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.

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Accepted Answer

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 [latex]\longleftarrow \boxed{ 3g = 29.4 }[/latex] 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

[latex]\left( \begin{matrix} F \ 0 \end{matrix} ight) +\left( \begin{matrix} 0 \ R \end{matrix} ight) + \left( \begin{matrix} -29.4 sin 30° \ -29.4 cos 30° \end{matrix} ight) = \left( \begin{matrix} 0 \ 0 \end{matrix} ight)[/latex]

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 .

Question 5.1: A brick of mass 3 kg is at rest on a rough plane inclined at...

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