Question

Two blocks $A$ of 6 $kg$ and $B$ of 4 $kg$ are placed in contact with each other as shown in figure.

There is no friction between $A$ and ground and between both the blocks. The coefficient of friction between $B$ and ground is 0.5. A horizontal force $F$ is applied on $A$. Find the minimum and maximum values of $F$, which can be applied so that both the blocks can move combinely without any relative motion between them.

• A. 10 $N$, 50 $N$
• B. 12 $N$, 50 $N$
• C. 12 $N$, 75 $N$
• D. None of these

Answer 1

The correct option is C 12 $N$, 75 $N$

$F-N\sin 37°=6a\Rightarrow F-\left(\frac{3N}{5}\right)=6a\left(i\right)$


$N_2=4g-N\cos 37°=40-\frac{4N}{5}\left(ii\right)$
$N\sin 37°-f=4a\left(iii\right)$
From Eqs. ($i$) and ($iii$):
$F-f=10a$
$\Rightarrow F-\mu N_2=10a\left(iv\right)\Rightarrow F-\mu \left(40-\frac{4N}{5}\right]=10a\left(v\right)$
Put the value of $\mu$ from Eq. ($i$) in ($v$) and also put the value of $\mu$ to get $a=\frac{5F-60}{42}$
Now to start motion : $a>0$

⇒ $F>12$$N$

So the minimum force $F$ to just start the motion is $12$$N$.
Now maximum $F$ will be when $N_2$ just becomes zero.
Then from Eq. $\left(ii\right):N=50$$N$.
From $Eqs.\left(i\right)and\left(iv\right)$, we get $F=75$$N$ (by putting $N_2=0$)
If we apply $F>75N$, then B will start sliding up on A.