Question

Determine the magnitude of frictional force and acceleration of the block in each of the following cases :

Answer 1

in Case I

$N_1=100N$, $mg=50N$,

$f_k={\mu }_k{N}_1=0.3\times 100=30N$

$f_s={\mu }_s{N}_1=0.4\times 100=40N$.

As the downward force $mg$ is greater than the force of static friction ${\mu }_s{N}_1$

Hence, the block will not be able to stay at rest. it will accelerate downwards. But, when it starts slipping, then kinetic friction will come into action. Now,

$a=\frac{mg-f_k}{m}$ $=\frac{50-30}{5}=4m/s^2$.

So, in this case, $f=f_k=30N$ and $a=4m/s^2\left(downwards\right).$

in Case II

$N_2=500N$, $mg=500N$,

$f_k={\mu }_k{N}_2=0.3\times 500=150N$

$f_s={\mu }_s{N}_2=0.4\times 500=200N$.

Here, $f_1$ is greater than $mg\left(drivingforce\right).$

Hence,

Block will not move. So, in this case $a=0$, $f=mg=50N$

in case III,

$N_3=500cos{37}^o=400N$,

$mg=120N$,

$f_k={\mu }_k{N}_1=0.3\times 400=120N$

$f_s={\mu }_s{N}_1=0.4\times 400=160N$.

Here $drivingforce=500sin{37}^o-120=180N$ (in upward direction).

The block will move upward as the diving force is greater than the static friction.

Therefore,

$a=\frac{mg-f_k}{m}$ $=\frac{180-120}{12}=5m/s^2$.

So, in thiscase, $f=f_k=30N$ and $a=5m/s^2\left(upwards\right).$