Question

Figure shows a block of mass m kept on inclined plane with inclination 0. The tension in the string is:
m=1 kg
T30° M=0.8​

Answer 1

Answer:

0 N

Explanation:

1) Let the Normal reaction of wedge on block be "N".

Making components of 'mg' along the wedge and perpendicular to the wedge.

Then,

$N=mgcos(30^{\circ})$

2) Static friction acting on 'm' is given by

$f_{s,max}=\mu*N=0.8*mgcos(30^{\circ})\\ \\=0.8*1*10*\frac{\sqrt{3}}{2}=6.93N$

3) Now,

Since,static friction can act upto its  maximum value if force opposite to it is lesser to it.

f_s,max > mgsin(30)

=> f_s=mgsin(30)

T+f_s=mgsin(30)

=>T+f_s=f_s

=>T=f_s-f_s=0

Hence,

T = 0 as f_s balances force along the wedge.

Answer 2

i have solved it below