Question 5.1: A brick of mass 3 kg is at rest on a rough plane inclined at...
A brick of mass 3 kg is at rest on a rough plane inclined at an angle of 30° to the horizontal. Find the friction force F N, and the normal reaction R N of the plane on the brick.
The diagram shows the forces acting on the brick.
Take unit vectors i and j parallel and perpendicular to the plane as shown.
Since the brick is in equilibrium the resultant of the three forces acting on it is zero.
Resolving in the i direction: F – 29.4 sin 30° = 0 \longleftarrow \boxed{ 3g = 29.4 } ①
F = 14.7
Resolving in the j direction: R – 29.4 cos 30° = 0 ②
R = 25.5
Written in vector form the equivalent is
\left( \begin{matrix} F \\ 0 \end{matrix} \right) +\left( \begin{matrix} 0 \\ R \end{matrix} \right) + \left( \begin{matrix} -29.4 sin 30° \\ -29.4 cos 30° \end{matrix} \right) = \left( \begin{matrix} 0 \\ 0 \end{matrix} \right)
or alternatively
F i + R j – 29.4 sin 30° i – 29.4 cos 30° j = 0 .
Both these lead to the equations ① and ② .
