Question

Determine the magnitude of frictional force f in each of the following cases:

Answer 1

(a) REF. Image.

N = 100 N (due to action & reaction force of 100 N)

$f_{max}=\mu N$

$=0.2\times 100=20N$

$mg=5\times 10=50N$

$f_{applied}=20N$

(b) REF. Image.

N = 500 N

$f_{max}=\mu N$

$=0.2\times 500$

$=100N$

$mg=5\times 10=50N$

f applied = 50N

because friction is resist

the force

(C) REF.Image.

$f=\mu N$

$e{q}^n$ in vertical direction

$f+50=50mg$

if means that friction will act on

downward direction so

(C) REF.Image.

$f_{max}=\mu N=0.2\times 50\sqrt{3}=10\sqrt{3}N$

$50+f=mg$

$50+f=50$

$f=0N$ No friction in it.