Question

a block of mass m is placed on a horizontal surface. If the block is pulled by applying a force of magnitude $F=5mg$ at an angle $\theta ={37}^0$ with horizontal as shown in $Fig.6.29$, find the acceleration of the block at the given instant.

Answer 1

Three forces act on the block (a) $mg\downarrow$ (b) $F=5mg\nearrow$ (c) $N\uparrow$

Resolving the forces along x-and y-axis, we have equation of motion

$\sum F_x=m{a}_x$

$4mg=m{a}_x$

This gives $a_x=4g$.......(i)

$\sum F_y=m{a}_y$

$3mg-mg+N=m{a}_y$

$2mg+N=m{a}_y$.......(iii)

Let $a_y=0$. Then $N=-2mg$

The negative result signifies that N will be directed down (opposite to the assumed direction), but the ground cannot pull the block down. Hence, the block will lose contact with the ground.

or $N=0$

Hence, from (ii), $a_y=2g\uparrow$

Hence, the net acceleration, $|\overrightarrow{a}|=\sqrt{a_x^2+a_y^2}$

Where $a_x=4g$ from Eq. (i) and $a_y=2g$ from Eq. (ii)

This gives $a=\sqrt{a_x^2+a_y^2}=\sqrt{{\left(4g\right)}^2+{\left(20\right)}^2}=2\sqrt{5}g$

Analysis of Newton's laws of motion in connected bodies : Problems based on normal reaction