Question

A cart is moving with a constant acceleration of $2\frac{m}{s^2}$ as shown in the figure. $A$ person of mass $6kg$ is pulling a block of mass $3kg$ with the help of a string. If the coefficient of friction between the cart and the block is $0.5$, then the minimum value of force applied by the person to slide the block on the cart is given by $\frac{a}{\sqrt{1.25}}$ $N$. Find the value of a.

Answer 1

Draw $FBD$ of block with respect to cart frame

Let $T,N$ and $f$ be tension in string, normal reaction on block and friction acting on block respectively and let the angle made by string be$\theta$
Given:$a=2\frac{m}{s^2},m_{man}=6kg,m_{block}=3kg,\mu =0.5$
From non-inertial frame of reference pseudo force will also act on block as shown in $FBD$ towards left

Find Tension acting in the string required to pull the block
From force equation along perpendicular direction
$N+T\sin \theta =mg$
$\therefore N=mg-T\sin \theta \dots \dots \left(i\right)$
Now force equation along the plane, to find tension to make block slide we will use limiting value of friction
$ma+T\cos \theta =f\mu N\dots \dots \left(ii\right)$
Substituting value of normal reaction from equation $\left(i\right)$ in $\left(ii\right)$
$ma+Tcos\theta =\mu (mg-T\sin \theta )$
$\therefore T=\frac{\mu mg-ma}{\cos \theta +\mu \sin \theta }$

Find minimum force applied by man on string
$T=\frac{\mu mg-ma}{\cos \theta +\mu \sin \theta }$
To find minimum value of tension, $\cos \theta +\mu \sin \theta$ should be maximum
So, maximum value of $\cos \theta +\mu \sin \theta$ is$\sqrt{1^2+{\mu }^2}$
Hence,$T_{m}in=\frac{m(\mu g-a)}{\sqrt{1^2+{\mu }^2}}$
$=\frac{3(0.5g-2)}{\sqrt{1+{0.5}^2}}$
$=\frac{9}{\sqrt{1.25}}$
now compare with given answer we get $a=9$