Question

A man of 72kg weight is standing on a board inclined at ${{20}^{\circ }}$ with the horizontal find the components of man weight (i) perpendicular to the board (ii) parallel to the plane of time board

Answer 1

Hint: First draw body diagram of the man inclined on the board at 20 degree. That is, a free body diagram is a schematic diagram representing the information of forces acting on the body including the magnitude and direction of the forces. From that find the parallel and perpendicular components acting on the body. The weight of the body is the product of mass and acceleration due to gravity which acts downwards. Then find the parallel and perpendicular forces acting on that man. Thus we get the components of man weight in perpendicular and parallel to the board.

Complete answer:
Given that the weight of man is,
m= 72kg
And the standing board inclines at $\theta ={{20}^{\circ }}$ with the horizontal.
The free body diagram for the given data can be shown as,

From the figure,
Thus the component of man’s weight perpendicular to the board is,
${{F}_{\bot }}=mg\cos \theta $
$\Rightarrow {{F}_{\bot }}=72\cos ({{20}^{\circ }})$ $=67.66N$
Then the component of man’s weight parallel to board is,
${{F}_{ll}}=mg\sin \theta $
$\Rightarrow {{F}_{ll}}=72\sin \left( {{20}^{\circ }} \right)$
$\Rightarrow {{F}_{ll}}=24.63N$

Note:
To handle a typical problem in mechanics a schematic diagram should be drawn showing various parts of the body and forces acting on it. This schematic diagram is known as the free body diagram. In a free body diagram, we should include information about forces, that is their magnitude as well as direction.