Question

In the arrangement shown in the figure $[\sin {37}^{\circ }=\frac{3}{5}]$

• A. direction of force of friction is up the plane.
• B. the magnitude of force of friction is zero.
• C. the tension in the string is $20\text{N}$.
• D. magnitude of force of friction is $56\text{N}$

Answer 1

The correct option is A direction of force of friction is up the plane.


By balancing force on $10\text{kg}$ mass in perpendicular direction to the incline we get
$N=mg\cos {37}^{\circ }$
$\Rightarrow N=100\times \frac{4}{5}\Rightarrow N=80\text{N}$
Now we know friction is given by
$f=\mu N=0.7\times 80\Rightarrow f=56\text{N}$
By balancing force on $4\text{kg}$ mass we get
$T-40=4a$ .....(i)
where a is the acceleration with which it is going upward
Now balancing force on $10\text{kg}$ block in direction along the incline we get
$100\sin {37}^{\circ }-T-f=10a$ ....(ii)
Acceleration of both the blocks will be same.
From (i) and (ii)
$f=20-14a$ .....(iii)

From (iii) we can conclude that
as we increase the magnitude of $a$, the value of friction is decreasing.
So
$f\le 20$
Hence maximum value of friction can be $20$
And $T$ can be calculated from force balancing of $4\text{kg}$ block
$T=40\text{N}$
So only option (a) is correct as friction acts in the direction opposite to the tendency of slipping