Question

A mass of 0.273 kg is placed on an incline of 38.382 degrees. It is attached by a rope over a pulley to a mass of 0.254 kg. Find the coefficient of friction and the force of friction

Answer 1

here tension in the string is counter balanced by weight of block of mass m1

so we can say

$T = m_1g$

$T = 0.254 \times 9.8 = 2.49 N$

now on the other side the block which is placed on the inclined plane

we can say that component of weight of the block and friction force is counter balanced by tension force

$m_2g sin\theta + F_f = T$

now we can plug in all values to find the friction force

$0.273 \times 9.8 sin38.382 + F_f = 2.49$

$1.66 + F_f = 2.49$

$F_f = 0.83 N$

so it will have 0.83 N force on it due to friction

now to find the friction coefficient

$F_f = \mu \times F_n$

here we know that

$F_n = m_2gcos\theta$

$F_n = 0.273 \times 9.8 \times cos38.382 = 2.1 N$

now from above equation

$0.83 = \mu \times 2.1$

$\mu = 0.38$

so friction coefficient will be 0.38