Question

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?

Answer 1

(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\ cos 30^@`.
• the component of weight `=-mg\ sin30^@`
• the force of friction, `-f_k`.

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

`sumF_x = 50\ cos30^@ - mg\ sin30^@ - f_k = ma_x`

Inserting given values we get

`50\ sqrt3/2 - 3xx9.81xx 1/2 - f_k = 3xx2`
`=> f_k = 50\ sqrt3/2 - 3xx9.81xx 1/2 -6`
`=> f_k = 22.6\ N`

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

• Component of applied force `=50\ sin 30^@`.
• the component of weight `=mg\ cos30^@`
  `:.N = mg\ cos30^@+50\ sin 30^@`

Inserting various values we get

`N = 3xx9.81xxsqrt3/2+50xx1/2`
`N = 50.49\ N`

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

`f_k = mu_kN`
`=>mu_k = (f_k)/N`

Plugging in known values we get

`mu_k = (22.6)/50.49=0.45`