Question

a block of 15 kg mass is resting on a rough inclined plane as shown in the figure the block is tied up by a horizontal string which has a tension of 15 Newton calculate the minimum coefficient of friction between the block and the incline plane​

Answer 1

Answer: 2/3

Explanation:

Refer attachment

Answer 2

ANSWER:

The minimum friction coefficient is 0.81

EXPLANATION:  

After finding the components along the horizontal and perpendicular plane shown in the above figure.

At Equilibrium, the perpendicular forces can be written as,      

15gcos45 + 15sin45 = N ......................(1)

At Equilibrium, the horizontal forces can be written as,      

15gsin45 = μN+15cos45 ..................(2)

Calculating the value of N from the first equation we get

N = 116.69 N

By substituting N in eqn.(2), we can find the coefficient of friction from eqn.(2).  

we get the coefficient of the friction

$15{gsin}45=\muN+15\cos45$

$15 g \sin 45=\mu(21.21)+15 \cos 45$

$\frac{150}{\sqrt{2}}=\mu(116.69)+\frac{15}{\sqrt{2}}$

$\frac{135}{\sqrt{2}}=\mu(116.69)$

$\frac{95.47}{116.69}=\mu$

$\mu=0.81$