Question

A 2.14 kg block is held in equilibrium on an incline of angle q = 67.7o by a horizontal force, F.


If the coefficient of static friction between block and incline is ms = 0.314, determine the minimum value of F.
Determine the normal force of the incline on the block.

Answer 1

x: m•g•sinα-F(fr) –F•cosα = 0
y: N-m•gvcosα –Fvsinα = 0.
N = m•gvcosα +F•sinα,
F(fr) = μ•N=μ• (m•gvcosα +F•sinα).
m•g•sinα - μ• (m•g•cosα +Fvsinα) – F•cosα = 0
m•g•sinα - μ•m•g•cosα + μ•F•sinα – F•cosα = 0
F• (cosα – μ•sinα) =m•g• (sinα – μ•cosα)
F= m•g• (sinα – μ•cosα)/ (cosα – μ•sinα)=
= 2.14•9.8•(0.93- 0.314•0.38)/0.38-0.314•0.93) =
=20.97(0.81)/0.088=193.06 N.
N = m•gvcosα +F•sinα =
=2.14•9.8•0.38 + 193.16•0.93 =
=187.52 N