Question

A block of mass $4\text{kg}$ is pressed against a wall by two perpendicular horizontal forces as shown in the figure. $F_1$ is perpendicular to the wall and $F_2$ is parallel to the wall. ${\mu }_s=0.8,{\mu }_k=0.5$ and $g=10{\text{ms}}^{-2}$

For $F_1=75N$ and $F_2=30N$, the block remains at rest on the wall. Then, the force of friction (in $N$) is

Answer 1

As we know that the normal force due to the wall is balanced by external force $F_1$.
So, we have
$F_1=75N$
$\text{N}=75N$
Now, maximum possible static friction force on the block is given as
$F_S={\mu }_s\text{N}$
$F_S=0.8\left(75\right)$
$F_S=60N$
Now, along vertical downward direction weight of the block acts, which is given as
$W=mg$
$W=4\left(10\right)=40N$
Horizontal force on the block applied externally is given as –
$F=30N$
Now, net force parallel to the wall is given as $F=\sqrt{{30}^2+{40}^2}$
$F=50N$
So here, friction force will balance this net force.
So, we have $F_f=50N$