Horizontal force of 50N accelerates a block of mass 3.0 kg up an incline with an A=2.0ms^-2. The angle with horizontal is 30^o. Find a) force of friction between the box and the incline b) the coefficient of kinetic friction between box and the incline?

May 20, 2018

(a) Let us take the positive $x$-axis as directed up the incline, and positive $y$-axis as perpendicular to the incline.

The forces acting on the block along the $x$-axis are

• Component of applied force $= 50 \setminus \cos {30}^{\circ}$.
• the component of weight $= - m g \setminus \sin {30}^{\circ}$
• the force of friction, $- {f}_{k}$.

Therefore, net force equation along the $x$-axis is

$\sum {F}_{x} = 50 \setminus \cos {30}^{\circ} - m g \setminus \sin {30}^{\circ} - {f}_{k} = m {a}_{x}$

Inserting given values we get

$50 \setminus \frac{\sqrt{3}}{2} - 3 \times 9.81 \times \frac{1}{2} - {f}_{k} = 3 \times 2$
$\implies {f}_{k} = 50 \setminus \frac{\sqrt{3}}{2} - 3 \times 9.81 \times \frac{1}{2} - 6$
$\implies {f}_{k} = 22.6 \setminus N$

(b) The forces acting on the block along the $y$-axis which make up normal reaction $N$ are

• Component of applied force $= 50 \setminus \sin {30}^{\circ}$.
• the component of weight $= m g \setminus \cos {30}^{\circ}$
$\therefore N = m g \setminus \cos {30}^{\circ} + 50 \setminus \sin {30}^{\circ}$

Inserting various values we get

$N = 3 \times 9.81 \times \frac{\sqrt{3}}{2} + 50 \times \frac{1}{2}$
$N = 50.49 \setminus N$

Now he equation relating the friction force and the coefficient of kinetic friction is

${f}_{k} = {\mu}_{k} N$
$\implies {\mu}_{k} = \frac{{f}_{k}}{N}$

Plugging in known values we get

${\mu}_{k} = \frac{22.6}{50.49} = 0.45$