\sin {37}^{\circ }$
$\sum F_x=0$
$\Rightarrow f_1=T\cos {37}^{\circ }----\left(1\right)$
In limiting case,
$f_1={\mu }_1{N}_1=0.8\times (10g-T\sin {37}^{\circ })$
$\Rightarrow f_1=0.8\times (100-\frac{3T}{5})$
$\Rightarrow f_1=(80-\frac{12T}{25})----\left(2\right)$

From (1) & (2), we get
$\frac{4T}{5}=80-\frac{12T}{25}\Rightarrow \frac{4T}{5}+\frac{12T}{25}=80$
$\therefore T=62.5\text{N}$

Now, from the FBD of $5kg$ block, we have at equilibrium
$F=f_1+f_2----\left(3\right)$
and $N_2=N_1+5g=10g-T\sin {37}^{\circ }+5g$
$\Rightarrow N_2=15g-62.5\times \frac{3}{5}$
$\Rightarrow N_2=150-37.5$
$\Rightarrow N_2=112.5\text{N}$
In limiting case,
$f_2={\mu }_2{N}_2=0.8\times 112.5=90\text{N}$

And, from eq. (2),
$f_1=(80-\frac{12T}{25})=(80-\frac{12\times 62.5}{25})=50\text{N}$

On putting the values of $f_1$ and $f_2$ in eq. (3), we get
$F=f_1+f_2=50+90=140\text{N}$