Discussion Forum

A uniform wooden stick of mass 1.6 kg of length / rests in an inclined manner on a smooth, vertical wall of height h (<l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30⁰  with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio  $\frac{h}{l}$ and the frictional force  f at the  bottom of the stick are (g = 10 ms^-2

Answer:

Explanation:

$\sum F_{x}=0$ ,  $N_{1}\cos 30^{0}-f=0$ ........(i)

$\sum F_{y}=0$,   $N_{1}\sin 30^{0}+N_{2}-mg=0$........(ii)

$\sum \tau_{0}=0$,  $mg\frac{l}{2}\cos 60^{0}-N_{1}\frac{h}{\cos 30^{0}}=0$........(iii)

Also given N₁=N₂         ..........(iv)

Soving eqs (i),(ii),(iii) and (iv)

 we have 

$\frac{h}{l}=\frac{3\sqrt {3}}{16}$, $f=\frac{16\sqrt {3}}{3}N$